Quantum electrodynamics

Learning goals:
- Understanding what is quantized when the electromagnetic field is quantized
- The basic concept of coherent and squeezed states
- Origin of spontaneous emission resulting from the quantized em field
- Concept of hybridization, avoided crossing

The first part of the QM II course contained a discussion of the interaction between a classical electromagnetic field with charged particles. Here we lift the assumption of the classical field and see what happens when the electromagnetic field is quantized. The idea is

1. Quantize the electromagnetic field.

2. Study the properties of the quantized field, and express the coherent field in terms of the quantum description of the field. Coherent field is the closest analogue to a classical field, and it is a useful starting point for describing physical fields within the quantum realm. In addition, the path integral theory for bosons is most conveniently described in terms of the coherent fields (however, we do not discuss it here).

3. Study the interaction of the quantized electromagnetic field and a two-level atom. For weak interaction, we find the spontaneous emission as the clearest quantum effect. For stronger coupling, the two systems hybridize, which can be seen in the coupled response.

The wider field describing the properties of the quantized electromagnetic field is called quantum optics. One exhaustive online source on quantum optics is for example D.A. Steck: Quantum and Atom Optics.

Quantization of the electromagnetic field

In what follows, we introduce the common approach to quantize an electromagnetic field. The idea is first to define some eigenmodes of the field in some confined region, express the general field as a superposition of those eigenmodes, and identify the amplitude of the eigenmodes as a bosonic variable. We then discuss two alternative ways to describe the quantized field: first in terms of the Fock states of the bosonic field, and the second in terms of the coherent states, defined as the fields with a minimum uncertainty between different quadrature operators (corresponding to the generalized position and momentum of the harmonic oscillator fields). This notion can then be generalized to introduce squeezed states, which retain the minimum uncertainty, but make a difference between the two quadratures.

This discussion follows partially that in the book Quantum Optics by D.F. Walls and G.J. Milburn (2nd Edition, Springer 2010).

The most natural way to start describing an electromagnetic field is from Maxwell's equations describing a free radiation field, i.e., a field without source or sink terms. It satisfies \[ \begin{aligned} \nabla \cdot {\mathbf B} &= 0\\ \nabla \times {\mathbf E} &= -\frac{\partial {\mathbf B}}{\partial t}\\ \nabla \cdot {\mathbf E} &= 0\\ \nabla \times {\mathbf B} &= \frac{1}{c^2} \frac{\partial {\mathbf E}}{\partial t}, \end{aligned} \] where \(c=1/\sqrt{\epsilon_0 \mu_0}\) is the speed of light in free space.

Instead of using the magnetic field \({\mathbf B}\) and electric field \({\mathbf E}\), we concentrate on the vector potential \({\mathbf A}({\mathbf r},t)\). In optics the most often used gauge is the Coulomb gauge, where both \({\mathbf B}\) and \({\mathbf E}\) can be determined from the vector potential: \[ {\mathbf B} = \nabla \times {\mathbf A}, \quad {\mathbf E} = -\frac{\partial {\mathbf A}}{\partial t}. \] with the gauge condition \[ \nabla \cdot {\mathbf A} = 0. \] Substituting to the fourth Maxwell equation, we find that \({\mathbf A}\) satisfies the wave equation, \[ \nabla^2 {\mathbf A}({\mathbf r},t) = \frac{1}{c^2} \frac{\partial^2 {\mathbf A}}{\partial t^2}. \]

This wave equation has solutions of the form \({\mathbf A} = {\mathbf A}^{+}({\mathbf r}) e^{-i\omega t} + {\mathbf A}^{-}({\mathbf r}) e^{i\omega t}\) with some frequency \(\omega>0\). Due to the reality of the physical field, we have \(A^{-}=(A^{+})^*\).

In the same way as when we calculate some physical observable in an electronic system, it is convenient to consider a finite region of space and impose some boundary conditions to the field. These boundary conditions can also describe real boundaries, such as in the case of optical cavities. Nevertheless, such a procedure leads to an orthonormal set of eigenmodes \({\mathbf u}_{\mathbf k}\) of the wave equation so that we can write \[ {\mathbf A}^+({\mathbf r},t) = \sum_{\mathbf k,\lambda} a_{\mathbf k} {\mathbf u}_{\mathbf k,\lambda}(\mathbf{r}) e^{-i\omega_k t}. \] Here \({\mathbf k}\) denotes the wave vector of the eigenmode, \(\lambda\) denotes the polarization, and the coefficients \(a_{\mathbf k,\lambda}\) are position independent coefficients. The mode functions satisfy the boundary conditions, the wave equation in Fourier space \[ \left(\nabla^2+\frac{\omega_{\mathbf k}^2}{c^2}\right) u_{\mathbf k,\lambda}({\mathbf r})= 0, \] and the Coulomb gauge condition \[ \nabla \cdot u_{\mathbf k,\lambda} = 0. \] In the case of a cubic volume with side length \(L\), the modes are \[ u_{\mathbf k,\lambda}({\mathbf r}) = L^{-3/2} {\mathbf e}_{\lambda} e^{i{\mathbf k} \cdot {\mathbf r}} \] and \({\mathbf e}_{\lambda}\) is a unit vector perpendicular to \({\mathbf k}\) due to the gauge condition. In 3D space, this can be satisfied with two different polarization directions, so \(\lambda\) denotes these two possibilities. Moreover, \({\mathbf k}=\frac{2\pi}{L}(n_x,n_y,n_z)\) with \(n_i \in {\mathbb Z}.\)

Fig: Eigenmodes of the wave equation with periodic boundary conditions. Grey lines are identified as the same point. Blue (red) is the real (imaginary) part of the wavefunction.

Let us now choose \[a_{\mathbf k} = \left(\frac{\hbar}{2\omega_k \epsilon_0}\right)^{1/2} b_{\mathbf k}.\] The physical unit of the vector potential is (from the defining equations) Tesla\(\cdot\)meter or Newton/Ampere (unit of an electric field is N/As). Quick check for example with this shows that this results into a dimensionless \(b_{\mathbf k}\).

We can thus write the vector potential in the free space but within the confined volume as \[ {\mathbf A}(\mathbf r,t) = \sum_{\mathbf k,\lambda} \left(\frac{\hbar}{2\omega_k \epsilon_0}\right)^{1/2} \left[b_{\mathbf k,\lambda} {\mathbf u}_{\mathbf k,\lambda}({\mathbf r})e^{-i\omega_k t} + b_{\mathbf k,\lambda}^* {\mathbf u}_{\mathbf k,\lambda}^*({\mathbf r})e^{i\omega_k t}\right]. \]

Since the polarization direction does not have any particular role in the quantization of the field, in what follows we suppress the index \(\lambda\), but should remember to re-insert it when necessary.

Now we are ready to quantize the electromagnetic field. At this point it is straightforward: instead of treating \(b_{\mathbf k}\) as some complex numbers, we replace them by bosonic operators. In this case we must write \(b_{\mathbf k}^\dagger\) instead of a complex conjugate. In other words, \(b_{\mathbf k}^{(\dagger)}\) annihilates (creates) a photon with wave number \({\mathbf k}\).

These operators hence satisfy the bosonic commutation relations \[ \begin{aligned} [b_{\mathbf k},b_{\mathbf k'}]=[b_{\mathbf k}^\dagger,b_{\mathbf k'}^\dagger]=0\\ [b_{\mathbf k},b_{\mathbf k'}^\dagger]=\delta_{\mathbf k,\mathbf k'}. \end{aligned} \]

Let us then derive the Hamiltonian describing the electromagnetic field. We can start from the classical Hamiltonian yielding the energy of the electromagnetic field, \[ H= \frac{\epsilon_0}{2}\int d^{(3)} {\mathbf r} ({\mathbf E}^2+c^2 {\mathbf B}^2) \] Using the above definition of the fields with respect to the vector potential, and the vector potential with respect to the bosonic operators \(b_{\mathbf k}\), \(b_{\mathbf k}^\dagger\), you will show in the exercises that this can be expressed as \[ H= \sum_{\mathbf k,\lambda} \hbar \omega_{\mathbf k} \left(b_{\mathbf k,\lambda}^\dagger b_{\mathbf k,\lambda} + \frac{1}{2}\right). \] In other words, we get a set of Hamiltonians for free harmonic oscillators, so that the total energy is the sum of the mode energies \(\hbar \omega_{\mathbf k}\) times the number of photons in each mode.

In addition, there is the vacuum energy of \(\hbar \omega_{\mathbf k}/2\) from each mode. Note that since the sum over modes is not limited from above, this vacuum energy is in fact infinite! This as such does not have practical consequences, since no process can extract the vacuum energy directly. However, it is a relevant concept when considering for example (finite) changes in this energy due to changing boundary conditions as in the static or dynamic Casimir effect. These phenomena are however not considered here.

Fock or number states

The most natural way to characterize the states of the quantized electromagnetic field are the eigenstates of the number operator \(\hat n_{\mathbf k}=b_{\mathbf k}^\dagger b_{\mathbf k}\), which then are also eigenstates of the Hamiltonian. In other words, we would hence describe the states via \[ |\{n_{\mathbf k}\}\rangle = |n_{\mathbf k_1} n_{\mathbf k_2} \dots\rangle = \prod_{\mathbf k_i} (b_{\mathbf k_i}^\dagger)^{n_{\mathbf k_i}}|0\rangle. \] where \(n_{\mathbf k}\) denotes the number of photons in the momentum state \({\mathbf k}\). Note that since these are number states of bosons, the order of the states in the ket vector does not matter.

Description of the field in terms of number states is somewhat convenient, because they form an orthonormal set of vectors, and the algebra with them is somewhat straightforward. However, in practice it is very hard to bring the field into some specific Fock state except the vacuum state.

In fact, generating pure single- or two-photon states of electromagnetic field was achieved only quite recently, see Varcoe, et al., Nature 403, 743 (2000) and more general pure Fock states in M. Hofheinz, et al., Nature 454, 310 (2008).

More often, the electromagnetic radiation fields are in superpositions (pure state) or a mixture of number states (mixed state). Next, we discuss a useful way of describing a rather generic set of superposition states, those of coherent states.

Coherent states

The name 'coherent state' refers to the concept of optical coherence which is related with photon correlations. Historically, Roy Glauber constructed the theory of coherence (Phys. Rev. 130, 2529 (1963)) and, in doing so, investigated the coherent states (Phys. Rev. 131, 2766 (1963)).

There are three equivalent definitions of the coherent states and multiple ways to derive them. Let us take the one mentioned in the introduction: coherent states as minimum uncertainty states. In the following, we drop all the subscripts to simplify notation, consider a single mode, and denote this bosonic mode by \(a\). Then, we define the dimensionless quadrature operators \[ q = \frac{1}{\sqrt{2}}(a^\dagger + a), p = \frac{i}{\sqrt{2}}(a^\dagger - a), \] which have the commutation relations \([q,p] = i\). Note that this corresponds to the canonical commutation relation of position and momentum if \(\hbar = 1\).

The uncertainty relation in its general form follows from the Cauchy-Schwarz inequality which reads in terms of the operators \(P = p - \langle p \rangle\) and \(Q = q - \langle q \rangle\) \[ \langle \psi| Q^2 |\psi \rangle \langle \psi| P^2 |\psi \rangle \geq |\langle \psi| Q P |\psi \rangle|^2 \geq \frac{1}{4}|\langle \psi| [Q, P] |\psi \rangle|^2 = \frac{1}{4}. \] The Heisenberg uncertainty principle, the latter inequality, follows by writing the operator \(Q P\) as a sum of the commutator and the anticommutator between \(Q\) and \(P\) and then noting that the anticommutator term gives only a positive contribution. However, if we suppose that the lower bound of the Cauchy-Schwarz inequality is satisfied then necessarily \(Q\ket{\psi} = \mu P \ket{\psi}\) where \(\mu \in \mathbb{C}\) is a constant. We can solve \(\mu\) by using the commutator relation \([Q,P] = [q,p] = i\) \[\begin{align} \sigma_{q}^2 \equiv \langle \psi| Q^2 |\psi \rangle = \langle \psi| \mu Q P |\psi \rangle = \mu\langle \psi| ([Q,P] + P Q) |\psi \rangle = i\mu + \mu^2 \sigma_{p}^2. \end{align}\] The quantity \(\langle \psi| Q^2 |\psi \rangle\) is the variance of \(q\) which is why we denote it by \(\sigma^2_q\). If we now set \(\sigma_{q}^2 = \sigma_{p}^2 = \frac{1}{2}\) we find that \(\mu = - i\). Rearranging the equation \(Q \ket{\psi} = -i P\ket{\psi}\) and using the definitions of the quadrature operators, we find the eigenvalue equation \[ a \ket{\psi} = \langle a \rangle \ket{\psi}. \] This tells us that the minimum uncertainty states are, in fact, the eigenstates of the annihilation operator \(a\).

The value of \(\mu\) can be found even if \(\sigma_{q} \neq \sigma_{p}\). These solutions correspond to the squeezed states, which we discuss in detail below. Note that it is possible to squeeze the uncertainty in one quadrature (with the expense of the other quadrature) even below the value of vacuum uncertainty, e.g. \(\sigma_p^2 < \frac{1}{2} = \bra{0}p^2\ket{0}\).

Fig. Graph showing where the coherent states (red circle) and the squeezed states (black line) lie in -plane. The black line also represents the lower limit of Heisenberg uncertainty relation, i.e., . The white region is thus forbidden, while the blue region represents states for which one quadrature is squeezed below zero-point fluctuations.

As in the case of the harmonic oscillator and Fock states, we may now relabel our states based on the eigenvalue. Here, we denote \(\langle a\rangle = \alpha \in \mathbb{C}\) and \[ a \ket{\alpha} = \alpha \ket{\alpha}. \]

How does \(\ket{\alpha}\) look in the Fock space? To answer this, we operate with \(\bra{n}\) from the left which gives \[ \sqrt{n+1}\langle n+1 | \alpha \rangle =\alpha \langle n | \alpha \rangle. \] From this recursion relation we find that \[ \langle n | \alpha \rangle = \frac{\alpha}{\sqrt{n}} \langle n-1 | \alpha \rangle = \dots = \frac{\alpha^n}{\sqrt{n!}}\langle 0 | \alpha \rangle. \] Therefore \[ \ket{\alpha} = \sum_n \ket{n}\langle n | \alpha \rangle = \langle 0 | \alpha \rangle \sum_n \frac{\alpha^n}{\sqrt{n!}}\ket{n}. \] The prefactor is fixed by demanding that the state is normalized \(\langle \alpha | \alpha \rangle = 1\) giving in the end \[ \ket{\alpha} = e^{-\frac{|\alpha|^2}{2}} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}\ket{n}. \] The probability to observe \(n\) photons in the coherent state \(\ket{\alpha}\) is thus \(P_n = |\langle n | \alpha \rangle|^2 = e^{-|\alpha|^2}\frac{|\alpha|^{2n}}{n!}\). The probability distribution is thus Poissonian with the mean value \(|\alpha|^2\).

Alternatively, we may generate the coherent states with a displacement operator \[ \ket{\alpha} = D(\alpha)\ket{0} \equiv e^{\alpha a^\dagger - \alpha^* a}\ket{0}. \] To show that this coincides with the earlier definition, one should use a simplified version of the Baker-Campbell-Hausdorff formula \[ e^A e^B = e^{A + B + \frac{1}{2}[A,B]} \] which holds when \([A,[A,B]] = [B,[A,B]] = 0\).

The displacement operator has the following properties \[ \begin{aligned} D^{-1}(\alpha) = D^\dagger(\alpha) = D(-\alpha), \\ D^\dagger(\alpha) a D(\alpha) = a + \alpha, \\ D^\dagger(\alpha) a^\dagger D(\alpha) = a^\dagger + \alpha^*. \end{aligned} \] The two latter equations can be shown to be true with the help of the formula \(e^A B e^{-A} = \sum_{n=0}^\infty \frac{1}{n!} [A,B]_n\) where \([A,B]_0 = B\) and \([A,B]_{n+1} = [A,[A,B]_n]\). In other words, \([A,B]_n\) is a nested commutator with \(n\) amount of \(A\)'s.

Some properties of the coherent states

Non-orthogonality:

First, we derive the multiplication property of displacement operators \[ D(\beta) \ket{\alpha} = D(\beta) D(\alpha) \ket{0} = e^{i \mathrm{Im}(\alpha^* \beta)}D(\alpha + \beta)\ket{0} =e^{i \mathrm{Im}(\alpha^* \beta)}\ket{\alpha + \beta}. \] Thus, the operation of the displacement operator \(D(\beta)\) to the state \(\ket{\alpha}\) displaces the state in the complex plane from \(\alpha\) to \(\alpha + \beta\). Consequently, \[ \begin{aligned} &\langle \beta | \alpha \rangle = e^{-i\mathrm{Im}(\alpha^*\beta)}\langle 0 | \alpha-\beta \rangle = e^{-\frac{1}{2}(|\alpha|^2 + |\beta|^2) + \alpha \beta^*},\\ &|\langle \beta | \alpha \rangle|^2 = e^{-|\beta - \alpha|^2}, \end{aligned} \] which means that the coherent states are not orthogonal, unlike the Fock states.

Completeness:

The field coherent states form a complete set. They are in fact overcomplete since the label of coherent states \(\alpha\) is continuous and uncountable, although the underlying number state basis is countable. This means that the completeness relation is not unique. One useful expression of the completeness relation is \[ 1 = \sum_n |n\rangle\langle n| = \frac{1}{\pi}\int d^2{\alpha} |\alpha\rangle\langle \alpha|. \] The integration measure is defined as \(d^2{\alpha} = d(\mathrm{Re} \alpha)d(\mathrm{Im} \alpha)\) and the integration is over the whole complex plane. This can be proven by brute force, by inserting the Fock state representation of the coherent states and calculating the resulting integrals in polar coordinates.

Classicality:

The coherent states are in a sense 'classical states' of quantum harmonic oscillator which is described by a Hamiltonian \(H = \hbar \omega a^\dagger a\). If the initial state is a coherent state \(\ket{\alpha}\), the state at a later time is given by the Schrödinger equation \[\begin{equation} e^{-i H t/\hbar}\ket{\alpha} = e^{-i \omega a^\dagger a t}\ket{\alpha} = e^{-\frac{|\alpha|^2}{2}}\sum_n \frac{(\alpha e^{-i\omega t})^n}{\sqrt{n!}} \ket{n} = \ket{\alpha e^{-i\omega t}}. \end{equation}\] The state remains as a coherent state during time evolution. Hopefully this clarifies what is meant by 'a classical state': coherent state follows a classical trajectory with \(\langle q \rangle = q_0 \cos(\omega t + \phi)\) and \(\langle p \rangle = p_0 \sin(\omega t + \phi)\). Besides the trajectory, the state is as localized (both in space and in momentum) as possible and this property does not change over time.

Summary of coherent states

• Minimum uncertainty states with \(\sigma_q=\sigma_p\)
• Eigenvalues of the annihilation operator: \(a \ket{\alpha} = \alpha \ket{\alpha}\) with \(\alpha \in {\mathbb C}\)
• "Displaced vacuum states", \(|\alpha\rangle = D(\alpha) |0\rangle\)
• Form an overcomplete set

Squeezed states

How to generate states that have \(\sigma^2_p \neq \sigma^2_q\)? We can guess that there must be a squeezing operator \(S(\zeta)\) in the same way that there is a displacement operator \(D(\alpha)\). Indeed, we have the squeezing operator \[ S(\zeta) = \exp \left(\frac{\zeta^*}{2}a^2 - \frac{\zeta}{2}a^{\dagger2}\right), \] which defines the squeezed coherent states \[ \ket{\alpha,\zeta} = D(\alpha)S(\zeta)\ket{0}. \] That is, we first squeeze the vacuum and then displace the resulting state. The order of operations matters here, but this is the usual convention.

By utilizing the \(e^A B e^{-A}\) formula, we find \[ \begin{aligned} S^\dagger(\zeta) a S(\zeta) &= a \cosh r - e^{i \theta} a^\dagger \sinh r, \\ S^\dagger(\zeta) a^\dagger S(\zeta) &= a^\dagger \cosh r - e^{-i \theta} a \sinh r, \end{aligned} \] where \(\zeta = r e^{i\theta}\). Here \(r\) is the squeeze factor which determines how much a quadrature is squeezed. The phase \(\theta\) determines the axis of squeezing which does not have to be either \(p\) or \(q\).

We then look at the squeezed coherent states and their properties; especially we want to see the squeezing in effect. Thus, we work our way back to \(\sigma_q^2\) and \(\sigma_p^2\). We have \[ \begin{aligned} \bra{\alpha,\zeta}a\ket{\alpha,\zeta} &= \bra{0} S^\dagger D^\dagger a D S \ket{0} = \bra{0} (\alpha + a \cosh r - e^{i \theta} a^\dagger \sinh r) \ket{0} = \alpha, \\ \bra{\alpha,\zeta}a^2\ket{\alpha,\zeta} &= \dots = \alpha^2 - e^{i\theta}\cosh r \sinh r, \\ \bra{\alpha,\zeta}a^\dagger a\ket{\alpha,\zeta} &= |\alpha|^2 + \sinh^2 r. \end{aligned} \] Now, we can plug in these results to get \[ \begin{aligned} \sigma^2_q &= \frac{1}{2}\bra{\alpha,\zeta}[(a+a^\dagger)^2 - (\alpha + \alpha^*)^2]\ket{\alpha,\zeta} = \frac{1}{2}[\cosh 2r + \sinh 2r \cos \theta] \\ \sigma^2_p &= \frac{1}{2}\bra{\alpha,\zeta}[-(a-a^\dagger)^2 + (\alpha - \alpha^*)^2]\ket{\alpha,\zeta} = \frac{1}{2}[\cosh 2r - \sinh 2r \cos \theta]. \end{aligned} \] If we now set the squeezing angle \(\theta = 0\), then \(\sigma^2_q = \frac{1}{2} e^{2r}\) and \(\sigma^2_p = \frac{1}{2} e^{-2r}\). For non-zero \(r\) then \(\sigma_p^2 < \frac{1}{2} < \sigma_q^2\) which implies squeezing. In a general case, we find the same exponential squeezing along some other quadratures \(q'\) and \(p'\) defined by a relation \(q' + i p' = (q+ ip)e^{-i \theta/2}\). This is illustrated below.

Fig. The effect of squeezing. The shapes represent uncertainties, or variances, in -plane. The solid gray circle shows the uncertainty of a coherent state and the ellipse outlines the uncertainty of a squeezed state . Note that the squeezing axes are and which are obtained by a rotation of .

Similar to coherent states, squeezed states remain as such in a free electromagnetic field (i.e. \(H = \omega a^\dagger a\)). This can be shown either by finding the Fock state representation of a squeezed coherent state or by developing the theory of phase-space distributions. Both of these approaches are beyond this course, so we simply state the result: the time-evolution simply rotates the above figure around the origing, giving \(\ket{\alpha(t),r e^{i\theta(t)}} = \ket{\alpha_0 e^{-i\omega t}, r e^{i (\theta_0 - 2 \omega t)}}\). The code below allows you to see how this affects, for instance, the expectation value of \(q\) (or after suitable rotations a component of the electic field \(\mathbf{E}\)).

Interaction of radiation with a two-level atom

Now, we can generalize the description of an atom in a radiation field to a quantum electromagnetic field. This leads to

• spontaneous emission
• hybridization, avoided crossing.

In fact, only the first one is a 'true' quantum effect, the latter results from assuming the electromagnetic field to be constrained on a finite region of space. However, the details of the hybridized state spectrum can only be explained with the quantum model.

We start by deriving the Hamiltonian of this interaction.

Derivation of the Hamiltonian

A full treatment of QED is beyond this course, but, fortunately, only the low energy (non-relativistic) theory is needed to explain many experiments in quantum optics. Thus, we may fix the gauge as \(\nabla \cdot \mathbf{A} = 0\) and we can start from the minimal coupling Hamiltonian \[ H = \frac{1}{2m}[\mathbf{p} - e \mathbf{A}(\mathbf{r},t)]^2 + e V(\mathbf{r}) + H_\mathrm{rad}, \] where now \(\mathbf{A}\) is the quantized vector potential and \(H_\mathrm{rad}\) is the Hamiltonian of the free radiation field.

In the second quantization formalism, the Hamiltonian is \[ H = H_\mathrm{rad} + H_\mathrm{el} + H_\mathrm{int}, \] where the electronic and the interaction contributions are \[ \begin{aligned} H_\mathrm{el} &= \int d^3 \mathbf{r} \Psi^\dagger(\mathbf{r}) \left[ \frac{\mathbf{p}^2}{2m} + e V(\mathbf{r}) \right]\Psi(\mathbf{r}), \\ H_\mathrm{int} &= \int d^3 \mathbf{r} \Psi^\dagger(\mathbf{r}) \frac{e}{2m} \left[ e \mathbf{A}^2 - \mathbf{A} \cdot \mathbf{p} - \mathbf{p} \cdot \mathbf{A} \right]\Psi(\mathbf{r}), \end{aligned} \] respectively. Unless dealing with very intense fields, the second order term \(\mathbf{A}^2\) may be neglected from the interaction; this is what we now assume. Also, the anticommutative structure may be simplified using the gauge condition as \[ \mathbf{p} \cdot \mathbf{A} \Psi = -i \nabla \cdot (\mathbf{A} \Psi) = -i (\underbrace{\nabla \cdot \mathbf{A}}_{=0}) \Psi - i \mathbf{A} \cdot \nabla \Psi = \mathbf{A} \cdot \mathbf{p} \Psi. \]

Next, we rewrite the field operators in another basis, which represents the single-particle eigenstates of the Hamiltonian without radiation, \(\Psi(\mathbf{r}) = \sum_\nu \varphi_\nu(\mathbf{r}) c_\nu\) with \(H_\mathrm{el} = \sum_n E_\nu c_\nu^\dagger c_\nu\) and \((-\frac{\hbar^2}{2m}\nabla^2 + e V(\mathbf{r}))\varphi_\nu = E_\nu\varphi_\nu\). The interaction term can be rearranged to \[ H_\mathrm{int} = \hbar \sum_{\mathbf{k},\lambda}\sum_{\nu \mu} (g_{\mathbf{k}\lambda,\nu\mu} b_{\mathbf{k},\lambda} e^{-i\omega_\mathbf{k}t} + g^*_{\mathbf{k}\lambda,\mu\nu} b_{\mathbf{k},\lambda}^\dagger e^{i\omega_\mathbf{k}t})c_\nu^\dagger c_\mu \] where \[ g_{\mathbf{k}\lambda,\nu\mu} = - \frac{e}{m}\sqrt{\frac{1}{2\epsilon_0\hbar\omega_\mathbf{k} V}} \int d^3 \mathbf{r} \varphi_\nu^*(\mathbf{r}) \left[e^{i \mathbf{k}\cdot \mathbf{r}} \mathbf{e}_\lambda \cdot \mathbf{p} \right] \varphi_\mu(\mathbf{r}) \] is the light-matter coupling constant. Note the indices of \(g\); the Hamiltonian must be Hermitian. Note also that here the polarization vector \({\mathbf e}_\lambda\) plays a role.

There are two simplifying approximations: one at the level of the coupling constant \(g_{\mathbf{k}\lambda,\nu\mu}\), called the dipole approximation, and the other at the level of \(H_\mathrm{int}\), called the rotating wave approximation.

Dipole approximation: If we consider the electronic system to be small, e.g. an atom, the extent of wavefunctions \(\varphi_\nu(\mathbf{r})\) is much smaller than the optical wavelength. We can then treat the atom as point-like, located at \(\mathbf{r} = \mathbf{r}_0\), and replace the oscillatory term \(e^{i \mathbf{k} \cdot \mathbf{r}} \approx e^{i \mathbf{k} \cdot \mathbf{r_0}}\) inside the integral. Then, we would like to do something to \(\mathbf{p}\) inside the integral. We can use the Schrödinger equation to simplify the integral; to this end we employ the relation \[ \mathbf{p} = -i\frac{m}{\hbar} \left[\mathbf{r},\underbrace{\frac{\mathbf{p}^2}{2m} + e V(\mathbf{r})}_{H_{\rm el}^{(1)}}\right]. \] which you can easily check. Thus, we have effectively changed \(\mathbf{p}\) to \(-i m\frac{E_\mu - E_\nu}{\hbar} \mathbf{r}\) inside the integral. This allows us to write the light-matter coupling in terms of the electric dipole moment \(\mathbf{d}_{\nu\mu} = \bra{\psi_\nu}e \mathbf{r}\ket{\psi_\mu}\). In the end, the coupling reads \[ g_{\mathbf{k}\lambda,\nu\mu} = i e^{i \mathbf{k} \cdot \mathbf{r_0}} \sqrt{\frac{1}{2\epsilon_0\hbar\omega_\mathbf{k} V}} \frac{E_\mu - E_\nu}{\hbar} (\mathbf{e}_\lambda\cdot\mathbf{d}_{\nu\mu}). \] Thus, when the dipole approximation holds, we have \(g_{\mathbf{k}\lambda,\nu\nu} = 0\), i.e., no terms containing the number operators \(c_\nu^\dagger c_\nu\). Also, without loss of generality, we can always set the complex phase of \(g_{\mathbf{k}\lambda,\nu\mu}\) for some \(\nu,\mu\). From now on, we can again suppress the polarization index \(\lambda\).

In the simplest case where we consider only two electronic levels, we have \(\nu,\mu \in \{0,1\}\). The dipole approximated Hamiltonian is \[ H = E_1 c_1^\dagger c_1 + E_0 c_0^\dagger c_0 + \sum_\mathbf{k}\hbar \omega_k b_\mathbf{k}^\dagger b_\mathbf{k} + \hbar\sum_\mathbf{k} (g_{\mathbf{k}} b_\mathbf{k} + g_{\mathbf{k}}^* b_\mathbf{k}^\dagger)( c_1^\dagger c_0 + c_0^\dagger c_1), \] where we have set the phase so that \(g_\mathbf{k} \equiv g_{\mathbf{k},10} = g_{\mathbf{k},01}\), and we absorbed the exponential phase factor to the definition of \(b_\mathbf{k}\).

Further simplification is obtained by the pseudo-spin representation (which is a special case of Jordan-Schwinger map). The algebra of the two-level system can be presented explicitly via operators \[ \begin{aligned} \sigma_z &= c_1^\dagger c_1 -c_0^\dagger c_0 & 1 &= c_1^\dagger c_1 + c_0^\dagger c_0\\ \sigma_+ &= \sigma_-^\dagger = c_1^\dagger c_0 & 2\sigma_\pm &= \sigma_x \pm i \sigma_y. \end{aligned} \] These operators are then related to the spin-half operators by \(S_j = \frac{\hbar}{2}\sigma_j\), obeying the \(\mathfrak{su}(2)\)-like commutation relations \([\sigma_i,\sigma_j] = \sum_k 2i\epsilon_{ijk}\sigma_k\) (\(i,j,k \in \{x,y,z\}\)). The ladder operators are \(\sigma_\pm\), they move the electron from state 0 to 1 (\(\sigma_+\)) or from 1 to 0 (\(\sigma_-\)). Naturally, we then have \(\sigma_+^2 = \sigma_-^2 = 0\) and it is easy to see that \(\sigma_+ \sigma_-\) works as a number operator. Denoting the transition frequency by \(\hbar\omega_a = E_1 -E_0\), we get \[ H = \frac{\hbar \omega_a}{2} \sigma_z + \sum_\mathbf{k}\hbar \omega_k b_\mathbf{k}^\dagger b_\mathbf{k} + \hbar\sum_\mathbf{k} (g_{\mathbf{k}} b_\mathbf{k} + g_{\mathbf{k}}^*b_\mathbf{k}^\dagger)\sigma_x. \]

Rotating wave approximation: The form we arrived at looks fairly simple but is, in fact, difficult to solve. Even recently, articles that try to provide insight to this problem are published, e.g., Phys. Rev. Lett. 123, 133603 (2019).

To even further simplify the Hamiltonian, we have to go back to the form of interaction Hamiltonian \[ H_\mathrm{int} = \hbar\sum_\mathbf{k} (g_{\mathbf{k}} b_\mathbf{k} e^{-i\omega_\mathbf{k}t} + g_{\mathbf{k}}^* b_\mathbf{k}^\dagger e^{i\omega_\mathbf{k}t})( c_1^\dagger c_0 + c_0^\dagger c_1). \] Note that the photon terms \(b_{\mathbf k}\) are written in the interaction picture, i.e., they contain an explicit time dependence. Let us do the same for the electron operators \(c_i\). Recall that the interaction picture operators \(\hat V_I\) are obtained from the Schrödinger picture operators \(\hat V_S\) via \(\hat V_I(t) = e^{i\hat H_0 t/\hbar} \hat V_S(t) e^{-i\hat H_0 t/\hbar}\). Using \(\hat H_0=E_0 c_0^\dagger c_0 + E_1 c_1^\dagger c_1\) we get for the interaction picture operators \[ (c_j^\dagger)_S = e^{iE_j t/\hbar} (c_j^\dagger)_I, \quad (c_j)_S = e^{-iE_j t/\hbar} (c_j)_I. \] In other words, the term \((c_1^\dagger c_0)_S = (c_1^\dagger c_0)_I e^{i(E_1-E_0)t/\hbar} \equiv (c_1^\dagger c_0)_I e^{i\omega_a t}\). In the interaction picture the interaction Hamiltonian thus goes to the form \[ \tilde H_\mathrm{int} = \hbar\sum_\mathbf{k} (g_{\mathbf{k}} b_\mathbf{k} e^{-i\omega_\mathbf{k}t} + g_{\mathbf{k}}^* b_\mathbf{k}^\dagger e^{i\omega_\mathbf{k}t})( c_1^\dagger c_0 e^{i \omega_a t} + c_0^\dagger c_1 e^{-i \omega_a t}). \] If we look at the exponentials after taking the product, we have all the possible combinations of signs \(\exp(\pm i(\omega_\mathbf{k} \pm \omega_a)t)\). In the case that the optical frequencies match the transition frequency, \(\omega_\mathbf{k} \approx \omega_a\), we have two terms that are almost time-dependent as \(\omega_\mathbf{k} - \omega_a \approx 0\) while we also have two terms that rotate at high frequency, roughly at \(2\omega_a\). These high frequency terms we can neglect, if the field is neither very intense nor evolving on the time scale \(1/\omega_a\). Thus, pulses do not fit into this assumption. This assumption is called the rotating wave approximation. The resulting Hamiltonian is \[ H = \frac{\hbar \omega_a}{2} \sigma_z + \sum_\mathbf{k}\hbar \omega_k b_\mathbf{k}^\dagger b_\mathbf{k} + \hbar \sum_\mathbf{k} (g_{\mathbf{k}} b_\mathbf{k}\sigma_+ + g_{\mathbf{k}}^*b_\mathbf{k}^\dagger \sigma_-). \] The neglected terms \(b_\mathbf{k} \sigma_-\) and \(b_\mathbf{k}^\dagger \sigma_+\) are called counter-rotating terms. As implicitly presented here, the rotating wave approximation is often done together with the dipole approximation, even though they are not connected in any way.

Two-level atom in a radiation field

Now let us see the consequencies of treating the atom-field interaction fully quantum mechanically. Our idea is therefore to generalize the results in obtained in Quantum Mechanics II A to the case of a quantized electromagnetic field. We start from the Hamiltonian derived above, \[H=\sum_{\vec{k}} \hbar \omega(\vec{k}) b_{\vec{k}}^\dagger b_{\vec{k}}+ \frac{\hbar \omega_a}{2} \sigma_z + \sum_{\vec{k}} \hbar [g_{\vec{k}}b_{\vec{k}}\sigma_+ + g_{\vec{k}}^* b_{\vec{k}}^\dagger \sigma_-].\] Note that this Hamiltonian is characterized by a conserved quantity equal to the number of excitations in the system, i.e., \[\hat n_{\rm tot}=\sum_{\vec{k}}b_{\vec{k}}^\dagger b_{\vec{k}}+ \sigma_+ \sigma_-.\] Using the usual bosonic commutation relations \([b_{\vec{k}},b_{{\vec{k}}'}^\dagger]=\delta_{{\vec{k}},{\vec{k}}'}\) and the fact that \([\sigma_+,\sigma_-]=\sigma_z\), it is straightforward to show (Ex!) that \([H,\hat n_{\rm tot}]=0\). This means that the state of the system can be characterized via the quantum \(n_{\rm tot}\) corresponding to the expectation value of \(\hat n_{\rm tot}\).

Let us first assume that the coupling is a small perturbation to the system, i.e., consider the first two terms in \(H\) as the zeroth-order Hamiltonian \(H_0\), and the latter two terms as the perturbation \(\hat V_S=\sum_{\vec{k}}\hbar [g_{\vec{k}}b_{\vec{k}}\sigma_+ + g_{\vec{k}}^* b_{\vec{k}}^\dagger \sigma_-]\). Now let us do the same calculation as in the earlier half of the course for the lowest-order transition rate. We consider two transitions: either the atom makes a transition from its ground state \(|g\rangle\) to the excited state \(|e\rangle\), or from the excited state \(|e\rangle\) to the ground state \(|e\rangle\). Denote the rates of these transitions by \(W_{eg}\) and \(W_{ge}\), respectively. In the calculation made in QM II A, we got \(W_{eg}=W_{ge}\).

Fig: Electron system and transitions due to absorption and emission of photons.

We thus look at these rate of transitions in the electron system, driven by the (quantized) electromagnetic field.

We can proceed in two ways: In the first approach we include the change in the final state energy of the photon system, and consider only elastic transitions, where the initial and final state energies are the same. In the second, we transform the field into the interaction picture, obtaining a field that oscillates in time, in which case we ignore the change in the final energy of the photon system, and consider the transitions from an oscillating field. The results should of course be the same, but the latter one is closer to the case of the classical field. In the following, we only consider the first case.

The total rate of transitions is \[W_{ge}= 2\pi \sum_n \int d \omega \rho(\omega) P_{\omega,n} |g_{\omega}|^2 (n+1) \delta(\omega_a-\omega).\] We can make the corresponding calculation for the rate of transitions from the ground state to the excited states. In that case the matrix element is \[\begin{aligned} &T_{eg}\delta(E_i-E_f)=\langle f_e | \hat V_S | i_g \rangle \delta(E_i-E_f)\\&= \hbar \sum_{\vec{k}}\langle f_e | g_{\vec{k}}\hat b_{\vec{k}}\sigma_+ + g_{\vec{k}}^* \hat b_{\vec{k}}^\dagger \sigma_- |i_g\rangle \delta(E_i-E_f) \\&= \hbar \sum_{\vec{k}}\langle f_e | g_{\vec{k}}\hat b_{\vec{k}}\sigma_+ |i_g\rangle \delta(E_i-E_f),\end{aligned}\] where the second term vanishes because \(\sigma_-|e\rangle = 0\). In this case the final state contains one less photon than in the initial state and we have to calculate \(\hat b|n\rangle = \sqrt{n} |n-1\rangle\). The total rate becomes \[W_{eg}= 2\pi \sum_n \int d \omega \rho(\omega) P_{\omega,n} |g_{\omega}|^2 n \delta(\omega_a-\omega).\] Comparing these results to what was found in the A part of the course, we can identify that the amplitude of the electromagnetic field, encoded by \(A_0^2\), is here characterized by the average number of photons in the field. However, there is a difference between the rates of relaxation \(W_{ge}\) and excitation \(W_{eg}\): the previous is finite even when \(n=0\), i.e., when the field amplitude vanishes. This spontaneous emission is the main effect of the quantized electromagnetic field, and it becomes important when the typical number of photons is low, which is the case in many experiments done with fields at optical frequencies (or in other words at frequencies \(\omega_a\) satisfying \(\hbar \omega_a \gg k_B T\)).

Now let us assume we can tune the cavity frequency, for example by moving one of the mirrors. Assume that the cavity field is coupled to the atom (or an ensemble of them). What happens to the absorption dip when \(\omega_c\) becomes resonant with the atomic transition?

The states of the system can be characterized by \(|g/e n\rangle\), i.e., ground or excited state of the atom and \(n\) states in the field. It is straightforward to show that \(|g0\rangle\) is the ground state of the system, since the coupling terms evaluate to zero on this state. The energy of this state is \(0\), and hence the energies of the excited states become energy differences to the ground state. Now we can compute the action of the Hamiltonian on an arbitrary state \(|g/e n\rangle\) except the ground state. For example, \[\begin{aligned} H |e n-1\rangle &= (\hbar \omega_c (n-1) + \hbar \omega_a) |e n-1\rangle + \hbar g^* \sqrt{n} |g n\rangle\\ H |g n\rangle & = \hbar \omega_c n |g n\rangle + \hbar g \sqrt{n} |e n-1\rangle.\end{aligned}\] These two states are hence pairwise coupled. Note that the total number of excitations \(n_{\rm tot}=n\) is the same between these two states. Expressed in terms of the matrix elements between these states, the Hamiltonian thus goes to a block diagonal form, where the blocks with different \(n_{\rm tot}\) are not coupled. Diagonalizing the Hamiltonian thus becomes a question of diagonalizing each block separately. For the \(n\)'th block we hence have \[H_n = n \hbar \omega_c + \hbar \begin{pmatrix} \delta \omega & g \sqrt{n} \\ g^* \sqrt{n} & -\delta \omega \end{pmatrix},\] where \(\delta \omega=\omega_a-\omega_c\). The corresponding normalized eigenstates of the \(n\)'th block are hybrid states of atoms and electromagnetic field, \[|n b/a\rangle = \sqrt{\frac{\omega_g \pm \delta \omega}{2\omega_g}} |g n\rangle \mp \sqrt{\frac{\omega_g \mp \delta \omega}{2 \omega_g}} |e n-1\rangle,\] where \(\omega_g=\sqrt{4n g^2+\delta \omega^2}\). These are called dressed states. They have the eigenenergies \[E_{b/a} = \hbar \left(n \omega_c + \frac{\delta \omega}{2} \pm \frac{\omega_g}{2}\right).\] We can hence plot the eigenenergy spectrum:

Let us return back to the initial experiment and consider the case of a fixed number of photons in the driving pulse (for example, a state with most is also possible for a low-amplitude coherent state). As the cavity frequency crosses the atomic spectrum, the total spectrum of the system shows avoided crossings of this form:

This is hence a way to study the coupling. Such an avoided crossing as such is a property of strongly coupled systems, and not a genuine quantum effect. On the other hand, the dependence of the spectrum on the number of photons is a quantum effect.