9.3: Electromagnetic field

One can anticipate many, but not all, consequences of the Hamiltonian, eq. (9.2.9), on our description of the field wavefunction from what we already know about quantum harmonic oscillators. Harmonic oscillators have equidistant energy levels, and the state of the oscillator can be specified by the number of excitations in the oscillator \(\left| n \right\rangle\). Therefore, the state of a single electromagnetic mode \(k\) can be specified by the number of excitations (quanta) in that mode, \(N_{k}\). Since the electromagnetic modes form an orthonormal basis, the total wavefunction of the light \(\left| \varphi _{L} \right\rangle\) can be expressed in product states for the number of excitations in each of the modes:

\[\left| \varphi _{L} \right\rangle = \left| N_{1},N_{2},N_{3}, \ldots \right\rangle %DOCeq18 %LTeq01\]

The ground state of the electromagnetic field, the vacuum state, where all \(N_{k}\) = 0.

\[\left| 0 \right\rangle = \left| 0_{1},0_{2},0_{3}, \ldots \right\rangle %DOCeq19 %LTeq02\]

Wavefunctions of this form are referred to as Fock states. In this framework, the creation and annihilation operators act to raise or lower the number of excitations in a specific mode:

\[\begin{aligned}
& \hat a_k^{\dagger }\left| N_{1},N_{2}, \ldots N_{k} \ldots \right\rangle = \sqrt {N_{k} + 1} \left| N_{1},N_{2}, \ldots N_{k} + 1 \ldots \right\rangle \\
& {\hat a}_{k}\left| N_{1},N_{2}, \ldots N_{k} \ldots \right\rangle = \sqrt {N_{k}} \left| N_{1},N_{2}, \ldots N_{k} - 1 \ldots \right\rangle
\end{aligned}%DOCeq20 %LTeq03\]

The occupation of each mode can be related to the ground state through repeated application of the creation operator:

\[\left| N_{k} \right\rangle = \frac{\left( \hat a_k^{\dagger } \right)^{N_{k}}}{\sqrt {N_{k}!} }\left| 0_{k} \right\rangle %DOCeq21 %LTeq04\]

One can define a single mode number operator

\[\hat N_{k,\sigma } = \hat a_{k,\sigma}^{\dagger }\hat a_{k,\sigma } %DOCeq22 %LTeq05\]

that gives you the number of excitations in mode \(k\):

\[\hat N_{k,\sigma }\left| \varphi _{L} \right\rangle = N_{k,\sigma }\left| \varphi _{L} \right\rangle %DOCeq23 %LTeq06\]

That is, \(N_{k,\sigma }\) is the eigenvalue for the number operator \(\hat N_{k,\sigma }\), and can take on integer values such that \(N_{k,\sigma } = 0,1,2, \ldots \). The total number of excitations in the field is then given by the total number operator

\[\hat N = \sum_{k,\sigma } {\hat N}_{k,\sigma } %DOCeq24 %LTeq07\]

The energy associated with the excitations of a specific mode is the energy eigenvalue \(E_{k,\sigma }\), obtained from solving \(H_{L}\left| \varphi _{L} \right\rangle = E_{L}\left| \varphi _{L} \right\rangle \).

\[E_{k,\sigma } = \hbar \omega _{k}\left( N_{k,\sigma } + \frac{1}{2} \right) %DOCeq25 %LTeq08\]

The energy depends on the number of quanta occupying the mode, multiplied by the energy of each quantum, plus a ground state zero-point energy of \(\hbar \omega _{k}/2\). The total electromagnetic energy of the light is obtained from the expectation value of the Hamiltonian, which involves a sum over all modes:

\[E_{L} = \left\langle H_{L} \right\rangle = \sum_{k,\sigma } \left( N_{k,\sigma } + \frac{1}{2} \right)\hbar \omega _{k} %DOCeq26 %LTeq09\]

This does lead to a problem—if there are no bounds on the number of modes, a light field with no photons has infinite energy! That is, \(E_{0} = \sum\nolimits_{k,\sigma } \frac{1}{2}\hbar \omega _{k} \to \infty \). Since we are unable to measure the absolute energy of the ground state and can only ascertain the differences of energy between states, we won’t view this as a serious problem. For our purposes, it will serve to set the ground state energy to zero. There are ways of dealing with this divergence problem through renormalization, but that is beyond our scope. It’s good to remember that this is still just a model.

To describe the combined state of the light and matter, the total Hamiltonian given by eq. (9.1.2) can be expressed as a joint light-matter system, \(H_{0} = H_{L} + H_{M}\), and the interaction between the two:

\[\begin{aligned}
H_{\text{tot}} &= H_{M} + H_{L} + H_{LM} \\
&= H_{0} + H_{LM} \\
\end{aligned} %DOCeq27 %LTeq10\]

This is a useful partitioning when the light-matter interactions are weak and lends itself to working in the interaction picture. If the matter is described by the wavefunction \(\left| \psi _{M} \right\rangle \), where \(H_{M}\left| \psi _{M} \right\rangle = E_{M}\left| \psi _{M} \right\rangle \), then the total state of the joint light-matter system \(\left| \Psi \right\rangle \) can be expressed in terms of product states

\[\left| \Psi \right\rangle = \left| \psi _{M}{\kern 1pt} \varphi _{L} \right\rangle %DOCeq28 %LTeq11\]

Each state \(\left| \Psi \right\rangle \) is an eigenstate of \(H_{0}\), and the energy of the system is the sum of the energies for the light and matter, \(E = E_{L} + E_{M}\).

Now consider how to describe changes to the state of the system that are driven by light-matter interactions. For any transition in the matter from an initial state \(\left| \ell \right\rangle \) to a final state \(\left| m \right\rangle\)—both eigenstates of \(H_{M}\)—we expect an associated transition in the light field of equal and opposite energy. For such a transition, we can express the initial and final states of the total light-matter system as:

\[\begin{align}\begin{aligned}
\left| i \right\rangle &= \left| \ell ;N_{1},N_{2},N_{3} \ldots ,N_{k}, \ldots \right\rangle \\
\left| f \right\rangle &= \left| m;N_{1},N_{2},N_{3}, \ldots ,N_{k} \pm 1, \ldots \right\rangle \\
\end{aligned}\end{align} %DOCeq29 %LTeq12\]

The change in the light field is observed in the occupation of mode \(k\), which transition from \(N_{k}\) to \(N_{k} \pm 1\). In the final state \(N_{k} \pm 1\), a “\(+\)” refers to emission (an increase in the number of photons in the field), and “\(-\)” refers to absorption (a decrease in the number of photons). The energies of these the initial and final states of the system are obtained by summing over the total energy of the matter and electromagnetic field.

\[\begin{align}E_{i} = \left\langle i \right|H_{0}\left| i \right\rangle &= E_{\ell } + \sum_{j} N_{j}\hbar \omega _{j} \\ %DOCeq31 %LTeq13
E_{f} = \left\langle f \right|H_{0}\left| f \right\rangle &= E_{m} + \sum_{j} N_{j}\hbar \omega _{j} \pm \hbar \omega _{k} \end{align}%DOCeq32 %LTeq14\]

The total energy change of the matter-light system is \(E_{f} - E_{i} = E_{m} - E_{\ell } \pm \hbar \omega _{k}.\) The Golden Rule description of transition rates enforces the resonance condition \(E_{f} - E_{i} = 0\) . Thus, the energy change in the matter is the opposite to the change of energy of the light: \(E_{m} - E_{\ell } = \mp \hbar \omega _{k}\).

---