9.2: Hamiltonian for quantized electromagnetic field

\( \let\bibref\eqref %make-command_bibref-equivalent-to-eqref\)The energy density is the starting point for connecting the energy in an electromagnetic wave to its fields.

\[\mathcal{U} = \frac{1}{2}\varepsilon _{0}\boldsymbol{\mathcal{E}}^{2} + \frac{1}{2\mu _{0}}\boldsymbol{\mathcal{B}}^{2} = \frac{1}{2}\varepsilon _{0}\left( \boldsymbol{\mathcal{E}}^{2} + c^{2}\boldsymbol{\mathcal{B}}^{2} \right) %DOCeq05 %LTeq01\]

The energy of the electromagnetic field can then be obtained by integrating over a volume, which gives the classical Hamiltonian of the electromagnetic field.

\[H_{L} = \frac{1}{2}\varepsilon _{0}\int d^{3}r\left( \boldsymbol{\mathcal{E}}^{2} + c^{2}\boldsymbol{\mathcal{B}}^{2} \right) %DOCeq06 %LTeq02\]

This Hamiltonian can be expressed in the vector potential basis set, using the expressions that connect \(\mathbf{A}\) with the fields that we derived earlier: \(\boldsymbol{\mathcal{E}} = - \dot{\mathbf{A}} \) and \(\boldsymbol{\mathcal{B}} = \nabla \times \mathbf{A}\).

\[H_{L} = \frac{1}{2}\varepsilon _{0}\int d^{3}r\left( {\mathbf{\dot A}}^{2} + c^{2}\left| \vphantom{A^2} \nabla \times \mathbf{A} \right|^{2} \right) %DOCeq07 %LTeq03\]

The second term in the integral can be further simplified if we work with plane wave solutions such as eq. (9.1.4). Then, for a single wave mode, \(\smallint d^{3}r\) \(\left| \nabla \times \mathbf{A} \right|^{2}\) can be evaluated to give \(k^{2}\smallint d^{3}r \, A^{2}\). This leads to our working expression for the classical Hamiltonian in terms of the electromagnetic vector potential:

\[H_{L} = \int d^{3}r \, \frac{1}{2}\varepsilon _{0}\left[ {\mathbf{\dot A}}_{k}^{2} + \omega _k^{2}\mathbf{A}_{k}^{2} \right] %DOCeq08 %LTeq04\]

This integral can be evaluated for a large cubic volume \(V\), assuming periodic boundary conditions that allow an integral number of half wavelengths within the walls. The resulting energy in the single mode wave, averaged over time, is

\[\bar E_{k} = 2\varepsilon _{0}\omega _k^{2}\left| \mathbf{A}_{k} \right|^{2} = 2\varepsilon _{0} | \boldsymbol{\mathcal{E}}_{k} |^{2} %DOCeq09 %LTeq05\]

Note the similarity of the Hamiltonian in eq. (9.2.4) to that for a continuum of harmonic oscillators with varying frequency:

\[\begin{aligned}
H_{\text{HB}} &= \sum_{j} \frac{1}{2} \frac{p_j^{2}}{m_{j}} + \frac{1}{2}m_{j}\omega _j^{2}x_j^{2} \\
&= \sum_{j} \frac{1}{2} \left( \dot q_j^{2} + \omega _j^{2}q_j^{2} \right) \\
\end{aligned} %DOCeq10 %LTeq06\]

The abbreviation \(H_{\text{HB}}\) is applied to this Hamiltonian because it is identical in form to a harmonic bath Hamiltonian. We will use the harmonic bath later to describe the continuum states of a condensed matter environment in a statistical manner. In mapping eq. (9.2.4) to eq. (9.2.6), \(\mathbf{A}\) plays the role of the reduced oscillator coordinate \(q = \hat x/\sqrt {m} \), \(\mathbf{\dot A}\) is its conjugate momentum \(m\dot q = \sqrt {m} \dot x\), and the integral over volume replaces the sum over modes that vary in frequency. It therefore follows quite naturally that one can express the quantum Hamiltonian for an electromagnetic field in terms of a continuum of harmonic oscillator modes with frequencies of \(\omega _{k}\). Going even further, we can use second quantization expressions for the harmonic oscillator to express the vector potential as

\[A_{k} = \sqrt {\frac{\hbar }{2\omega _{\text{k}}}} \left( \hat a_{k,\sigma} ^{\dagger } + \hat a_{k,\sigma} \right) %DOCeq11 %LTeq07\]\[\dot A_{k} = \mathrm{i} \sqrt {\frac{\hbar \omega _{\text{k}}}{2}} \left( \hat a_{k,\sigma}^{\dagger } - \hat a_{k,\sigma} \right) %DOCeq12 %LTeq08\]

and the Hamiltonian becomes

\[H_{L} = \sum_{k,\sigma } {\hbar } \omega _{k}\left( \hat a_{k,\sigma} ^{\dagger}{\hat a}_{k,\sigma } + \frac{1}{2} \right) %DOCeq13 %LTeq09\]

Here \(\hat a_{k,\sigma}^{\dagger }\) and \(\hat a_{k,\sigma }\) are creation and annihilation operators for the light quanta (photons) in the mode with wavevector \(k\) and polarization state \(\sigma\). These obey the commutation relation \(\left[ {\hat a}_{k',\sigma '},\hat a_{k,\sigma} ^{\dagger } \right] = \delta _{k,k'}\delta _{\sigma ,\sigma '}\).

The connection between the classical and quantum Hamiltonians for the electromagnetic field is presented quite loosely here, but eq. (9.2.9) can be rigorously derived from eq. (9.2.4) using the vector potential for a continuum of plane waves.\(\bibref{Cite1} %in-text-citation\) Within this formalism, the one can define a vector potential operator that satisfies eq. (9.2.4) as

\[\mathbf{A}(\mathbf{r},t) = \sum_{k,\sigma} \sqrt {\frac{\hbar }{2\varepsilon _{0}V\omega _{k}}} \left( {\hat \varepsilon }_{k,\sigma }{\hat a}_{k,\sigma }e^{\mathrm{i} \left( \mathbf{k} \cdot \mathbf{r} - \omega t \right)} + \hat \varepsilon _{k,\sigma}^{ * }\hat a_{k,\sigma}^{\dagger }e^{ - \mathrm{i} \left( \mathbf{k} \cdot \mathbf{r} - \omega t \right)} \right) %DOCeq14 %LTeq10\]

\(V\) is the quantization volume, and the factor of \(1/\sqrt {V} \) ensures the orthonormality of the plane wave basis set when integrating over a volume:

\[\int d^{3}r \, \frac{e^{ - \mathrm{i}\mathbf{k} \cdot \mathbf{r}}}{\sqrt {V} }\frac{e^{\mathrm{i} \mathbf{k'} \cdot \mathbf{r}}}{\sqrt {V} } = \delta _{k,k'} %DOCeq15 %LTeq11\]

In eq. (9.2.10), \(\mathbf{A} \! ^{2}\) is an energy density which must then be integrated over \(V\).

The operators for the vector potential described in eq. (9.2.7), \(\mathbf{A}(\mathbf{r})\), and the Hamiltonian are both time-independent. However, in the Heisenberg or interaction pictures, these operators adopt the conventional time-dependence of the harmonic oscillator raising and lowering operators when evaluating \(\mathbf{A}(\mathbf{r},t) = e^{ - \mathrm{i}H_{L}t/\hbar }\mathbf{A}(\mathbf{r})e^{\mathrm{i}H_{L}t/\hbar }\). This results in the expected time-dependence in eq. (9.2.10). We also find that the operators corresponding to the electric and magnetic fields are

\[\begin{aligned}
\boldsymbol{\mathcal{E}}(\mathbf{r},t) & = \mathrm{i} \sum_{k,\sigma } \sqrt {\frac{\hbar \omega _{k}}{2\varepsilon _{0}V}} \left( {\hat \varepsilon }_{k,\sigma }{\hat a}_{k,\sigma }e^{\mathrm{i}\left( \mathbf{k} \cdot \mathbf{r} - \omega _{k}t \right)} - \hat \varepsilon _{k,\sigma} ^{ * }\hat a_{k,\sigma} ^{\dagger }e^{ - \mathrm{i}\left( \mathbf{k} \cdot \mathbf{r} - \omega _{k}t \right)} \right) \\
\boldsymbol{\mathcal{B}} (\mathbf{r},t) &= \mathrm{i} \sum_{k,\sigma } \sqrt {\frac{\hbar }{2\varepsilon _{0}V\omega _{k}}} \left( \mathbf{k} \times {\hat \varepsilon }_{k,\sigma } \right)\left( {\hat a}_{k,\sigma }e^{\mathrm{i}\left( \mathbf{k} \cdot \mathbf{r} - \omega _{k}t \right)} - \hat a_{k,\sigma} ^{\dagger }e^{ - \mathrm{i}\left( \mathbf{k} \cdot \mathbf{r} - \omega _{k}t \right)} \right) \\
\end{aligned} %DOCeq16 %LTeq12\]

Substituting these relationships into the Hamiltonian for the classical radiation field, eq. (9.2.2), leads directly to the quantum version, eq. (9.2.9).

Along similar lines to the calculation of the energy of the field in this section, one can also compute the momentum associated with the electromagnetic wave. The classical momentum is obtained from the Poynting vector using \(\mathbf{p}_{L} = \varepsilon _{0}\smallint d^{3}r \, \boldsymbol{\mathcal{E}} \times \boldsymbol{\mathcal{B}}\). Expressing this in terms of the vector potential and evaluating this for the plane wave modes gives:\(\bibref{Cite1} %in-text-citation\)

\[\mathbf{p}_{L} = \sum_{k,\sigma } \hbar \mathbf{k}\hat a_{k,\sigma}^{\dagger }{\hat a}_{k,\sigma } %DOCeq17 %LTeq13\]

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References

\[\tag {1} \label {Cite1}\]

Townsend, J. S. A Modern Approach to Quantum Mechanics, 2nd ed.; University Science Books, 2012.

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