In-Depth Quantum Mechanics, Part II

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We continue our exploration of quantum mechanics in detail in this second part of our series on quantum mechanics. Here, we'll cover the quantum harmonic oscillator, spin, quantization of angular momentum, perturbation theory, and more!

Chapter guide for Quantum Mechanics

• Part 1 covers the basic ideas of quantum mechanics and its fundamental formalism.
• Part 2 covers applications of quantum mechanics as well as more advanced techniques for solving quantum-mechanical systems. You are reading this part right now.

Note: this section is currently incomplete; some topics have not yet been added, including discussion of the hydrogen atom, completion of the section on angular momentum, and a mention of the variational principle as well as applications of perturbation theory (Zeeman effect, Stark effect, fine/hyperfine structure). Finally, advanced quantum theory, scattering (in the Born approximation), and the path integral are not covered. Despite this, it is hoped that the existing content will be of educational interest.

Introduction to intrinsic spins

To start off, let's get (back) into quantum mechanics by examining the quantum phenomenon of spin (more accurately termed intrinsic spin, although it is common to just call it "spin" for short).

Intrinsic spin is one of the most important and most fundamentally quantum phenomena, which cannot be explained in classical terms. It refers to the fact that there is a mysterious form of angular momentum that is a fundamental property of subatomic particles, like protons and electrons. This means that certain quantum particles behave like tiny spinning magnets, just like classical rotating charged spheres. A full explanation of what spin is in a physical sense, however, is very difficult, since it is so far from any sort of everyday intuition that trying to explain it in terms of concepts of rotation that are familiar to us would be a gross oversimplification.

Historically, spin was accidentally discovered by the Stern-Gerlach experiment, first proposed by German physicist Otto Stern and then experimentally conducted by Walther Gerlach.

Note: Despite its name, spin does not correspond to the concept of "spinning" particles. The quantum notion of particles - which have no well-defined volume and are essentially zero-dimensional points - means that the very idea of "spinning" quite nonsensical. Even if quantum particles could spin, theoretical calculations quickly show that they would spin faster than the speed of light, which of course is unphysical. In essence, the name "spin" was coined as a historical accident and unfortunately has stuck around to confuse every generation of physicists afterwards.

For particles like electrons (which are called spin-1/2 particles for complicated reasons), there are precisely two basis states of the spin operators, which are usually called "spin-up" and "spin down" and notated with $|\uparrow \rangle$ and $\langle \downarrow|$ respectively. Thus, we can write out the state-vector of a spin-1/2 system as:

$$ |\Psi\rangle = \alpha |\uparrow\rangle + \beta |\downarrow\rangle $$

Where $\alpha, \beta$ are the probability amplitudes of measuring the spin-up and spin-down states. Note that the two spin states are orthonormal (that is, $\langle \uparrow|\downarrow\rangle = 0$ and $\langle \uparrow|\uparrow\rangle = \langle \downarrow|\downarrow\rangle = 1$. The spin operator $\mathbf{\hat S} = (\hat S_x, \hat S_y, \hat S_z)^T$, which we have already seen, have a very important physical meaning: they give the spin angular momentum of a spin-1/2 particle. Specifically, they predict that all spin-1/2 particles have an additional angular momentum associated with their intrinsic spin, called the spin angular momentum. This is usually represented by $\mathbf{S} = (S_x, S_y, S_z)$, which is a vector of the spin angular momentum of the particle in each coordinate direction.

To accommodate this decidedly non-classical behavior, physicists invented a new operator $\mathbf{\hat S}$, the spin operator. The spin operator also has components, which are given by $\mathbf{\hat S} = (\hat S_x, \hat S_y, \hat S_z)$. This gives us three eigenvalue equations, one each for each component of the spin operator:

$$ \begin{align*} \hat S_x|\psi\rangle = S_x|\psi\rangle \\ \hat S_y|\psi\rangle = S_y|\psi\rangle \\ \hat S_z|\psi\rangle = S_z|\psi\rangle \end{align*} $$

This tells us that $S_x, S_y, S_z$ are the eigenvalues of the spin operators along the $x$, $y$, and $z$ axes (respectively). A remarkable result that defies all classical intuition is that these eigenvalues are constrained to only one of two values: $\hbar/2$ or $-\hbar/2$. That is to say:

$$ S_x = \pm \dfrac{\hbar}{2}, \quad S_y = \pm \dfrac{\hbar}{2}, \quad S_z = \pm \dfrac{\hbar}{2} $$

Note: We will rarely refer to the spin operator $\mathbf{\hat S}$ and usually just discuss its components $\hat S_x, \hat S_y, \hat S_z$. Thus, if we say "spin operator along $x$" we mean $\hat S_x$, not $\mathbf{\hat S}$.

Meanwhile, the direction of the spin angular momentum depends on the specific eigenstate. For instance, the spin-up eigenstate of the $\hat S_z$ operator tells us that the particle's spin angular momentum points along $+z$. Similarly, the spin-down eigenstate of the $\hat S_z$ operator tells us that the particle's spin angular momentum points along $-z$. It is important to note that the direction of the spin angular momentum vector has no correspondence with a particle's actual orientation. It only tells us information about where the particle's spin angular momentum points and is (usually) only relevant for understanding a particle's interaction with magnetic fields.

Note: It is often the case that we just say "spin" as opposed to "spin angular momentum", although technically the latter is the correct terminology. However, in the interest of simplicity, we will call both "spin" wherever convenient.

To make things more concrete, the spin operators in the matrix representation are given by:

$$ \begin{align*} \hat S_x = \frac{\hbar}{2} \sigma_x, \quad \sigma_x &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ \hat S_y = \frac{\hbar}{2} \sigma_y, \quad \sigma_y &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \\ \hat S_z = \frac{\hbar}{2} \sigma_z, \quad \sigma_z &= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{align*} $$

Note: $\sigma_x, \sigma_y, \sigma_z$ are the Pauli matrices, which have eigenvalues $\pm 1$. Since the spin operators are just the Pauli matrices multiplied by a factor of $\hbar/2$, the eigenvalues of the all three spin operators are just a factor of $\hbar/2$ multiplied by the eigenvalues of the Pauli matrices (which are $\pm 1$). It is also useful to note that all three have determinant of $-1$ and zero trace. Other information can be found on its wikipedia page

Meanwhile, the eigenstates of the spin operators are called spinors (or eigenspinors). They are often written as $|\uparrow_x\rangle$ and $|\downarrow_x \rangle$ to indicate whether they are spin-up or spin down eigenstates (indicated by the direction of the arrows), which corresponds directly to what direction the spin angular momentum points towards ($x$, $y$, or $z$).

Note: Another common notation is to use $|+\rangle_x, |-\rangle_x$ for spin-up and spin-down states respectively, though this can clutter things up so we'll avoid it.

The explicit forms of the spinors associated with the spin operators $\hat S_x, \hat S_y, \hat S_z$ are given by:

$$ \begin{align*} |\uparrow_x\rangle &= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad |\downarrow_x\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix} \\ |\uparrow_y\rangle &= \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix}, \quad |\downarrow_y\rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -i \end{pmatrix} \\ |\uparrow_z\rangle &= \begin{pmatrix} 1 \\ 0 \end{pmatrix},\quad\qquad |\downarrow_z\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ \end{align*} $$

For calculations, it is frequently useful to reference certain identities of the Pauli matrices and spin operators. For instance, a very useful one is that:

$$ \sigma_x^2 = \sigma_y^2 = \sigma_z^2 = \hat I $$

Where $\hat I$ is the identity matrix. The spin operators and Pauli matrices also satisfy the below commutation relations:

$$ \begin{align*} [\sigma_x, \sigma_y] &= 2i\sigma_z \\ [\sigma_j, \sigma_k] &= 2i\varepsilon_{jkl}\sigma_l \\ [\hat S_x, \hat S_y] &= i\hbar \hat S_z \\ [\hat S_y, \hat S_z] &= i\hbar S_x \\ [\hat S_z, \hat S_x] &= i\hbar \hat S_y \end{align*} $$

In addition, some other useful identities are:

$$ \begin{gather*} \sigma_j \sigma_k + \sigma_k \sigma_j = 2\delta_{jk} \\ [\hat S^2, \hat S_x] = [\hat S^2, \hat S_y] = [\hat S^2, \hat S_z] = 0 \end{gather*} $$

Where $\hat S^2 = \mathbf{\hat S}\cdot\mathbf{\hat S}$ is the squared spin operator, which will be very important when we discuss more types of angular momentum in quantum mechanics. Finally, we have the all-important Pauli identity:

$$ \sigma_i \sigma_j = \delta_{ij} + i\varepsilon_{ijk}\sigma_k, \quad i,j \in (x, y, z) $$

Where $\delta_{ij}$ (as we've seen before) is the Kronecker delta, defined as:

$$ \delta_{ij} = \begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases} $$

And $\varepsilon_{ijk}$ is called the Levi-Civita symbol (or Levi-Civita tensor), and is given by:

$$ \varepsilon_{ijk}= \begin{cases}+1&{\text{if }}(i,j,k){\text{ is }}(1,2,3),(2,3,1),{\text{ or }}(3,1,2),\\-1&{\text{if }}(i,j,k){\text{ is }}(3,2,1),(1,3,2),{\text{ or }}(2,1,3),\\\;\;\,0&{\text{if }}i=j,{\text{ or }}j=k,{\text{ or }}k=i \end{cases} $$

Note: A very useful property of the Levi-Civita symbol is that it can be used to define the cross product of two vectors $\mathbf{A} \times \mathbf{B}$ via $(\mathbf{A} \times \mathbf{B})_i = \varepsilon_{ijk} A_i B_j$. There is also the so-called permutation definition of the Levi-Civita symbol, although we will not cover that here.

While the eigenstates of the $\hat S_x, \hat S_y, \hat S_z$ operators are distinct, it is possible to express eigenstates in one particular spin basis as a superposition of eigenstates in another spin basis. For instance, the eigenstates of the $\hat S_x$ operator can be written in terms of the eigenstates of the $\hat S_z$ operator:

$$ \begin{align*} |\uparrow_x\rangle &= \frac{1}{\sqrt{2}} \left(|\uparrow_z\rangle + |\downarrow_z\rangle\right) \\ |\downarrow_x\rangle &= \frac{1}{\sqrt{2}} \left(|\uparrow_z\rangle - |\downarrow_z\rangle \right) \end{align*} $$

Likewise, the eigenstates of the $\hat S_y$ operator can be written in terms of the eigenstates of the $\hat S_z$ operator:

$$ \begin{align*} |\uparrow_y\rangle &= \frac{1}{\sqrt{2}} \left(|\uparrow_z\rangle + i |\downarrow_z\rangle\right) \\ |\downarrow_y\rangle &= \frac{1}{\sqrt{2}} \left(|\uparrow_z\rangle - i |\downarrow_z\rangle\right) \end{align*} $$

Historical note: Interestingly, Wolfgang Pauli, for which the Pauli matrices are named, did not initially even like matrices (or linear algebra for that matter) being there in quantum mechanics! He once said of Schrödinger (to Max Born), "Yes, I know you are fond of tedious and complicated formalism. You are only going to spoil Heisenberg's physical idea by your futile mathematics". Funnily enough, he would eventually be most remembered for the his contribution to matrix mechanics and in describing spin, something he had once furiously railed against! (See the first comment on this Physics SE answer).

Spin and the Stern-Gerlach experiment

The fact that the spin operator tells us that spin-1/2 particles have additional angular momentum not predicted by classical physics has a profound implication: it means that any two otherwise identical spin-1/2 particles (for instance, electrons) will behave differently in a magnetic field. The magnetic force exerted on a classical charged particle with magnetic moment $\vec{\boldsymbol{\mu}}$ is given by $\mathbf{F}_B = -\nabla(\vec{\boldsymbol{\mu}} \cdot \mathbf{B})$, where $\mathbf{B}$ is the magnetic field, and $\vec{\boldsymbol{\mu}}$ is given by:

$$ \vec{\boldsymbol{\mu}} = g\dfrac{q}{2m} \mathbf{S}, \quad g \approx 2 $$

Note: The magnetic moment $\vec{\boldsymbol{\mu}}$ is the vector that measures the orientation of the magnetic field associated with an electric charge. A nonzero magnetic moment causes a charge to align with (or against) an external magnetic field, an effect that can be measured very precisely and is used in a variety of applications.

Since two randomly-chosen electrons would most likely have (and in some cases, must have) different spins, they would have opposite magnetic moments, and thus be deflected in different ways due to the magnetic force. Unknowingly taking advantage of the phenomenon, Stern and Gerlach conducted an experiment (shown in the diagram below) that showed a beam of silver atoms would split into two in a magnetic field, experimentally confirming the prediction of spin from quantum theory. This, again, is because electrons with different spins are deflected in opposite directions by the magnetic field, leading to the beam splitting into two (one of spin-up electrons and one of spin-down electrons).

A diagram of the Stern-Gerlach experiment. Source: The Information Philosopher

Repeated measurements of spin

We saw previously that all three spin operators don't commute with each other; for instance, $[\hat S_x, \hat S_y] = i\hbar \hat S_z$. This is not just a mathematical peculiarity! In quantum mechanics, remember that any two operators that do not commute represent observables that cannot be simultaneously measured to arbitrary precision. In formal terms, the uncertainty principle tells us that for two non-commuting operators $\hat A, \hat B$, where $[\hat A, \hat B] = i\hat C$, then:

$$ \Delta A \Delta B \geq \dfrac{|\langle \hat C\rangle|}{2} $$

For instance, in the case of the position and momentum operators (where $[\hat x, \hat p] = i\hbar$), this reduces to the famous Heisenberg uncertainty principle:

$$ \Delta x \Delta p \geq \frac{\hbar}{2} $$

The consequence of the uncertainty principle is that if you measure one observable $A$ and then measure another observable $B$, if $\hat A, \hat B$ do not commute, then:

1. They cannot both be measured simultaneously to perfect accuracy, and
2. A measurement of one observable tells you nothing about the other observable

Let's break down what this means. Suppose you had an electron which initially is know to be spin-up along $z$. Mathematically, you know the precise value of the $\hat S_z$ operator (spin operator in the z direction): there is 100% probability that the electron is in the spin-up state along the $z$ axis. What happens if you measure the electron's spin along $x$? Well, the electron is equally likely to be spin-up or spin-down along the $x$ axis because of rule (2). We already saw that $[\hat S_z, \hat S_x] \neq 0$ (that is, the spin operators along $z$ and $x$ do not commute), so knowing $S_z$ (the electron's spin along $z$) tells you nothing about $S_x$.

Now, what happens if you measure the electron's spin along $z$ again? "This is a pointless question!", you may say, "since I already measured the electron to be in the spin-up state along $z$, so it must certainly be in that same state!" However, things are not quite so simple. Remember that since the spin operators along $z$ and $x$ do not commute, measuring $S_x$ tells you nothing about $S_z$ (and vice-versa). This means that if you know $S_x$, you would not know anything about $S_z$ prior to measuring it. Since you had measured $S_x$, if you then choose to measure $S_z$, it would be a contradiction if the electron's spin along $z$ could also be precisely known to be in one state. Thus we are left with the profound and puzzling conclusion that upon measuring $S_z$ after $S_x$, the electron is again equally likely to be spin-up or spin-down along the $z$ axis. The previous information you had about the electron - namely, that it was in the spin-up state along $z$ - has now been erased!

Larmor precession

If we think about a spinning top rotating on a table, it will rotate round and round in a circle tilted at an angle (until it inevitably falls). This phenomenon is known as precession, and is shown in the animation below:

Source: Wikimedia Commons

The reason why everyday precession happens is that the Earth's gravity produces a torque on the spinning top. But since it is also spinning, the top has angular momentum which resists the torque, leading to the gyroscope rotating at an angle. In physics, the gyroscope is said to precess. But angular momentum is not just found in the classical world: it is also found in the quantum world, and it leads to an effect called Larmor precession.

To understand Larmor precession, recall that we earlier noted that spin-1/2 particles have an intrinsic angular momentum that comes from their spin. We have notated this angular momentum as $\mathbf{S}$, as it is the spin angular momentum, and we know it must have a magnitude of $\pm \hbar/2$ along any axis (so long as it is a spin-1/2 particle). This angular momentum creates a magnetic moment $\vec{\boldsymbol{\mu}}$ associated with the particle, given by:

$$ \vec{\boldsymbol{\mu}} = \gamma \mathbf{S} $$

Here, $\gamma$ is known as the gyromagnetic ratio, and it is a constant that can be calculated from the mass, charge, and other characteristics of the particle in question. From the magnetic moment, we can construct a Hamiltonian for a spin-1/2 particle placed in an applied magnetic field $\mathbf{B}$ with the spin operator $\mathbf{\hat S}$, given by:

$$ \hat H = -\vec{\boldsymbol{\mu}} \cdot \mathbf{B} = -\gamma(\mathbf{\hat S} \cdot \mathbf{B}) $$

It is usually easiest to consider a magnetic field aligned along the $z$-direction, and thus we have:

$$ \hat H = -\gamma \hat S_z B_z $$

By solving for the eigenvalues and eigenstates of the Hamiltonian, we get two spin eigenstates, which are respectively given by:

$$ |\uparrow_z\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix},\quad |\downarrow_z\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

Meanwhile, the energy eigenvalues are given by:

$$ E_{\pm z} = \pm \frac{1}{2}\hbar \gamma B_z $$

This tells us that we have two distinct energy eigenstates: the first one with higher energy, and the second one with lower energy. The energy difference between the two energy levels can be written in terms of the Larmor frequency $\omega_0$ via:

$$ \begin{align*} \Delta E &= E_+ - E_- \\ &= \hbar \gamma B_z \\ &= \hbar \omega_0, \quad \omega_0 \equiv |\gamma B_z| \end{align*} $$

Physically, when a magnetic field is applied, we say that the magnetic moments of a spin-1/2 particle precess, since they rotate (just like a gyroscope) to line up along or against the magnetic field. This type of precession is called Larmor precession, and it is very similar to the classical precession of a gyroscope, with some notable differences being that there is nothing physically "rotating" at an angle; rather, it is the magnetic moment vector that becomes tilted off-axis, which is why we term it as precession.

A diagram of Larmor precession. Source: Wikimedia Commons

Precise measurements of Larmor precession are essential in many areas of science, especially in nuclear magnetic resonance (NMR) spectroscopy, which is used in the medical sciences, as well as biotechnology and biochemistry. Additionally, many electronic devices (like hard disk storage) relies on exploiting the effects of spin, with an entire field of [spintronics](https://en.wikipedia.org/wiki/Spintronics devoted to research in this area. With so many observations proving the existence of spin, we know for certain that even if spin runs counter to our intuitions (pun intended!), it is nevertheless a very real aspect of the quantum world.

Generalized two-level systems

We will close our introductory discussion of spin and the ways to analyze spin by quickly discussing the generalization of a spin-1/2 system: two-level systems. Two-level systems (that is, a quantum system with two degrees of freedom) are found throughout quantum mechanics. Since spin-1/2 particles can either be spin-up or spin-down, which are the only two possible states in a spin-1/2 system, it is the prototypical two-level system. However, it is by no means unique: there are other two-level systems out there, and they can often be modelled using the same tools. For instance, it is very common to see Hamiltonians of the form:

$$ \hat H \propto \sigma_i $$

Where $\sigma_i \in [\sigma_x, \sigma_y, \sigma_z]$ is a Pauli matrix, and they obey:

$$ \begin{align*} [\sigma_x, \sigma_y] = \sigma_z \\ [\sigma_y, \sigma_z] = \sigma_x \\ [\sigma_z, \sigma_x] = \sigma_y \end{align*} $$

The Bloch sphere

A common generalization of the spin-1/2 system we have discussed at length is known as the Bloch sphere. The Bloch sphere is a formulation of a generalized two-level system with two states, which we notate as $|0\rangle$ and $|1\rangle$. Here, $|0\rangle$ is the ground state, so it is associated with a lower energy, whereas $|1\rangle$ is the excited state, so it is associated with a higher energy.

The Bloch sphere (illustrated in the diagram below) allows us to visualize the abstract space in which these states "live", much like the complex plane can be drawn as a 2D coordinate grid. The state space of the system is parametrized in spherical coordinates by the angles $(\theta, \phi)$ - it is important to note that these aren't physical locations in space, but are rather located in the abstract Hilbert space of the two-state system.

An illustration of the Bloch sphere. Source: Wikipedia

On the Bloch sphere, $|0\rangle$ "lives" at the north pole of the Bloch sphere, and $|1\rangle$ "lives" at the south pole. Other states of the system lie somewhere between the two poles. The general state-vector of the system can thus be written as:

$$ |\psi\rangle = \cos\left( \frac{\theta}{2} \right) |0\rangle + e^{i\phi} \sin\left( \frac{\theta}{2} \right) |1\rangle $$

Where $e^{i\phi}$ is the phase of the state-vector, and where the basis states are given by:

$$ |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$

Note: in the standard convention, the $x$ axis represents the real part of the phase, whereas the $y$ axis represents the imaginary part of the phase.

In the case that the two-level system models a spin-1/2 particle (a very common but not universal case), then we have:

$$ |\psi\rangle = \cos\left( \frac{\theta}{2} \right) |\uparrow\rangle + e^{i\phi} \sin\left( \frac{\theta}{2} \right) |\downarrow\rangle $$

Note: the choice is mathematically-arbitrary, so it does not matter whether we call $|0\rangle$ the spin-up or spin-down state, as long as our choice is consistent. However, $|\uparrow\rangle$ is usually associated with $|0\rangle$ in physics and engineering, and likewise $|\downarrow\rangle$ is usually associated with $|1\rangle$.

Along the "equator" of the Bloch sphere, we have $\theta = \pi/2$, and thus the state-vector takes the simpler form:

$$ |\psi\rangle = \frac{1}{\sqrt{ 2 }}\big(|0\rangle + e^{i\phi}|1\rangle\big) $$

Note that this corresponds to a state with 50% probability of measuring $|0\rangle$ and $|1\rangle$ since the complex phase factor has unit magnitude (you can check it for yourself by computing $P_0 = |\langle 0|\psi\rangle|^2$ and $P_1 = |\langle 1|\psi\rangle|^2$). However, the phase is still physically-relevant

Hamiltonian of a generalized two-level system

A very common basic Hamiltonian for a generalized two-level system is given by:

$$ \hat{H} = \frac{\hbar \omega}{2} \sigma_{z} $$

The energy eigenvalues are then $E = \pm \frac{1}{2} \hbar \omega$, with an energy difference of $\Delta E = \hbar \omega$ between the ground state and the excited state. Thus, the general time-dependent state-vector of the system, valid for all $t$, is given by:

$$ |\psi(t)\rangle = \cos\left( \frac{\theta}{2} \right) |0\rangle e^{-i \omega t/2} + e^{i\phi} \sin\left( \frac{\theta}{2} \right) |1\rangle e^{i \omega t/2} $$

This comes from tacking on a factor of $e^{-i E t/\hbar}$ to each of the eigenstates, and substituting in our known values of $E = \pm \frac{1}{2}\hbar\omega$. It is standard to factor out the phase factor of $e^{-i\omega t/2}$, giving us:

$$ \begin{align*} |\psi(t)\rangle &= e^{-i \omega t/2}\left[\cos\left( \frac{\theta}{2} \right) |0\rangle + e^{i\phi} \sin\left( \frac{\theta}{2} \right) |1\rangle e^{i \omega t} \right] \\ &= e^{-i \omega t/2}\left[\cos\left( \frac{\theta}{2} \right) |0\rangle + e^{i(\phi + \omega t)} \sin\left( \frac{\theta}{2} \right) |1\rangle\right] \end{align*} $$

Since the factor $e^{-i\omega t/2}$ is a phase factor, its magnitude is one, and thus it is not directly observable; thus, the physics of the system are identical if it is dropped. Defining $\phi_{r}(t) = \phi + \omega t$ as the relative phase of the system (since it is a phase that comes from the difference in energy of the two states, which is a relative quantity), we can rewrite the state-vector as:

$$ |\psi(t)\rangle = \cos\left( \frac{\theta}{2} \right) |0\rangle + e^{i\phi_{r}(t)} \sin\left( \frac{\theta}{2} \right) |1\rangle $$

As long as the system is undisturbed and the Hamiltonian is time-independent, the relative phase $\phi_r$ changes by a constant rate $\omega$ (the Larmor frequency) as the state-vector precesses along the Bloch sphere - note this is independent of the $\theta$ coordinate! If we imagine the state-vector of the system to be represented by an arrow, the change in the relative phase can be visualized as the rotation of this "arrow" as it precesses (i.e. rotates) around the Bloch sphere.

Note: For an application of the Bloch sphere in quantum optics, please see these lecture notes.

Conclusion to two-level systems

Two-level systems are the fundamental model behind a vast variety of quantum systems, including qubits in quantum computing, optically-pumped lasers, and molecular ions, as well as playing an important role in understanding the emission and absorption of radiation at the quantum level. We will discuss more examples of them throughout this guide - stay tuned!

The quantum harmonic oscillator

One of the simplest, but most important quantum systems is the quantum harmonic oscillator. At face value, it describes a particle in a harmonic potential well. To start, let us recall that the classical harmonic potential is given by:

$$ V(x) = \dfrac{1}{2} kx^2 = \dfrac{1}{2} m\omega^2 x^2, \quad \omega \equiv \sqrt{k/m} $$

The classical solutions to the (classical) harmonic oscillator are in terms of sinusoidal functions, which is why it is indeed called the harmonic oscillator. However, on a basic level, a harmonic potential is nothing more than a basic quadratic potential, one that we show in the below diagram:

In quantum mechanics, we retain the same form of the harmonic potential, except we perform the substitution $x \to \hat x$, where $\hat x$ is the position operator. Thus the Hamiltonian is given by:

$$ \hat H = \dfrac{\hat p^2}{2m} + V(x) = \dfrac{\hat p^2}{2m} + \dfrac{1}{2} m\omega^2 \hat x^2 $$

Note that we can also write this in the position representation (where $\hat x = x$ and $\hat p = -i\hbar \frac{d}{dx}$ in 1D) as:

$$ \hat H = -\frac{\hbar^2}{2m} \dfrac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 $$

The quantum harmonic oscillator is a very useful model in quantum mechanics, since it is one of the few problems that can be solved exactly. This does not mean it is trivial - the quantum harmonic oscillator finds numerous applications in molecular and atomic physics. The quantum harmonic oscillator can first be used as an approximation for a complicated potential. This is because the Taylor expansion of an arbitrary potential centered at $x = x_0$ is given by:

$$ V(x) = V_0 + V'(x-x_0) + \dfrac{1}{2} V''(x-x_0)^2 + \dfrac{1}{6} V'''(x-x_0)^3 + \dots $$

Another application is to describe the interaction of a charged (quantum) particle with an standing electromagnetic waves, something we will discuss later. When an electromagnetic field is trapped in some cavity, it decomposes into a series of modes, whose wavelength can only come in quantized values:

$$ \omega = \dfrac{2\pi c}{\lambda}, \quad \lambda = \dfrac{2L}{n}, \quad n = 1,2,3, \dots $$

Thus, any charged particle within such a cavity will interact with the standing waves of the electromagnetic field, leading to its energy levels being quantized. This is the origin of uniquely quantum phenomena such as the Stark effect, but we'll explain this in more detail later.

The third main application of the quantum harmonic oscillator is to describe the interaction of a particle with a quantized electromagnetic field, which is the realm of quantum electrodynamics and second quantization. While this can get complicated very quickly, the essence is to describe the quantized electromagnetic field as a series of coupled harmonic oscillators. We will cover this more at the very end of this guide.

To solve the quantum harmonic oscillator we begin with the same general methods as for essentially any quantum system - to write out the eigenvalue equation for the Hamiltonian:

$$ \hat H|\psi\rangle = E|\psi\rangle $$

This is the starting point, and there are several different ways to proceed from here. For instance, we can solve the eigenvalue equation in the position basis by taking the inner product with a position basis ket:

$$ \begin{gather*} \langle x|\hat H|\psi\rangle = \langle x|E|\psi\rangle \\ \langle x|\hat H|\psi\rangle = E\langle x|\psi\rangle \\ \left\langle x \left|-\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2} + \dfrac{1}{2} m\omega^2 \hat x^2\right|\psi\right\rangle = E\langle x|\psi\rangle \\ \Rightarrow -\dfrac{\hbar^2}{2m}\dfrac{d^2 \psi}{dx^2} + \dfrac{1}{2} m\omega^2 x^2 \psi(x) = E \psi(x) \end{gather*} $$

This is a differential equation that can indeed be solved, although it is not very easy to solve. Indeed, a much better approach is to use the so-called algebraic approach, which originated with the physicist Paul Dirac, which we'll now discuss.

The ladder operator approach

Dirac's key insight in solving the quantum harmonic oscillator is to "factor" the Hamiltonian by defining two new operators $\hat a$ and $\hat a^\dagger$, given by:

$$ \begin{align*} \hat a &= \dfrac{1}{\sqrt{2}}(\hat x'+ i\hat p') \\ \hat a^\dagger &= \dfrac{1}{\sqrt{2}}(\hat x' - i\hat p') \end{align*} $$

Where $\hat x', \hat p'$ are related to the position and momentum operators $\hat x, \hat p$ as follows:

$$ \hat x' = \sqrt{\dfrac{m\omega}{\hbar}} \hat x, \quad \hat p' = \dfrac{1}{\sqrt{m\hbar \omega}} \hat p $$

It is also useful to note that:

$$ \begin{align*} \hat x' &= \frac{1}{\sqrt{2}}(\hat a^\dagger + \hat a) \\ \hat p' &= \dfrac{i}{\sqrt{2}}(\hat a^\dagger - \hat a) \\ \end{align*} $$

$\hat a$ and $\hat a^\dagger$ are conventionally called the ladder operators. These two operators satisfy $[\hat a, \hat a^\dagger] = 1$, an important identity to keep in mind for later. While many steps avoid using this approach and write $\hat a$ and $\hat a^\dagger$ purely in terms of $\hat x$ and $\hat p$, by defining our new operators $\hat x', \hat p'$ we can non-dimensionalize the problem, making it easier to solve. This is essentially the same thing as a change of variables in a classical mechanics problem, only here we're using operators, not classical functions.

Note: While $\hat a$ and $\hat a^\dagger$ are indeed adjoints of each other, it is common to consider them essentially separate operators (for reasons we'll soon see). It is also common to use the notation $(\hat a_-, \hat a_+)$ instead of $(\hat a, \hat a^\dagger)$ (where $\hat a_- = \hat a$ and $\hat a_+ = \hat a^\dagger$) which is a completely equivalent notation.

Thus, with our operators $\hat x'$ and $\hat p'$, the Hamiltonian can be written as:

$$ \hat H = \hbar \omega \hat H', \quad \hat H' = \dfrac{1}{2}(\hat x'^2 + \hat p'^2) $$

These operators allow us to simplify the Hamiltonian down greatly, since we find that:

$$ \begin{align*} \hbar \omega\left(\hat a^\dagger \hat a + \frac{1}{2}\right) &= \hbar \omega\left(\frac{1}{2}(\hat x' + i\hat p')(\hat x' - i\hat p') + \frac{1}{2}\right)\\ &= \dfrac{1}{2}\hbar \omega \left(\hat x'^2 - i\hat x' \hat p' + i\hat x' \hat p' + \hat p'^2 + 1\right) \\ &= \dfrac{1}{2}\hbar \omega(\hat x'^2 + \hat p'^2 + i[\hat x', \hat p'] + 1) \\ &= \dfrac{1}{2}\hbar \omega(\hat x'^2 + \hat p'^2 -1 + 1) \\ &= \hbar \omega H' \\ &= \hat H \end{align*} $$

If one defines another new operator $\hat N$ (we will discuss what this means later), given by $\hat N = \hat a^\dagger \hat a$, then the Hamiltonian takes the form:

$$ \hat H = \hbar \omega \left(\hat N + \dfrac{1}{2}\right) $$

Note: Be careful of the order of the $\hat a$ and $\hat a^\dagger$ operators! This is because $\hat N = \hat a^\dagger \hat a$, but $\hat a^\dagger \hat a \neq \hat a \hat a^\dagger$ so $\hat N \neq \hat a \hat a^\dagger$! Indeed we find that $\hat a \hat a^\dagger = 1 + \hat N$, which can be derived from the commutation relation $[\hat a, \hat a^\dagger] = 1$.

The genius of using this operator-based "algebraic" approach to solving the quantum harmonic oscillator is that the $\hat a, \hat a^\dagger$ operators satisfy:

$$ \hat a|\psi_0\rangle = 0, \quad \hat a^\dagger |\psi_0\rangle = |\psi_1\rangle $$

In fact, in the general case, we find that for the $n$-th eigenstate $|\psi_n\rangle$ we have:

$$ \hat a|\psi_n\rangle = \sqrt{n}|\psi_{n - 1}\rangle, \quad \hat a^\dagger|\psi_n\rangle = \sqrt{n + 1}~|\psi_{n + 1}\rangle $$

It is common convention to indicate the $n$-th eigenstate of the quantum harmonic oscillator with $|n\rangle = |\psi_n\rangle$, in which case one may write the more elegant expression:

$$ \hat a|n\rangle = \sqrt{n}~|n-1\rangle, \quad \hat a^\dagger|n\rangle = \sqrt{n+1}~|n + 1\rangle $$

Thus we again find that $\hat a|0\rangle = \hat a|\psi_0\rangle = 0$ and $\hat a^\dagger |0\rangle = \hat a|\psi_0\rangle = |\psi_1\rangle$. These are indeed the chief identities of the quantum harmonic oscillator, because if we substitute in the definitions of the $\hat a, \hat a^\dagger$ operators, we have:

$$ \begin{align*} \hat a^\dagger \hat a|n\rangle &= \hat a^\dagger(\sqrt{n}~|n+1\rangle) \\ &= \sqrt{n^2} |(n + 1) - 1\rangle \\ &= n|n\rangle \end{align*} $$

But we already know that $\hat a^\dagger \hat a$ is just the $\hat N$ operator, so we therefore have:

$$ \hat N|n\rangle = n|n\rangle $$

In addition, we have:

$$ \begin{align*} \hat a^\dagger\hat a|n\rangle &= \hat N|n\rangle = n|n\rangle \\ \hat a \hat a^\dagger|n\rangle &= (\hat N + \hat I)|n\rangle = (n + 1)|n\rangle \end{align*} $$

The $\hat N$ operator is often called the number operator since it returns the index $n$ corresponding to the nth eigenstate. For instance, for the first eigenstate $|0\rangle$ (in more traditional notation, this can be written as $|\psi_0\rangle$, which is equivalent), $\hat N|0\rangle = 0$, which tells us that (as we expect) the first eigenstate is labelled with index $n = 0$. Likewise, for the eigenstate $|3\rangle$ then $\hat N|3\rangle = 3 |n\rangle$, which indeed returns its index $n = 3$. What makes it particularly special is how the number operator can be defined solely in terms of the $\hat a$ and $\hat a^\dagger$ operators, which is a very non-trivial result. Thus, if we substitute $\hat N|n\rangle = n|n\rangle$ into our Hamiltonian, we have:

$$ \hat H|n\rangle = \hbar \omega \left(\hat N + \dfrac{1}{2}\right)|n\rangle = \underbrace{\hbar \omega \left(n + \dfrac{1}{2}\right)}_{E_n}|n\rangle $$

Where by comparison our expression with the Hamiltonian's eigenvalue equation $\hat H|n\rangle = E_n|n\rangle$, we find that:

$$ E_n = \hbar \omega\left(n + \dfrac{1}{2}\right) $$

Thus, we have found the energy eigenvalues without needing to solve any differential equation, which is quite an enormous feat! Additionally, since the energies are only dependent on $n$, the energies are non-degenerate, meaning that each energy eigenvalue is associated with a distinct eigenstate. This is incredibly important because it's uncommon to encounter fully non-degenerate systems in quantum mechanics, where each eigenstate can be labelled by a single eigenvalue (think about our previous discussion of CSCOs).

In addition, we can also calculate the eigenstates in the position (or momentum) bases, so the algebraic formalism helps us get the wavefunctions too. To do so, let us recognize that if we take the $n = 0$ state (often called the ground state, and notated as $|0\rangle$), we have $\hat a |0\rangle = 0$. Keep this in mind! Now, recall that we defined our relevant operators to take the following forms:

$$ \begin{align*} \hat a &= \dfrac{1}{\sqrt{2}}(\hat x'+ i\hat p') \\ &= \sqrt{\dfrac{m\omega}{2\hbar}}\hat x + \frac{i}{\sqrt{2m\omega \hbar}}\hat p \\ \hat a^\dagger &= \dfrac{1}{\sqrt{2}}(\hat x' - i\hat p') \\ &= \sqrt{\dfrac{m\omega}{2\hbar}}\hat x + \frac{i}{\sqrt{2m\omega \hbar}}\hat p \end{align*} $$

Thus, the explicit forms of $\hat a$ and $\hat a^\dagger$ can be found in the position basis by substituting in $\hat x = x$ and $\hat p = -i\hbar \dfrac{d}{dx}$, giving us:

$$ \begin{align*} \hat a &= \sqrt{\frac{m\omega}{2\hbar}} \left(x + \frac{\hbar}{m\omega} \dfrac{d}{dx}\right) \\ \hat a^\dagger &= \sqrt{\frac{m\omega}{2\hbar}} \left(x - \frac{\hbar}{m\omega} \dfrac{d}{dx}\right) \end{align*} $$

Now substituting these the explicit form of $\hat a$ in the position basis, we have:

$$ \begin{gather*} \hat a|0\rangle = 0 \\ \Rightarrow ~ \hat a\langle x|0\rangle = \hat a\psi_0(x) = 0 \\ \Rightarrow ~ \left(x + \dfrac{\hbar}{m\omega} \dfrac{d}{dx}\right)\psi_0(x) = 0 \end{gather*} $$

Note: It is also possible to do this in the momentum basis but the calculations become much more hairy. See this Physics StackExchange answer if interested.

This gives us a differential equation to solve, albeit a much easier one that can be solved explicitly by the standard methods of solving 1st-order differential equations. The solution is a Gaussian function, and in particular:

$$ \psi_0(x) = C e^{-m\omega x^2 / (2\hbar)} $$

And applying the normalization condition, the undetermined constant $C$ can be found to be $C = \left(\frac{m\omega}{\pi \hbar}\right)^{1/4}$, and thus we have:

$$ \psi_0(x) = \left(\dfrac{m\omega}{\pi \hbar}\right)^{1/4} e^{-m\omega x^2 / (2\hbar)} $$

Note: Since this is a Gaussian function, it is thus symmetric about $x = 0$, and thus we can infer that the expectation value of $\psi_0(x)$ (the quantum harmonic oscillator in its ground state) is $\langle x\rangle = 0$, which is also true for the classical harmonic oscillator.

We can then use the definition of the $\hat a^\dagger$ operator to find all of the wavefunctions for the higher-energy states, since:

$$ \begin{gather*} \hat a^\dagger|n\rangle = \sqrt{n + 1}~|n+1\rangle \\ \Rightarrow ~ \hat a^\dagger\langle x|n\rangle = \sqrt{n + 1}\langle x|n+1\rangle \\ \Rightarrow ~ \hat a^\dagger \psi_n(x) = (\sqrt{n + 1} )~\psi_{n + 1}(x) \end{gather*} $$

Thus, we have:

$$ \begin{align*} \psi_{n + 1}(x) &= \dfrac{1}{\sqrt{n + 1}} \hat a^\dagger \psi_n(x) \\ &= \psi_{n + 1}(x) \\ &= \ \sqrt{\frac{m\omega}{2(n+1)\hbar}} \left(x - \frac{\hbar}{m\omega} \dfrac{d}{dx}\right) \psi_n(x) \end{align*} $$

This allows us to recursively construct all the eigenstates of the system. For instance, we have:

$$ \begin{align*} \psi_1(x) &= \dfrac{1}{\sqrt{0 + 1}}\hat a^\dagger \psi_0(x) \\ &= \ \sqrt{\frac{m\omega}{2\hbar}} \left(x - \frac{\hbar}{m\omega} \dfrac{d}{dx}\right) \left[\left(\frac{m\omega}{\pi \hbar}\right)^{1/4} e^{-m\omega x^2 / (2\hbar)}\right] \\ &= \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \sqrt{\frac{2m\omega}{\hbar}} \, x \, e^{-\frac{m\omega}{2\hbar}x^2} \\ &= \left(\dfrac{4m^3\omega^3}{\pi \hbar^3}\right)^{1/4} x e^{-m\omega x^2 / (2\hbar)} \end{align*} $$

In general, we have:

$$ \begin{gather*} |n\rangle = \dfrac{(\hat a^\dagger)^n}{\sqrt{n!}}|0\rangle, \\ \psi_n(x) = \langle x|n\rangle = \sqrt{\frac{m\omega}{2\hbar n!}} \left(x - \frac{\hbar}{m\omega} \dfrac{d}{dx}\right)^n \psi_0(x) \end{gather*} $$

In the same way, we can get $\psi_2$ from applying the $\hat a^\dagger$ operator on $\psi_1$, then get $\psi_3$ from $\psi_2$, then get $\psi_4$ from $\psi_3$, and so on and so forth. With some clever mathematics (that we won't show here), this recursive formula can be solved in closed-form to yield a generalized formula for the nth eigenstate's wavefunction representation:

$$ \psi_n(x) = \left(\dfrac{m\omega}{\pi \hbar}\right) \dfrac{1}{\sqrt{2^nn!}}H_n\left(\sqrt{\dfrac{m\omega}{\hbar}}x\right)\exp \left(-\dfrac{m\omega x^2}{2\hbar}\right) $$

Here, $H_n(x)$ is a Hermite polynomial of order $n$, defined as:

$$ H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} $$

Where the first three Hermite polynomials are given by:

$$ \begin{align*} H_0(x) &= 1 \\ H_1(x) &= 2x \\ H_2(x) &= 4x^2 - 2 \end{align*} $$

Using these definitions, we show a plot of the ground-state wavefunction $\psi_0(x)$ and several excited states' wavefunctions below:

Source: Wikimedia Commons

We can also keep things abstract by noting that we can express any of the wavefunctions of each state $\psi_n(x)$ as $\psi_n(x) = \langle x|n\rangle$, where:

$$ |n\rangle = \dfrac{(\hat a^\dagger)^n}{\sqrt{n!}}|0\rangle, \quad \psi_0(x) = \langle x|0\rangle $$

The ladder operator approach to the quantum harmonic oscillator is a powerful technique, one that carries over to relativistic quantum mechanics and allows us to skip solving the Schrödinger equation entirely. In addition, solving problems using operators alone will give us the tools to understand the Heisenberg picture of quantum mechanics that we'll soon see.

Note for the advanced reader: The algebraic approach works mathematically because the eigenvalues of $\hat H$ are positive, and that the $\hat p^2$ operator is semi-positive definite. Additionally, the eigenspectrum (energy spectrum) of the system is discrete and non-degenerate (that is, all eigenstates have unique eigenvalues).

Expectation values of the quantum harmonic oscillator

In any quantum system, it is very useful to find their expectation values, and the same is true for the quantum harmonic oscillator. In particular, we are interested in calculating $\langle x\rangle$ and $\langle p\rangle$, the expectation values of the position and momentum. To do so, we will follow an approach from Brilliant wiki's authors. We start with the fact that:

$$ \begin{align*} \hat x' &= \frac{1}{\sqrt{2}}(\hat a^\dagger + \hat a) \\ \hat p' &= \dfrac{i}{\sqrt{2}}(\hat a^\dagger - \hat a) \\ \end{align*} $$

Now, we can express $\hat x'$ and $\hat p'$ in terms of $\hat x$ and $\hat p$ by simply rearranging the definitions of $\hat x'$ and $\hat p'$ from earlier:

$$ \begin{align*} \hat x' &= \sqrt{\dfrac{m\omega}{\hbar}} \hat x \\ \hat p' &= \dfrac{1}{\sqrt{m\hbar \omega}} \hat p \\ \end{align*} \quad \Rightarrow \quad \begin{align*} \hat x &= \sqrt{\dfrac{\hbar}{m\omega}}\hat x' \\ \hat p &= \sqrt{m\hbar \omega}\, \hat p' \end{align*} $$

Thus we have:

$$ \begin{align*} \hat x &= \sqrt{\frac{\hbar}{2 m \omega}} (a^{\dagger} + a)\\ \hat p &= i \sqrt{\dfrac{m \hbar \omega}{2}} (\hat{a}^{\dagger} - a) \end{align*} $$

From here, we can easily calculate the expectation values of $x$ and $p$. We'll start with the expectation value of $x$:

$$ \begin{align*} \langle x \rangle &= \langle n|\hat x|n\rangle \\ &= \langle n| \sqrt{\frac{\hbar}{2 m \omega}} (\hat a^{\dagger} + \hat a)|n\rangle \\ &= \sqrt{\frac{\hbar}{2 m \omega}} \big[\langle n|\hat a^\dagger|n\rangle + \langle n|\hat a |n\rangle\big] \\ &= \sqrt{\frac{\hbar}{2 m \omega}} \big[\sqrt{n + 1} \langle n|n+1\rangle + \sqrt{n}\langle n|n-1\rangle] \\ & = 0 \end{align*} $$

Where we used our definitions $\hat a^\dagger = \sqrt{n + 1} |n+1\rangle$, $\hat a = \sqrt{n} |n-1\rangle$ and also utilized the fact that any two eigenstates are orthogonal, that is, $\langle n|m\rangle = \delta_{mn}$, so $\langle n|n+1\rangle$ and $\langle n|n-1\rangle$ are both automatically zero. We can do the same thing with the momentum operator:

$$ \begin{align*} \langle p \rangle &= \langle n|\hat p|n\rangle \\ &= \langle n| i \sqrt{\dfrac{m \hbar \omega}{2}} (\hat a^{\dagger} - \hat a)|n\rangle \\ &= i \sqrt{\dfrac{m \hbar \omega}{2}} \big[\langle n|\hat a^\dagger|n\rangle - \langle n|\hat a |n\rangle\big] \\ &= i \sqrt{\dfrac{m \hbar \omega}{2}} \big[\sqrt{n + 1} \langle n|n+1\rangle - \sqrt{n}\langle n|n-1\rangle] \\ & = 0 \end{align*} $$

Note: The result that $\langle x\rangle = \langle p\rangle = 0$ only holds true for single eigenstates. It does not necessarily hold true in a superposition of states.

The same methods of calculation can be used to establish that:

$$ \begin{align*} \langle \hat x^2\rangle &= \dfrac{\hbar}{2m\omega}(2n + 1) \\ \langle \hat p^2 \rangle &= \dfrac{\hbar m\omega}{2}(2n + 1) \end{align*} $$

From the formula for the uncertainty of an observable $\Delta A = \sqrt{\langle A^2\rangle - \langle A\rangle^2}$ this tells us that:

$$ \Delta x \Delta p = \dfrac{\hbar}{2}(2n + 1) $$

Where for the ground state ($n = 0$) we have the minimum uncertainty:

$$ \Delta x \Delta p = \dfrac{\hbar}{2} $$

The quantum harmonic oscillator in higher dimensions

It is also possible to solve the quantum harmonic oscillator in higher dimensions. Indeed, consider the quantum harmonic oscillator along a 2D plane or a 3D box, where we can use Cartesian coordinates. Here, we do not actually need to do much more solving at all. The only difference is that rather than a single integer $n$, we need one integer $n$ for each coordinate. That is to say, for 2D we need two integers $n_x, n_y$, and for 3D we need three integers $n_x, n_y, n_z$ to describe all the eigenstates of the system. Therefore, the respective energy eigenvalues are:

$$ \begin{align*} E_{n_x, n_y}^{(2D)} &= \hbar \omega \left(n_x + n_y + \dfrac{1}{2}\right) \\ E_{n_x, n_y, n_z}^{(3D)} &= \hbar \omega \left(n_x + n_y + n_z + \dfrac{1}{2}\right) \end{align*} $$

Note that this means that we have degenerate eigenenergies, losing one of the key distinguishing features of the 1D quantum harmonic oscillator. In addition, the excited-state wavefunctions are unfortunately much more complicated for the 2D and 3D cases; we will restrict our attention to just the ground-state wavefunction. In $K$ dimensions, the ground-state wavefunction is given by:

$$ \psi_0(\mathbf{r}) = \left(\dfrac{m\omega}{\pi \hbar}\right)^{K/4} \exp\left(-\dfrac{m\omega}{2\hbar}r^2\right), \quad r = |\mathbf{r}| $$

Note that in 2D and 3D, we also have different cases, such as the quantum harmonic oscillator across a disk or in a spherical region. In these cases, the ground-state wavefunction exhibits polar and spherical symmetry respectively, so the general solutions are quite different from the 1D case. This is very important in nuclear and molecular physics, although we will not discuss it further here.

Note for the interested reader: If you are interested in further applications of the quantum harmonic oscillator, it can be used to model diatomic molecules like $\ce{N2}$ or $\ce{O2}$ and describe atomic nuclei with the nuclear shell model, as well as serving an important role in second quantization of light - something we'll see more of later.

Time evolution in quantum systems

In all areas of physics, we're often interested in how systems evolve. A system that depends on time is usually called a dynamical system, and at different points in time, the state of the system changes. Now, if we know that at some initial time $t_0$ a system is in a particular state $A$, and at some arbitrary later time $t$ is in another state $B$, the time evolution of the system describes how the system "gets" from $A$ to $B$.

Consider a very simple example: a particle moving along a line. Its state is described by a single variable - position - which we describe with $x$. In physics, we would describe the motion of this particle with a function $x(t)$, which is a trajectory. This trajectory is the time evolution of the system, because from a certain initial time $t_0$, we can calculate the particle's position at any future time $t$ with $x(t)$.

In quantum mechanics, we also see quantum systems exhibit time evolution. For instance, the state-vector may have an initial state $|\psi(t_0)\rangle$ at time $t = t_0$, and at some future time $t$ have the final state $|\psi(t)\rangle$. The question is, how does that initial state become the final state? The answer to that question is the time-evolution operator $\hat U(t, t_0)$, which satisfies:

$$ |\psi(t)\rangle = \hat U(t, t_0)|\psi(t_0)\rangle $$

That is to say, the time-evolution operator maps the system's state at an initial time $t_0$ to its future state at time $t$. But how does this all work? This is what we'll explore in this section.

Unitary operators

Before we go more in-depth into the time-evolution operator, we need to introduce the idea of a unitary operator. An arbitrary unitary operator $\hat U$ (forget about the time-evolution operator for now) satisfies two essential properties:

1. $\hat U^{-1} = U^\dagger$, that is, its inverse is equal to its adjoint.
2. $\hat U \hat U^{-1} = \hat U^{-1} \hat U = 1$, that is, multiplying a unitary operator by its adjoint gives the identity matrix. Together with the first rule, this automatically means that $\hat U \hat U^\dagger = \hat U^\dagger \hat U = 1$.

Note on notation: We will frequently use the shorthand $\hat I = 1$, where $\hat I$ is the identity matrix, when discussing operators, but remember that matrix multiplication always gives another matrix (and not the scalar number 1), so this is just a shorthand!

Note that unitary operator is not necessarily Hermitian - in fact, it usually isn't! So why do we care about a non-Hermitian operator when most of the operators we use in quantum mechanics are Hermitian? Well, if we act a unitary operator $\hat U$ on a (normalized) state-vector, we find that:

$$ (\hat U |\psi\rangle)^\dagger(\hat U |\psi\rangle) = \langle \psi |\hat U^\dagger \hat U|\psi\rangle = \langle \psi|\psi\rangle = 1 $$

This is the most important property of a unitary operator - it preserves the normalization of the state-vector! That is to say, acting $\hat U$ on $|\psi\rangle$ does not change its normalization $\langle \psi|\psi\rangle = 1$.

The unitary time-evolution operator

Now let's return back to the time-evolution operator. It is no accident that we denoted the time-evolution operator as $\hat U(t, t_0)$ and a unitary operator as $\hat U$. This is because the time-evolution operator is a unitary operator. It is indeed common to call the time-evolution operator the unitary time-evolution operator for this very reason! So from this point on, anytime you see $\hat U$, that means the time-evolution operator (unless otherwise stated).

Here is where the unitary nature of the time-evolution operator truly makes sense. This is because by knowing $\hat U \hat U^\dagger = \hat U^\dagger \hat U = 1$, and that $|\psi(t)\rangle = \hat U|\psi\rangle$, we also know that:

$$ \langle \psi(t)|\psi(t)\rangle = \langle \psi(t_0)|\hat U^\dagger\hat U(t, 0)|\psi(t_0)\rangle = \langle \psi(t_0)|\psi(t_0)\rangle = 1 $$

That is, if a state-vector $|\psi\rangle$ is normalized at $t = t_0$, it will continue to be normalized for all future times $t$, satisfying the normalization condition. This means that the time-evolution operator $\hat U$ automatically guarantees conservation of probability in a dynamical quantum system (a system that changes with time). Furthermore, we also add the requirement that the time-evolution operator must satisfy:

$$ \hat U(t_0, t_0) = 1 $$

This means that:

$$ \hat U(t_0, t_0)|\psi(t_0)\rangle = |\psi(t_0)\rangle $$

Therefore operating $\hat U$ on the state-vector returns the system in its initial state at $t = t_0$. This makes sense because at the initial time $t_0$, the system hasn't had any time to evolve, so acting the time-evolution operator on it does nothing but tell you the initial state!

Note: Another name for the unitary time-evolution operator is the propagator, which is common in advanced quantum mechanics. Later on in this guide, when we cover the path integral formulation of quantum mechanics, we'll speak of $\hat U$ as the propagator. Remember that whether we call $\hat U$ the unitary time-evolution operator or the propagator, we are referring to the same thing!

Now, we've spoken a lot about what the time-evolution operator $\hat U$ does, but how do we express it in explicit form? To be able to start, let's write out the Schrödinger equation in a special form. The most general form of the Schrödinger equation - at least, in the form we've generally seen - is given by:

$$ i\hbar \dfrac{\partial}{\partial t}|\psi(t)\rangle = \hat H|\psi(t)\rangle $$

But since $|\psi(t)\rangle = \hat U|\psi\rangle$, this can also be written as:

$$ i\hbar \dfrac{\partial}{\partial t}\hat U|\psi(t_0)\rangle = \hat H(\hat U|\psi(t_0)\rangle) $$

Since $|\psi(t_0)\rangle$ does not depend on time, we can factor it out from both sides, giving us:

$$ i\hbar \dfrac{\partial}{\partial t} \hat U = \hat H \hat U $$

Which can be written more explicitly as:

$$ i\hbar \dfrac{\partial}{\partial t} \hat U(t, t_0) = \hat H \hat U(t, t_0) $$

This is the essential equation of motion for the unitary operator. We'll now do something that may defy intuition but is actually mathematically sound. First, we'll temporarily drop the operator hats and not write out the explicit dependence on $t$ and $t_0$, giving us:

$$ i\hbar \dfrac{\partial U}{\partial t} = HU $$

Now, dividing by $i\hbar$ from both sides gives us:

$$ \dfrac{\partial U}{\partial t} = \frac{1}{i\hbar}HU = -\frac{i}{\hbar} HU $$

(Here we use the fact that $1/i = -i$, and $\dot U = \frac{\partial U}{\partial t}$). This now looks like a differential equation in the form $\dot U = -\frac{i}{\hbar} H U$! Solving this differential equation (using separation of variables) along with our known property $U(t_0) = 1$ gives us:

$$ U = e^{-i H (t - t_0)/\hbar} $$

Now, we can restore the operator hats and we can write the most general form of the time evolution operator:

$$ \hat U(t, t_0) = \exp\left(-\dfrac{i}{\hbar} \hat H (t - t_0)\right) $$

If we adopt the convention of choosing $t_0 = 0$, this gives us:

$$ \hat U(t) = \exp\left(-\dfrac{i}{\hbar} \hat H t\right), \quad \hat U(t) \equiv \hat U(t, 0) $$

Perhaps you might be inclined to answer with "You're wrong! What in the world is the exponential function of a matrix??" The way of making sense of this is to recognize that the exponential function can be defined in terms of a power series:

$$ e^X = \exp(X) = \sum_{n = 0}^\infty \dfrac{X^n}{n!} $$

Taking powers of a matrix is a perfectly acceptable operation, and therefore a term like $\hat H^n$ would raise no alarms, since $\hat H^n = \underbrace{\hat H \hat H \dots \hat H}_{n \text{ times}}$. This allows us to write $\hat U(t)$ in the form:

$$ \begin{align*} \hat U &= \sum_{n = 0}^\infty \frac{1}{n!}\left(-\dfrac{i}{\hbar} \hat H t\right) \\ &= 1 -\frac{i}{\hbar} \hat H t - \frac{1}{2\hbar^2} \hat H^2 t^2 + \dots \end{align*} $$

Usually, applying this definition is quite cumbersome (summing infinite terms is hard!) but if we truncate the series to just a few terms, we can often find a good approximation to the full series. For instance, if we truncate the series to first-order, we have:

$$ \hat U \approx 1 -\frac{i}{\hbar} \hat H t $$

Using this approximation can allow us to calculate the future state of a time-dependent quantum system with only knowledge of the Hamiltonian and the initial state. Of course, since we truncated the series, this calculation can yield only an approximate answer, but in some cases an approximate answer is enough. Thus, the time-evolution operator is the starting-point for perturbative calculations in quantum mechanics, where we can make successively more accurate approximations to the future state of a quantum system by invoking the time-evolution operator in series form, and taking only the first few terms.

The Heisenberg picture

Introducing the time-evolution operator has an interesting consequence: it allows us to calculate the future state of any quantum system from a known initial state "frozen" in time. This is because the initial state of a quantum system has not had time to evolve yet, so it is independent of time. In fact, it is possible to dispense with time-dependence in calculations almost completely, because it turns out that there is also a way to calculate the measurable quantities of quantum systems at any future point in time without needing to explicitly calculate $|\psi(t)\rangle$. This approach is known as the Heisenberg picture in quantum mechanics.

Consider the position operator $\hat X$ (we will use an uppercase $X$ here for clarity). Normally, this is a time-independent operator, since we know it is defined by $\hat X|\psi_0\rangle = x|\psi_0\rangle$, where $|\psi_0\rangle$ is a stationary state and $x$ is a position eigenvalue: notice here that time does not appear at all as a variable. Taking the inner product of both sides with the bra $\langle \psi_0|$ gives us the expectation value of the position:

$$ \langle \psi_0|\hat X |\psi_0\rangle = \langle \psi_0| x|\psi_0\rangle $$

Now, we want to find a time-dependent version of the position operator, which we'll call $\hat x_H(t)$, which also satisfies an eigenvalue equation:

$$ \hat X_H(t)|\psi(t)\rangle = x(t)|\psi(t)\rangle $$

Notice how our position eigenvalue is now time-dependent, because as the state of the system changes, the positions $x(t)$ also change. Our challenge will be able to write $\hat X_H(t)$ in terms of $\hat X$. How can we do so? Well, recall that $|\psi(t)\rangle = \hat U|\psi(t_0)\rangle$, and $|\psi(t_0)\rangle$ is the same thing as $|\psi_0\rangle$. Thus we can write:

$$ \hat X_H(t)|\psi(t)\rangle = \hat X\hat U|\psi_0\rangle $$

Now, let us take its inner product with the bra $\langle \psi(t)|$, which gives us:

$$ \langle \psi(t)|\hat X_H(t)|\psi(t)\rangle = \langle \psi(t) |x\hat U|\psi_0\rangle $$

We'll now use the identity that:

$$ |\psi(t)\rangle = \hat U |\psi_0\rangle \quad \Leftrightarrow \quad |\psi_0\rangle = \hat U^\dagger |\psi(t)\rangle $$

You can prove this rigorously, but it can be intuitively understood by recognizing that $\hat U^\dagger = \hat U^{-1}$, meaning that just as $\hat U$ evolves the system forwards in time, $\hat U^\dagger$ evolves the system backwards in time (the "inverse" direction in time). Hence acting $\hat U^\dagger$ on a system at some time $t$ returns it to its original state at some past time $t_0$. With the same result, we note that:

$$ \langle \psi(t)|\hat X_H(t)|\psi(t)\rangle = \langle \psi(t) |x\hat U|\psi_0\rangle = \langle \psi_0|\hat U^\dagger x \hat U|\psi_0\rangle $$

Thus by pattern-matching we have:

$$ \hat X_H(t) = \hat U^\dagger x \hat U = \hat U^\dagger \hat X \hat U $$

Notice that the latter result holds for all time $t$! We have indeed arrived at our expression for the time-dependent version of the position operator $\hat X_H(t)$:

$$ \hat X_H(t) = \hat U^\dagger \hat X \hat U $$

It is also common to say that $\hat X_H$ is the position operator in the Heisenberg picture. Unlike the Schrödinger picture that we've gotten familiar working with, the Heisenberg picture uses time-dependent operators that operate on a constant state-vector $|\psi\rangle = |\psi_{0}\rangle$. It is completely equivalent to the Schrödinger picture, but it is sometimes more useful, since we can dispense with calculating the state-vector's time evolution as long as we know the $\hat U$ operator, which can simplify (some) calculations. In the most general case, for any operator $\hat A$, its equivalent time-dependent version $\hat A_H$ in the Heisenberg picture is given by:

$$ \hat A_H(t) = \hat U^\dagger \hat A \hat U $$

If we don't know $\hat U$, it is also possible to calculate $\hat A_H$ via the Heisenberg equation of motion, the analogue of the Schrödinger equation in the Heisenberg picture:

$$ \dfrac{d}{dt} \hat{A}_{H}(t) = \frac{i}{\hbar}[\hat{H}, \hat{A}_{H}(t)] $$

A particularly powerful consequence of the Heisenberg picture is how easily it maps classical systems into a corresponding quantum system. For instance, the classical harmonic oscillator follows the equation of motion $\dfrac{d^2 x}{dt^2} + \omega^2 x = 0$, which has the (classical) solution:

$$ \begin{align*} x(t) &= a e^{-i\omega t} + a^* e^{i\omega t} \\ p(t) &= m \dfrac{dx}{dt} = b^*e^{-i\omega t} + be^{i\omega t}, \quad b =i\omega ma^* \end{align*} $$

Where $a, a^*$ here are some amplitude constants that can be specified by the initial conditions, and for generality, we assume that they can be complex-valued. Now, the Heisenberg picture tells us that if we want to find the corresponding quantum operators $\hat X_H(t), \hat p(t)$, all we have to do is to change our constants $a, a^*$ to operators $\hat a, \hat a^\dagger$ (same with $b, b^*$), giving us:

$$ \begin{align*} \hat X_H(t) &= \hat ae^{-i\omega t} + \hat a^\dagger e^{i\omega t} \\ \hat p(t) &= \hat be^{-i\omega t} + \hat b^\dagger e^{i\omega t}, \quad \hat b = i\omega m a^\dagger \end{align*} $$

Indeed, we can then identify $\hat a, \hat a^\dagger$ as just the ladder operators we're already familiar with from studying the quantum harmonic oscillator! In addition, we can also show that $\hat x$ satisfies a nearly identical equation of motion as the classical case ($\frac{d^2 x}{dt^2} + \omega^2 x = 0$), with the exception that the position function $x$ is replaced by the operator $\hat X_{H}$:

$$ \dfrac{d^2 \hat X_H(t)}{dt^2} + \omega^2 \hat X_H(t) = 0 $$

Notice the elegance correspondence between the classical and quantum pictures. By doing very little work, we have quantized a classical system, taking a classical variable ($x(t)$, representing a particle's position) and turning it ("promoting it") into a quantum operator $\hat X_{H}(t)$, a process formally called first quantization. This method will be essential once we discuss second quantization, where we take classical field theories and use them to construct quantum field theories. But we've not gotten to there yet! We'll save a more in-depth discussion of second quantization for later.

Note for the advanced reader: In second quantization, we essentially do the same thing as first quantization, but rather than quantizing the position (by taking the classical variable $x(t)$ and promoting it to an operator $\hat X_{H}$) we are interested in taking a classical field $\phi(x, t)$ and promoting it to an quantum field operator $\hat \phi$. Just as in first quantization, second-quantized fields follow the same equations of motion as their classical field analogues. In particular, the simplest type of quantum field (known as the free scalar field) obeys the equation $\partial^2_{t}\hat \phi - \nabla^2 \phi + m^2\phi = 0$, which is very similar to the harmonic oscillator equation of motion.

The correspondence principle and the classical limit

As we have seen, the Heisenberg picture makes it easy to show the intricate connection between quantum mechanics and classical mechanics, which is also known as the correspondence principle. The correspondence principle is essential because it explains why we live in a world that can be so well-described by classical mechanics, even though we know that everything in the Universe is fundamentally quantum at the tiniest scales. A key part of the correspondence principle is Ehrenfest's theorem, which is straightforward to prove from the Heisenberg picture. We start by writing down the Heisenberg equation of motion (which we introduced earlier), given by:

$$ \dfrac{d}{dt} \hat{A}_{H}(t) = \frac{i}{\hbar}[\hat{H}, \hat{A}_{H}(t)] $$

The Heisenberg equations of motion for the position and momentum operators $\hat X_{H}(t)$, $\hat P_{H}(t)$ are therefore:

$$ \begin{align*} \dfrac{d \hat X_H(t)}{dt} = \frac{i}{\hbar}[\hat{H}, \hat X_H(t)] \\ \dfrac{d \hat P_H(t)}{dt} = \frac{i}{\hbar}[\hat{H}, \hat P_H(t)] \end{align*} $$

If we take the expectation values for each equation on both sides, we have:

$$ \begin{align*} \left\langle\dfrac{d \hat X_H(t)}{dt}\right\rangle = \frac{i}{\hbar}\langle[\hat{H}, \hat X_H(t)]\rangle \\ \left\langle\dfrac{d \hat P_H(t)}{dt}\right\rangle = \frac{i}{\hbar}\langle[\hat{H}, \hat P_H(t)]\rangle \end{align*} $$

Now making use of the fact that $[\hat H, \hat X_{H}(t)] = -i\hbar \frac{\hat{P}_{H}}{m}$ and $[\hat H, \hat P_{H}(t)] = i\hbar \nabla V$ (we won't prove this, but you can show this yourself by calculating the commutators with $\hat H = \hat P^2/2m + V$) we have:

$$ \begin{align*} \left\langle\dfrac{d \hat X_H(t)}{dt}\right\rangle = \frac{\langle \hat{P}_{H}\rangle}{m} \\ \left\langle\dfrac{d \hat P_H(t)}{dt}\right\rangle = \langle -\nabla V\rangle \end{align*} $$

The first equation tells us that the expectation value of the position is equal to the expectation value of the momentum, divided by the mass. In the classical limit, this is exactly $\dot x = p/m$, which comes directly from the classical definition of the momentum $p = mv = m\dot x$! Meanwhile, the second equation tells us that the expectation value of the momentum is equal to $\langle -\nabla V\rangle$. This is (approximately) the same as Newton's second law $F = \frac{dp}{dt} = -\nabla V$. Thus, Ehrenfest's theorem says that at classical scales, quantum mechanics reduces to classical mechanics; this is why we don't observe any quantum phenomena in our everyday lives!

The interaction picture

Using Heisenberg's approach to quantum mechanics is powerful, but it often comes at the cost of needing to compute a lot of operators. The physicist Paul Dirac looked at the Heisenberg picture, and decided that there was a better way that would simplify the calculations substantially, while preserving all of the physics of a quantum system. His equivalent approach is known as the interaction picture, although it is often also called the Dirac picture (obviously after him).

We will quickly go over the interaction picture for the sake of brevity. Essentially, it says that we can split a quantum system into two parts - a non-interacting part and an interacting part. When we say "non-interacting", we mean a hypothetical system that is completely isolated from the outside world and is essentially in a Universe of its own. To do this, we write the Hamiltonian of the system as the sum of a non-interacting Hamiltonian $\hat H_0$ and an interaction Hamiltonian $\hat W$:

$$ \hat{H} = \hat{H}_{0} + \hat{W} $$

As with the Schrödinger picture, the state-vector of the system $|\psi(t)\rangle$ will depend on time. But here is where the interaction picture begins to differ from the Schrödinger picture. First, let us consider the time-evolution operator $\hat U_0(t, t_{0}) = e^{-i\hat H_0 (t-t_{0})/\hbar}$. Strictly-speaking, this time-evolution operator is only valid for the non-interacting part of the system, since it comes from $\hat H_0$, the non-interacting Hamiltonian. We will now define a modified state-vector $|\psi_I(t)\rangle$, which is related to the original state-vector of the system $|\psi(t)\rangle$ by:

$$ |\psi_{I}(t)\rangle = \hat U_{0}^{\dagger} |\psi(t)\rangle = e^{i \hat{H}_{0} (t - t_{0})/\hbar} |\psi(t)\rangle $$

We can of course also invert this relation to write $|\psi(t)\rangle$ in terms of $|\psi_I\rangle$, as follows:

$$ |\psi(t)\rangle = \hat U_{0} |\psi_{I}(t)\rangle = e^{-i \hat{H}_{0} (t - t_{0})/\hbar} |\psi_{I}(t)\rangle $$

We can write an arbitrary operator $\hat A$ in its interaction picture representation, which we will denote with $\hat A_I$, via:

$$ \hat{A}_{I}(t) = \hat U_{0}^{\dagger} \hat{A} \hat U_{0} = e^{i \hat{H}_{0} (t - t_{0})/\hbar} \hat{A} e^{-i \hat{H}_{0} (t - t_{0})/\hbar} $$

In addition, an operator's representation in the interaction picture follows the equation of motion:

$$ i\hbar\frac{d \hat{A}_{I}}{dt} = [\hat{A}_{I}, \hat{H}_{0}] $$

Our modified state-vector $|\psi_I(t)\rangle$ then satisfies the following equation of motion:

$$ i\hbar \frac{d}{dt}|\psi_{I}(t)\rangle = \hat{W}_{I}|\psi_{I}(t)\rangle $$

Where $\hat W_I = \hat U_{0}^{\dagger} \hat{W} \hat U_{0}$ is the interaction picture representation of the interaction Hamiltonian $\hat W$. What this means is that using the interaction picture, we can isolate the interacting parts of the system from the non-interacting parts of the system - something that isn't possible to do in the Heisenberg or Schrödinger pictures! The interacting part of the system follow the equation of motion we already presented for $|\psi_I(t)\rangle$, whereas the non-interacting part satisfies the equation of motion for $\hat U_0$:

$$ i\hbar \dfrac{\partial}{\partial t} \hat U_{0}(t, t_0) = \hat H_{0} \hat U_{0}(t, t_0) $$

Since these two equations of motion are completely decoupled from each other, we can solve for the interacting and non-interacting parts separately. Once we have successfully solved for $|\psi_I(t)\rangle$ and $\hat U_0$, the state-vector of the full system is just a unitary transformation away, since:

$$ |\psi(t)\rangle = \hat U_{0} |\psi_{I}(t)\rangle $$

The interaction picture is powerful because it allows us to describe a quantum system that undergoes very complicated interactions as if those interactions were not present, and simply "layer" the interactions on top. This is an idea essential to solving very complicated quantum systems, especially once we get to the topic of time-dependent perturbation theory in quantum mechanics. As an added bonus, it turns out that under certain circumstances it is possible to write out an exact series solution to solve for the interacting part of a system. As long as we assume that interactions are reasonably "small", we can convert the equation of motion for the interacting part of the system into an integral equation:

$$ \begin{gather*} i\hbar \frac{d}{dt}|\psi_{I}(t)\rangle = \hat{W}_{I}|\psi_{I}(t)\rangle \\ \downarrow \\ |\psi_{I}(t) = |\psi_{I}(t_{0})\rangle + \frac{1}{i\hbar} \int_{t_{0}}^t dt' W_{I}(t')|\psi_{I}(t')\rangle \end{gather*} $$

One can then write out a series solution that solves the integral equation, which is given by:

$$ \begin{align*} |\psi_{I}(t) = \bigg\{1 &+ \frac{1}{i\hbar} \int dt_{1} W_{I}(t_{1}) + \frac{1}{(i\hbar)^2}\int dt_{1} dt_{2} W_{I}(t_{1})W_{I}(t_{2}) \\ &+ \dots + \frac{1}{(i\hbar)^n} \int dt_{1}dt_{2} \dots dt_{n} W_{I}(t_{1})W_{I}(t_{2}) \dots W_{I}(t_{n})\bigg\}|\psi_{I}(t_{0})\rangle \end{align*} $$

This is the Dyson series. Right now, the Dyson series is unimportant to us, but it has a great deal of importance in analyzing scattering. We have already seen scattering-state solutions to the Schrödinger equation, like the case of the rectangular potential barrier. But quantum-mechanical scattering is far more broad, and the Dyson series provides us with a way to calculate very complex scattering interactions in a solvable way. In fact, this technique is so general that it is even used in quantum field theory!

Summary of time evolution

We have seen that there are three equivalent approaches to understanding the time evolution of the quantum system: the Schrödinger picture, Heisenberg picture, and interaction (or Dirac) picture. In the Schrödinger picture, operators are time-independent but states are time-dependent; in the Heisenberg picture, operators are time-dependent but states are time-independent; and finally, in the interaction picture, both are time-dependent. Each of these approaches has their own strengths and weaknesses, and they are useful in different scenarios. The key idea is that having these different approaches to describing quantum systems gives us powerful tools to solve these systems, even if we don't have to use them all the time.

Angular momentum

In quantum mechanics, we are often interested in central potentials, that is, potentials in the form $V = V(r)$. For instance, the hydrogen atom can be modelled by a Coulomb potential $V(r) \propto 1/r$, and a basic model of the atomic nucleus uses a harmonic potential $V(r) \propto r^2$.

Note: In case it was unclear, in central potential problems, $r = \sqrt{x^2 + y^2 + z^2}$ is the radial coordinate.

Due to the symmetry of such problems, it is often convenient to use a radially-symmetric coordinate system, like polar coordinates (in 2D) or cylindrical/spherical coordinates (in 3D). This leads to an interesting result - the conservation of angular momentum. A rigorous explanation of why this is the case requires Noether's theorem, which is explained in more detail in the classical mechanics guide. There are a few differences, however. For instance, while classical central potentials lead to orbits around the center-of-mass of a system, the idea of orbits is somewhat vague in quantum mechanics since the idea of probability waves "orbiting" doesn't really make sense. However, for ease of visualization (and also due to some historical reasons), it is still common to say that central potential problems in quantum mechanics have "orbits", and thus we conventionally call this associated type of angular momentum the orbital angular momentum, denoted $\mathbf{L}$.

In addition, a classical spinning object also has angular momentum, and likewise a quantum particle also does - again, this is why we say that electrons (and other spin-1/2 particles) have spin, since they do have angular momentum in the form of spin angular momentum. Since we know the relationship between the magnetic moment $\boldsymbol{\mu}$ and the spin angular momentum $\mathbf{S}$ (it is proportional to a factor of $\gamma$, the gyromagnetic ratio), we can rearrange to find $\mathbf{S}$:

$$ \boldsymbol{\mu} = \gamma \mathbf{S} \quad \Rightarrow \quad \mathbf{S} = \frac{\boldsymbol{\mu}}{\gamma} $$

It is important to recognize that spin angular momentum $\mathbf{S}$ is different from the orbital angular momentum $\mathbf{L}$. They, however, share one important similarity - they are both conserved quantities. This means they obey some similar behaviors. Additionally, the study of orbital angular momentum is extremely important for understanding some of the most important problems in quantum mechanics, so we will explore it in detail.

Stationary perturbation theory

Perturbation theory exists when we come upon a problem that is too complicated to solve exactly. These problems are often (but not always) variations of existing problems. For instance, we know the solution of the hydrogen atom, since that can be solved exactly, but it turns out that for the helium atom, which has just one more electron than the hydrogen atom, there is no analytical solution! In such cases, we typically resort to one of two options:

1. Solve the system on a computer using numerical methods
2. Find an approximate analytical solution

The second option is what we'll focus on here, since numerical methods in quantum mechanics is a topic broad enough for an entire textbook on its own. This approach - making calculations using approximations - is known as perturbation theory, and it allows us to solve many kinds of problems that cannot be solved exactly.

Note: Perturbation theory, despite its association with quantum mechanics, is actually a far more general technique for solving complicated differential equations (even those describing classical systems). For more information, see this excellent article on a classical application of perturbation theory.

First off, we should mention that there are two general kinds of perturbation theory in quantum mechanics: stationary perturbation theory, which (as the name suggests) applies only for stationary (time-independent) problems, and time-dependent perturbation theory, which applies for problems that explicitly depend on time. Right now, we'll be focusing on stationary perturbation theory; we'll get to the time-dependent version later. While there are notable differences, both types of perturbation theory use the same general method: a complicated system is approximated as a simpler, more familiar system with some added corrections (called perturbations). By computing these correction terms, we are then able to find an approximate solution to the system, even if there is no exact analytical solution.

A description of perturbation theory from XKCD.

Non-degenerate perturbation theory

We will first review the simplest type of stationary perturbation theory, known as non-degenerate perturbation theory, which applies to quantum systems without degeneracy (meaning that each eigenstate is uniquely specified by an energy eigenvalue of the Hamiltonian). It turns out that this is in many cases an overly simplified assumption, but the methods we will develop here will be extremely useful for our later discussion of degenerate perturbation theory that accurately describes a variety of real-world quantum systems.

The starting point in perturbation theory is to assume that the Hamiltonian of a complicated system can be written as a sum of a Hamiltonian $\hat H_0$ with an exact solution and a small perturbation $\hat{W}$, such that:

$$ \hat{H} = \hat{H}_{0} + \lambda\hat{W} $$

Note: Here $\hat H_0$ is known as the unperturbed Hamiltonian or free Hamiltonian. Also, it is common to write $\hat W$ without the operator hat, and it is also common to denote it as $V$ (confusingly). Be aware that in all cases, $W$ is an operator, not a function!

For instance, $\hat H_0$ might be the Hamiltonian of a free particle, or of the hydrogen atom, or the quantum harmonic oscillator. The key commonality here is that $\hat H_0$ must be the Hamiltonian of a simpler system that can be analytically solved. On top of $\hat H_0$ we add the perturbation $\hat W$, which represents the deviations (also called perturbations) of the system's Hamiltonian as compared to the simpler system. This perturbation is assumed to be small, so we scale it by a small number $\lambda$ (where $\lambda \ll 1$), giving us a term of $\lambda \hat W$. If we write out the Schrödinger equation for the system, we have:

$$ \hat{H}|\varphi_{n}\rangle = E_{n} |\varphi_{n}\rangle \quad \Rightarrow \quad (\hat{H}_{0} + \lambda\hat{W})\varphi_{n}\rangle = E_{n} |\varphi_{n}\rangle $$

Note that when we take the limit $\lambda \to 0$, the perturbation vanishes, and the Hamiltonian is exactly the unperturbed Hamiltonian $\hat H_0$. This is why perturbation theory is an approximation; it assumes that the simpler system's Hamiltonian $\hat H_{0}$ is already close enough to the more complicate system's Hamiltonian $\hat H$ that $\hat H_0$ can be used to approximate $\hat H$.

The key idea of perturbation theory is that we assume a series solution for $\hat{H}|\varphi_{n}\rangle = E_{n}|\varphi_{n}\rangle$. More accurately, we assume that we can write the solution in terms of a power series in powers of $\lambda$. Now, this assumption doesn't always work - in fact there are some systems where it doesn't work at all - but using this assumption makes it possible to find an approximate solution using analytical methods, which is "good enough" for most purposes. Remember, in the real world, it is impossible to measure anything to infinite precision, so having an approximate answer to a problem that is close enough to the exact solution is often more than sufficient to make testable predictions that align closely with experimental data.

But let's get back to the math. For our solution to be expressed as a power series in $\lambda$, we would write:

$$ \begin{align*} |\varphi_{n}\rangle &= \sum_{m = 0}^\infty \lambda^m|\varphi_{n}^{(m)}\rangle \\ &=|\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots \end{align*} $$

Here, remember that $|\varphi_{n}\rangle$ is the exact solution to the system (representing all $n$ exact eigenstates of the complicated Hamiltonian $\hat H$), but $|\varphi_{n}^{(0)}\rangle, |\varphi_{n}^{(1)}\rangle, |\varphi_{n}^{(2)}\rangle, \dots$ are successive states whose sum converges to the exact eigenstates of the system. (For those who need a refreshed on power series please see the series and sequences guide). Be aware that the brackets $(1), (2), \dots$ are not exponents; rather they are labels for the successive sets of eigenstates (the first set of eigenstates, the second, the third, and so forth). By summing up infinitely many of these terms in the expansion of the Hamiltonian's eigenstates, we would in principle get the exact eigenstates of the complicated Hamiltonian.

In the same way, we assume that the system's energy eigenvalues $E_n$ can also be written as a power series in $\lambda$, given by:

$$ \begin{align*} E_{n} &= \sum_{m=0}^\infty \lambda^m E_{n}^{(m)} \\ &= E_{n}^{(0)} + \lambda E_{n}^{(1)} + \lambda^2 E_{{n}}^{(2)} + \dots \end{align*} $$

The first term in the expansion, $E_n^{(0)}$, as we'll see, are simply the energy eigenvalues of the unperturbed Hamiltonian $\hat H_0$. The subsequent terms $E_{n}^{(1)}$, $E_{n}^{(2)}$ are known as the first-order correction and second-order correction to the energy eigenvalues, since they respectively have coefficients of $\lambda^1$ and $\lambda^2$. By summing up infinitely many of these terms in the expansion of the energy, we would in principle get the exact energies.

Now, if we substitute our power series solution into the Hamiltonian's eigenvalue equation $\hat{H}|\varphi_{n}\rangle = E_{n}|\varphi_{n}\rangle$, we have:

$$ \begin{align*} (\hat{H}_{0} + \lambda \hat{W})(|\varphi_{n}^{(0)}\rangle &+ \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots) \\ &= (E_{n}^{(0)} + \lambda E_{n}^{(1)} + \lambda^2 E_{{n}}^{(2)} + \dots)(|\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots) \end{align*} $$

Distributing the left-hand side gives us:

$$ \begin{align*} \hat{H}_{0}\bigg(|\varphi_{n}^{(0)}\rangle &+ \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots\bigg) + \lambda \hat{W}\bigg(|\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots\bigg) \\ &= E_{n}^{(0)}\left(|\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots\right) \\ &\qquad+ \lambda E_{n}^{(1)}\left(|\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots\right)\\ &\qquad+ \lambda^2 E_{n}^{(2)}\left(|\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle + \dots\right) \end{align*} $$

If we do some algebraic manipulation to group terms by powers of $\lambda$, we get:

$$ \begin{align*} % LHS of equation \hat{H}_{0}|\varphi_{n}^{(0)}\rangle &+ \lambda \left(\hat{H}_{0}|\varphi_{n}^{(1)}\rangle + \hat{ W}|\varphi_{n}^{(0)}\rangle\right) + \lambda^2\left(\hat{H}_{0}|\varphi_{n}^{(2)}\rangle + \hat{W}|\varphi_{n}^{(1)}\rangle\right) + \dots \\ &= % RHS of equation E_{n}^{(0)}|\varphi_{n}^{(0)}\rangle + \lambda\left(E_{n}^{(0)}|\varphi_{n}^{(1)}\rangle + E_{n}^{(1)}|\varphi_{n}^{(0)}\rangle\right) + \lambda^2 \left(E_{n}^{(0)}|\varphi_{n}^{(2)}\rangle + E_{n}^{(1)}|\varphi_{n}^{(1)}\rangle + E_{n}^{(2)}|\varphi_{n}^{(0)}\rangle\right) + \dots \end{align*} $$

Notice how each term on the left-hand side of the equation now corresponds to a term on the right-hand side with the same power of $\lambda$. Thus, by equating the quantities in the brackets for every power of $\lambda$, we get a system of equations to solve for each order of $\lambda$:

$$ \begin{align*} \mathcal{O}(\lambda^0):& \quad E_{n}^{(0)}|\varphi_{n}^{(0)}\rangle \\ \mathcal{O}(\lambda^1):& \quad \lambda \left(\hat{H}_{0}|\varphi_{n}^{(1)}\rangle + \hat{W}|\varphi_{n}^{(0)}\rangle\right) = \lambda\left(E_{n}^{(0)}|\varphi_{n}^{(1)}\rangle + E_{n}^{(1)}|\varphi_{n}^{(0)}\rangle\right) \\ \mathcal{O}(\lambda^2):& \quad \lambda^2\left(\hat{H}_{0}|\varphi_{n}^{(2)}\rangle + \hat{W}|\varphi_{n}^{(1)}\rangle\right) = \lambda^2 \left(E_{n}^{(0)}|\varphi_{n}^{(2)}\rangle + E_{n}^{(1)}|\varphi_{n}^{(1)}\rangle + E_{n}^{(2)}|\varphi_{n}^{(0)}\rangle\right) \\ & \qquad\vdots \\ \mathcal{O}(\lambda^n): &\quad \lambda^n\left(\hat{H}_{0}|\varphi_{n}^{(n)}\rangle + \hat{W}|\varphi_{n}^{(n-1)}\rangle\right) = \lambda^n\left( E_{n}^{(0)}|\varphi_{n}^{(n)}\rangle + \sum_{j = 1}^n E_{n}^{(j)} \left|\varphi_{n}^{(n - j)}\right\rangle\right) \end{align*} $$

Note: The final, generalized expression for $\mathcal{O}(\lambda^n)$ comes from Dr. Moore's Lecture Notes from Michigan State University.

If we solve every single one of these equations and substituted our found values of the energy corrections $E_n^{(1)}, E_n^{(2)}, E_n^{(3)}, \dots$ and the corrections to the eigenstates $|\varphi_{n}^{(1)}\rangle, |\varphi_{n}^{(2)}\rangle, |\varphi_{n}^{(3)}\rangle, \dots$ we would in principle know the exact eigenstates and energies of the system.

However, in practice, we obviously wouldn't want to solve infinitely many equations, so we usually truncate the series to just a few terms to get an approximate answer to our desired accuracy. For the lowest-order approximation (also called the zeroth-order approximation) we keep only terms of order $\mathcal{O}(\lambda^0)$ - or in simpler terms, drop all terms containing $\lambda$. We are thus left with just the equation for $\mathcal{O}(\lambda^0)$, that is:

$$ \hat H_0|\varphi_n^{(0)}\rangle = E_n^{(0)}|\varphi_n^{(0)}\rangle $$

The result is trivial - this is just the eigenvalue equation of the unperturbed Hamiltonian, which we can solve exactly, and tells us nothing new. However, let's keep going, because the first-order approximation will be where we'll find a crucial result from perturbation theory. In the first-order approximation we include all terms up to first-order in $\lambda$, but no higher-order terms (i.e. ignoring $\lambda^2, \lambda^3, \lambda^4, \dots$ terms). This means that:

$$ |\varphi_{n}\rangle \approx |\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle, \quad E_n \approx E_{n}^{(0)} + \lambda E_{n}^{(1)} $$

We will thus also need to solve the second equation in the system of equations we previously derived, given by:

$$ (\hat{H}_{0} - E_{n}^{(0)})|\varphi_{n}^{(1)}\rangle + \hat{W}|\varphi_{n}^{(0)}\rangle = E_{n}^{(1)}|\varphi_{n}^{(0)}\rangle $$

Now, the trick is to take the inner product of the above equation with the bra $\langle \varphi_n^{(0)}|$. This gives us:

$$ \langle \varphi_n^{(0)}|(\hat{H}_{0} - E_{n}^{(0)})|\varphi_{n}^{(1)}\rangle + \langle \varphi_n^{(0)}|\hat{W}|\varphi_{n}^{(0)}\rangle = \langle \varphi_n^{(0)}|E_{n}^{(1)}|\varphi_{n}^{(0)}\rangle $$

Since $\hat H_0$ is a Hermitian operator, we know that for any two states $|\phi\rangle, |\psi\rangle$, it must be the case that $\langle \phi|\hat H_0|\psi\rangle = \big(\langle \phi|\hat H_0\big)\cdot|\psi\rangle$, meaning that:

$$ \langle \varphi_n^{(0)}|(\hat{H}_{0} - E_{n}^{(0)})|\varphi_{n}^{(1)}\rangle = \underbrace{ \bigg(\langle \varphi_n^{(0)}|\hat{H}_{0} - \langle \varphi_n^{(0)}|E_{n}^{(0)}U\bigg) }_{ \hat H_0|\varphi_n^{(0)}\rangle = E_n^{(0)}|\varphi_n^{(0)}\rangle }|\varphi_{n}^{(1)}\rangle = 0 $$

Thus the entire first term goes to zero, and we are simply left with:

$$ \langle \varphi_n^{(0)}|\hat{W}|\varphi_{n}^{(0)}\rangle = \langle \varphi_n^{(0)}|E_{n}^{(1)}|\varphi_{n}^{(0)}\rangle $$

But since our states are normalized, then it must be the case that the right-hand side reduces to:

$$ \begin{align*} \langle \varphi_n^{(0)}|E_{n}^{(1)}|\varphi_{n}^{(0)}\rangle &= E_{n}^{(1)} \underbrace{ \langle \varphi_n^{(0)}|\varphi_{n}^{(0)}\rangle }_{ 1 } = E_{n}^{(1)} \\ &\Rightarrow~\langle \varphi_n^{(0)}|\hat{W}|\varphi_{n}^{(0)}\rangle = \langle \varphi_n^{(0)}|E_{n}^{(1)}|\varphi_{n}^{(0)}\rangle = E_{n}^{(1)} \end{align*} $$

Finally, after fully simplifying our results, we come to a refreshingly-simple expression for the first-order correction to the eigenenergies:

$$ E_n^{(1)} = \langle \varphi_n^{(0)}|\hat W |\varphi_{n}^{(0)}\rangle $$

This is one of the most important equations in all of quantum mechanics and in most cases gives a good approximation to the exact eigenenergies of the system, at least where $\lambda$ is small. Note that the result is very general since it applies for all $n$ eigenstates of the system. Adding in the first-order corrections gives us the (approximate) eigenenergies of the system:

$$ \begin{align*} E_n &\approx E_{n}^{(0)} + \lambda E_{n}^{(1)} \\ &= E_{n}^{(0)} + \lambda \langle \varphi_n^{(0)}|\hat W |\varphi_{n}^{(0)}\rangle \end{align*} $$

We can use a similar process to get the first-order correction $|\varphi_n^{(1)}\rangle$ to the eigenstates of the system. We'll spare the derivation for now and just state the results - the first-order correction to the system's eigenstates are given by:

$$ \begin{align*} |\varphi_n^{(1)}\rangle &= \sum_{m\,(m \neq n)} \frac{E_{n}^{(1)}}{\left(\small E_{n}^{(0)} - E_{m}^{(0)}\right)}|\varphi_m^{(0)}\rangle \\ &= \sum_{m\,(m \neq n)} \frac{\langle \varphi_m^{(0)}|\hat W |\varphi_{n}^{(0)}\rangle}{\left(\small E_{n}^{(0)} - E_{m}^{(0)}\right)}|\varphi_m^{(0)}\rangle \end{align*} $$

In most cases, the first-order correction is sufficient to get a "good enough" answer. But we can go further to get a more accurate result! We'll now use a second-order approximation, where we include all terms up to second-order in $\lambda$, but no higher-order terms (i.e. ignoring $\lambda^3, \lambda^4, \lambda^5, \dots$ terms). This means that:

$$ \begin{align*} |\varphi_{n}^{(0)}\rangle &\approx |\varphi_{n}^{(0)}\rangle + \lambda|\varphi_{n}^{(1)}\rangle + \lambda^2|\varphi_{n}^{(2)}\rangle \\ E_{n} &\approx E_{n}^{(0)} + \lambda E_{n}^{(1)} + \lambda^2 E_{{n}}^{(2)} \end{align*} $$

We'll therefore need the third equation in the system of equations we derived at the start of this section, which is given by:

$$ \lambda^2\left(\hat{H}_{0}|\varphi_{n}^{(2)}\rangle + \hat{W}|\varphi_{n}^{(1)}\rangle\right) = \lambda^2 \left(E_{n}^{(0)}|\varphi_{n}^{(2)}\rangle + E_{n}^{(1)}|\varphi_{n}^{(1)}\rangle + E_{n}^{(2)}|\varphi_{n}^{(0)}\rangle\right) $$

Again, making some algebraic simplifications gives us:

$$ (\hat{H}_{0} - E_{n}^{(0)})|\varphi_{n}^{(2)}\rangle + \hat{W}|\varphi_{n}^{(1)}\rangle = E_{n}^{(1)}|\varphi_{n}^{(1)}\rangle + E_{n}^{(2)}|\varphi_{n}^{(0)}\rangle $$

Using our trick from before by taking the inner product with $\langle \varphi_n^{(0)}|$ and exploiting orthogonality, we get:

$$ \underbrace{ \langle \varphi_n^{(0)}|(\hat{H}_{0} - E_{n}^{(0)}) }_{ 0 }|\varphi_{n}^{(2)}\rangle + \langle \varphi_n^{(0)}|\hat{W}|\varphi_{n}^{(1)}\rangle = E_{n}^{(1)}\cancel{ \langle \varphi_n^{(0)}|\varphi_{n}^{(1)}\rangle }^0 + E_{n}^{(2)}\cancel{ \langle \varphi_n^{(0)}|\varphi_{n}^{(0)}\rangle }^1 $$

Where the first term again becomes zero since $\hat{H}_{0}|\varphi_{n}^{(0)}\rangle = E_{n}^{(0)}|\varphi_{n}^{(0)}\rangle$ and since $\hat H_0$ is Hermitian - this follows the same reasoning we explained for the first-order case. We thus have:

$$ E_{n}^{(2)} = \langle \varphi_n^{(0)}|\hat{W}|\varphi_{n}^{(1)}\rangle $$

But we previously found that $|\varphi_n^{(1)}\rangle$ is given by:

$$ |\varphi_n^{(1)}\rangle = \sum_{m\,(m \neq n)} \frac{\langle \varphi_m^{(0)}|\hat W |\varphi_{n}^{(0)}\rangle}{\left(\small E_{n}^{(0)} - E_{m}^{(0)}\right)}|\varphi_m^{(0)}\rangle $$

Thus substituting it into our expression for $E_n^{(2)}$ gives us an explicit expression for the second-order corrections to the eigenenergies of the system:

$$ \begin{align*} E_{n}^{(2)} &= \langle \varphi_n^{(0)}|\hat{W}|\varphi_{n}^{(1)}\rangle \\ &= \langle \varphi_{n}^{(0)}|\hat{W} \left(\sum_{m\,(m \neq n)} \frac{\langle \varphi_m^{(0)}|\hat W |\varphi_{n}^{(0)}\rangle}{\left(\small E_{n}^{(0)} - E_{m}^{(0)}\right)}|\varphi_m^{(0)}\rangle\right) \\ &= \sum_{m\,(m \neq n)} \frac{|\langle \varphi_m^{(0)}|\hat W |\varphi_{n}^{(0)}\rangle|^2}{\left(\small E_{n}^{(0)} - E_{m}^{(0)}\right)} \end{align*} $$

While we will not derive it here, one may show that the third-order corrections to the eigenenergies of the system are given by:

$$ E_{n}^{(n)} = \sum_{m~(m \neq n)}\sum_{l} \frac{V_{nl} V_{lm} V_{mn}}{\small (E_{n}^{(0)} - E_{l}^{(0)})(E_{n}^{(0)} - E_{m}^{(0)})} - V_{nn}\sum_{m\,(m \neq n)} \frac{|V_{nm}|^2}{\left(\small E_{n}^{(0)} - E_{m}^{(0)}\right)^2}|\varphi_m^{(0)}\rangle $$

Where here, $V_{ij} \equiv \langle \varphi_{i}^{(0)}|\hat{W}|\varphi_{j}^{(0)}\rangle$. Note that in the most general case, we can find the $k$-th order correction to the eigenenergies of the system via:

$$ E_{n}^{(k)} = \langle \varphi_{n}^{(0)}|\hat{W}|\varphi_{n}^{(k - 1)}\rangle $$

Note: For more in-depth discussion of the formulas for perturbation theory up to arbitrary order, see this Physics StackExchange post.

Advanced quantum theory

Relativistic wave equations and the Dirac equation

Thus, we arrive at the Dirac equation for a free particle:

$$ (i\hbar \gamma^\mu \partial_\mu - mc)\psi = 0 $$

The Dirac equation with the electromagnetic four-potential $A_\mu = (A_0, \mathbf{A}) = (\frac{1}{c} V, \mathbf{A})$ takes a very similar form, except the partial derivative $\partial_\mu$ is replaced by a new differential operator $D_\mu$:

$$ (i\hbar \gamma^\mu D_\mu - mc)\psi = 0, \quad D_\mu = \partial_\mu + \dfrac{ie}{\hbar} A_\mu $$

Second quantization and quantum electrodynamics

In this section, we will not analyze the full relativistic theory of quantum electrodynamics. For that, see my quantum field theory book. Rather, we will discuss the non-relativistic theory of quantum electrodynamics (often referred to as NRQED for short), which nonetheless has many applications, including quantum optics and quantum information theory.

The process of going from classical electrodynamics to quantum electrodynamics is called second quantization, a term to differentiate it from first quantization, where we take classical variables (e.g. position, momentum, and angular momentum) and translate them into quantum operators.

To start, we note that an arbitrary electromagnetic field with electric potential $\phi$ and magnetic potential $\mathbf{A}$ can be decomposed as a sum (or integral) of plane waves (called modes), each of different wavevector $\mathbf{k}$ (this is just the Fourier series):

$$ \begin{align*} \phi(\mathbf{r}, t) &= \sum_\mathbf{k}A_\mathbf{k} e^{i(\mathbf{k} \cdot \mathbf{r} + \omega t)} \\ \mathbf{A}(\mathbf{r}, t) &= \sum_\mathbf{k} \vec B_\mathbf{k} e^{i(\mathbf{k} \cdot \mathbf{r} + \omega t)} \\ \end{align*} $$

Where $A_\mathbf{k}, \vec B_\mathbf{k}$ are constant coefficients in the series expansion over all modes. To quantize the electromagnetic field, it is necessary to sum over all the modes of the system....

$$ \hat H = \sum_\mathbf{k}\hbar \omega_\mathbf{k} \left(\hat a_\mathbf{k}^\dagger \hat a_\mathbf{k} + \dfrac{1}{2}\right) $$

Note that since we have decomposed the electromagnetic field into modes, and each mode represents an exact momentum (by $\mathbf{p} = \hbar \mathbf{k}$), this means that by the Heisenberg uncertainty principle, photons are completely delocalized in space. Thus, the Fock states are states in the momentum basis, where particle states are plane waves of the form $e^{i\mathbf{p} \cdot \mathbf{x}}$.

One might ask, how do states in conventional quantum mechanics fit in to the quantum electrodynamics picture? For instance, if we had a hydrogen atom interacting with a quantized electromagnetic field, how could we model this? The answer is that as long as we're working with energies that are not high enough to require us to consider the effects of relativity (which we can assume to be true most of the time), we can just use the normal $|n, m, \ell\rangle$ states of the hydrogen atom. We know that ultimately, the hydrogen atom is made of elementary particles that come from quantum fields, but for our purposes, we can use the first-quantized hydrogen atom together with the second-quantized electromagnetic field.

Relativity, the Dirac equation, and the road to QFT

Unfortunately, the Dirac equation, despite its successes, has limited applicability. Why? Primarily, because at the relativistic energies it describes, new particles can be created from pure energy (remember Einstein's famous equation $E = mc^2$, this means that a particle of mass $m$ can be created from energy $E$ if $E/c^2 > m$). Additionally, particles can annihilate with each other and be destroyed, and particles can turn into new (and often different types of) particles. The number of particles is never constant - new particles are being created all the time, and old particles are getting annihilated or turn into new particles. This makes the utility of a quantum wave equation that describes fixed numbers of particles rather limited; after a few nanoseconds (or shorter still), the electron you were describing no longer exists, and its wavefunction also vanishes.

The emergence of field quanta

Quantum field theory tells us that all matter in the Universe is composed of quantum fields. These fields are said to be quantized as they can only oscillate between distinct states. Mathematically, this corresponds to quantum fields being operator-valued as opposed to classical fields, which are functions of space and time.

Free particles, which have a very small range of momenta, can be approximated as plane waves $\psi(x) = e^{\pm ipx}$.

The energy of the lowest excited state of a quantum field is given (relative to the ground state) by:

$$ E_\omega = \hbar \omega $$

In the case of massive fields (that is, fields describing particles with mass), this result can be written as:

$$ E_\omega = E_k = \sqrt{(pc)^2 + (mc^2)^2} = \sqrt{(\hbar kc)^2 + (mc^2)^2} $$

This is a special case of the quantized energies of a massive field (that is, fields describing particles with mass), which comes from the relativistic energy-momentum relation:

$$ E^2 = (pc)^2 + (mc^2)^2, \quad p = \hbar k $$

If we switch back to natural units, our expression reads:

$$ E_k = \sqrt{k^2 + m^2} $$

Note: For massless particles (like photons), this simplifies to $E_k = k = \omega$

We can show this result

$$ p^2 + m^2 = E^2 $$

The total energy is of course simply the sum of all of the modes (so sum over $\omega$ for the first and sum over $k$ for the second).

Note how both $\omega$ and $k$ describe oscillations - in fact, in natural units, we know that for massless particles, $\omega = k$. This tells us that stable particles are just long-lived vibrational modes in quantum fields. It is similar to how phonons in solid-state physics appear as quasiparticles from vibrational modes.

The total energy of a quantum field is given by summing over all the energy fluctuations of the field, which becomes an integral:

$$ E_\text{total} = \int d^3 \omega~ E_\omega = \sum_k E_k $$

Second quantization

Action principle, etc. and then also link to the advanced classical mechanics guide for an overview of tensors and the Euler-Lagrange equations.

To go from classical field theory to quantum field theory:

• Classical equations of motion become operator equations motion (that look the same but have very different properties)
• Classical plane-wave solutions to the field equations become single-particle states of the fields
• Fields have a nonzero energy even when in their lowest-energy state
• Classical oscillating fields become coupled quantum harmonic oscillators

The spin-statistics theorem