Field operators

Field operators definition§

From the discrete case,

$$\ket{\psi;f}=\sum_k f(k) a^\dagger(k)\ket{0}$$

We can easily see what we need, which is

$$\ket{\psi;f}=\int \mathrm{d}\omega\, f(\omega) E^\dagger(\omega)\ket{0}$$

This is the definition of $E$ we will take; it is the $a$ operator in the continuous limit.

Properties§

Since $\ket{\psi;f}$ should be normalized, and assuming $[E(\omega),E^\dagger(\omega')]$ is a c‑number, the normalization condition forces

$$[\hat{E}(\omega), \hat{E}^\dagger(\omega')]=\delta(\omega-\omega').$$

From the previous chapter, we noted that

$$[\hat{a}_k, \hat{a}_{k'}^\dagger]=\delta_{kk'}.$$

For a field with continuous modes, the equivalent relation is

$$[\hat{E}(\omega), \hat{E}^\dagger(\omega')]=\delta(\omega-\omega').$$

Rigor§

Although the interpretation makes it obvious why the above should be true, it is worth noting more physical detail via a rigorous derivation.

Setup§

We consider an one‑dimensional cavity of length $L$ with periodic boundary conditions. The classical electromagnetic field in Coulomb gauge can be expanded in normal modes. Quantization promotes the Fourier amplitudes to operators.

The full free‑field Hamiltonian is

$$\hat{H} = \frac{1}{2}\int_0^L \mathrm{d}x \left( \epsilon_0 \hat{E}^2(x) + \frac{1}{\mu_0}\hat{B}^2(x) \right).$$

A general expansion of the electric field consistent with the wave equation and the boundary conditions is

$$\hat{E}(x) = i\sum_k \mathcal{E}_k \left( \hat{a}_k e^{ikx} - \hat{a}_k^\dagger e^{-ikx} \right),$$

where $k = 2\pi n/L$ with $n\in\mathbb{Z}$, and $\mathcal{E}_k$ is a real normalization constant to be determined (it may depend on $|k|$).

The $i$ and the negative phase on $\hat{a}^\dagger$ here is by convention where $\hat{A}$ should be $\propto \hat{a}e^{ikx} + \hat{a}^\dagger e^{-ikx}$ then $\hat{E}=-\partial_t \hat{A}$ since $\hat{A}$ is the canonical variable that have the priority to be quantized.

The corresponding magnetic field for a plane‑polarized wave propagating along $x$ is

$$\hat{B}(x) = i\sum_k \frac{k}{|k|}\frac{\mathcal{E}_k}{c} \left( \hat{a}_k e^{ikx} - \hat{a}_k^\dagger e^{-ikx} \right), \qquad c = \frac{1}{\sqrt{\epsilon_0\mu_0}}.$$

The bosonic operators satisfy $[\hat{a}_k,\hat{a}_{k'}^\dagger] = \delta_{k,k'}$, $[\hat{a}_k,\hat{a}_{k'}]=0$.

Equate to find $\mathcal{E}_k$§

Insert the expansions into $\hat{H}$ and use the spatial integrals

$$\int_0^L e^{i(k+k')x}\mathrm{d}x = L\delta_{k,-k'},\qquad \int_0^L e^{i(k-k')x}\mathrm{d}x = L\delta_{k,k'}.$$

After straightforward algebra, the Hamiltonian becomes

$$\begin{aligned} \hat{H} &= \sum_k \epsilon_0 L \mathcal{E}_k^2 \Bigl(2\hat{a}_k^\dagger\hat{a}_k + 1 + \hat{a}_k\hat{a}_{-k} + \hat{a}_k^\dagger\hat{a}_{-k}^\dagger \Bigr) \\ &\qquad - \sum_k \frac{1}{\mu_0 c^2} L \mathcal{E}_k^2 \Bigl( \hat{a}_k\hat{a}_{-k} + \hat{a}_k^\dagger\hat{a}_{-k}^\dagger \Bigr), \end{aligned}$$

where we used $[\hat{a}_k,\hat{a}_k^\dagger]=1$ and kept only non‑vanishing terms.
Because $\epsilon_0\mu_0 c^2 = 1$, the coefficients in front of the non‑diagonal terms $\hat{a}_k\hat{a}_{-k}$ and $\hat{a}_k^\dagger\hat{a}_{-k}^\dagger$ cancel exactly. The remaining diagonal part is

$$\hat{H} = \sum_k 2\epsilon_0 L \mathcal{E}_k^2 \left( \hat{a}_k^\dagger\hat{a}_k + \frac{1}{2} \right).$$

We demand that this reproduces the known harmonic‑oscillator Hamiltonian $\hat{H} = \sum_k \hbar\omega_k (\hat{a}_k^\dagger\hat{a}_k + 1/2)$. Comparing the two expressions yields

$$2\epsilon_0 L \mathcal{E}_k^2 = \hbar\omega_k \quad\Longrightarrow\quad \mathcal{E}_k = \sqrt{\frac{\hbar\omega_k}{2\epsilon_0 L}}.$$

Thus the electric field operator is

$$\hat{E}(x) = i\sum_k \sqrt{\frac{\hbar\omega_k}{2\epsilon_0 L}} \left( \hat{a}_k e^{ikx} - \hat{a}_k^\dagger e^{-ikx} \right).$$

Infinite limit§

Let $\Delta k = \frac{2\pi}{L}$. In order to make the modes continuous, we take $L\to\infty$. Factor as follows:

$$\begin{align*} \hat{E}(x) &= i\sum_k \frac{2\pi}{L} \sqrt{\frac{\hbar\omega_k}{4\pi\epsilon_0}} \left( e^{ikx} \sqrt{\frac{L}{2\pi}}\,\hat{a}_k - e^{-ikx} \sqrt{\frac{L}{2\pi}}\,\hat{a}_k^\dagger \right) \\ &\xrightarrow[L\to\infty]{} i\int_{-\infty}^{\infty} \mathrm{d}k \, \sqrt{\frac{\hbar\omega_k}{4\pi\epsilon_0}} \, \hat{E}(k) e^{ikx} + \text{h.c.}, \end{align*}$$

where we defined the continuum annihilation operator

$$\hat{E}(k) = \sqrt{\frac{L}{2\pi}}\,\hat{a}_k, \qquad [\hat{E}(k),\hat{E}^\dagger(k')] = \delta(k-k').$$

$E(\omega)$ operator§

Because $\omega_k = c|k|$, we can restrict the integration to positive frequencies. For $\omega>0$, set $k = \omega/c$ (positive $k$) and $\mathrm{d}k = \mathrm{d}\omega/c$. Then

$$\hat{E}(x) = i\int_{0}^{\infty} \mathrm{d}\omega \, \sqrt{\frac{\hbar\omega}{4\pi\epsilon_0 c}} \, \hat{E}(\omega) e^{i\omega x/c} + \text{h.c.},$$

where we have absorbed the factor $1/\sqrt{c}$ from the integration measure into the definition of the continuous frequency operator

$$\hat{E}(\omega) \equiv \frac{1}{\sqrt{c}} \hat{E}(k\!=\!\omega/c).$$

With this definition, $[\hat{E}(\omega),\hat{E}^\dagger(\omega')] = \delta(\omega-\omega')$ because

$$[\hat{E}(\omega),\hat{E}^\dagger(\omega')] = \frac{1}{c} [\hat{E}(\omega/c),\hat{E}^\dagger(\omega'/c)] = \frac{1}{c} \delta\!\left(\frac{\omega-\omega'}{c}\right) = \delta(\omega-\omega').$$

Positive‑ and negative‑frequency parts (standard decomposition)§

We define the positive‑frequency part of the electric field operator as

$$\hat{E}^{(+)}(x) \equiv i\int_{0}^{\infty}\mathrm{d}\omega\, \sqrt{\frac{\hbar\omega}{4\pi\epsilon_0 c}}\,\hat{E}(\omega) e^{i\omega x/c},$$

and the negative‑frequency part as its Hermitian conjugate:

$$\hat{E}^{(-)}(x) \equiv \left[\hat{E}^{(+)}(x)\right]^\dagger = -i\int_{0}^{\infty}\mathrm{d}\omega\, \sqrt{\frac{\hbar\omega}{4\pi\epsilon_0 c}}\,\hat{E}^\dagger(\omega) e^{-i\omega x/c}.$$

Then the full (Hermitian) field operator is simply

$$\hat{E}(x) = \hat{E}^{(+)}(x) + \hat{E}^{(-)}(x).$$

This decomposition is standard in quantum optics because $\hat{E}^{(+)}(x)$ annihilates the vacuum and $\hat{E}^{(-)}(x)$ creates excitations.

Substitute back to the state expansion§

The state $|\psi;f\rangle$ transforms accordingly. Up to a re-scaling of $f$,

$$\begin{align*} \ket{\psi;f} &= \int\mathrm{d}k\,\sqrt{\frac{L}{2\pi}}f(k)\,\hat{E}^\dagger(k)\ket{0} \\ &= \int\mathrm{d}k\,\tilde{f}(k)\,\hat{E}^\dagger(k)\ket{0} \\ &= \int\mathrm{d}\omega\,f(\omega)E^\dagger(\omega)\ket{0}. \end{align*}$$

The commutation§

From the definition $\hat{E}(k)=\sqrt{L/(2\pi)}\,\hat{a}_k$ and $[\hat{a}_k,\hat{a}_{k'}^\dagger]=\delta_{kk'}$, we have

$$[\hat{E}(k),\hat{E}^\dagger(k')] = \frac{L}{2\pi}\delta_{kk'} \;\xrightarrow[L\to\infty]{}\; \delta(k-k').$$

In terms of frequency,

$$[E(\omega),E^\dagger(\omega')] = \delta(\omega-\omega'),$$

which matches the result obtained from normalization.

In 3D§

It is anyway a phase convention, we will drop the $i$ from here. Then, in 3D

$$\hat{\mathbf{E}}(\mathbf{r})=\sum_{\mathbf{k},\lambda}\mathcal{E}_k\left(\hat{a}_{\mathbf{k},\lambda}\,\hat{\mathbf{e}}_{\mathbf{k},\lambda}\,e^{i\mathbf{k}\cdot\mathbf{r}}+\hat{a}^\dagger_{\mathbf{k},\lambda}\,\hat{\mathbf{e}}^*_{\mathbf{k},\lambda}\,e^{-i\mathbf{k}\cdot\mathbf{r}}\right)$$

where

$$\mathcal{E}_k=\sqrt{\frac{\hbar \omega_k}{2\epsilon_0 V}}$$

The sum is over all wavevectors $\mathbf{k} = \frac{2\pi}{L}(n_x, n_y, n_z)$ with $n_i \in \mathbb{Z}$, and over two transverse polarizations $\lambda = 1,2$ satisfying $\hat{\mathbf{e}}_{\mathbf{k},\lambda} \perp \mathbf{k}$.

Continuous limit in 3D§

The discrete sum becomes an integral via the replacement

$$\sum_\mathbf{k} \longrightarrow \frac{V}{(2\pi)^3}\int \mathrm{d}^3k$$

which follows from the fact that each mode occupies a volume $(2\pi/L)^3 = (2\pi)^3/V$ in $\mathbf{k}$-space. Define the continuum operators

$$\hat{a}_{\mathbf{k},\lambda} \longrightarrow \sqrt{\frac{V}{(2\pi)^3}}\,\hat{a}(\mathbf{k},\lambda), \qquad [\hat{a}(\mathbf{k},\lambda),\hat{a}^\dagger(\mathbf{k}',\lambda')] = \delta^{(3)}(\mathbf{k}-\mathbf{k}')\delta_{\lambda\lambda'}$$

so that the commutation relation becomes a Dirac delta instead of a Kronecker delta. Note the same $1/\sqrt{V}$ cancellation as in 1D: $\mathcal{E}_k \propto 1/\sqrt{V}$ while the mode density grows as $V$, so all physical observables are $V$-independent.

Substituting:

$$\hat{\mathbf{E}}(\mathbf{r}) = \sum_\lambda \int \frac{\mathrm{d}^3k}{(2\pi)^3} \sqrt{\frac{\hbar\omega_k}{2\epsilon_0}} \left( \hat{a}(\mathbf{k},\lambda)\,\hat{\mathbf{e}}_{\mathbf{k},\lambda}\,e^{i\mathbf{k}\cdot\mathbf{r}} + \hat{a}^\dagger(\mathbf{k},\lambda)\,\hat{\mathbf{e}}^*_{\mathbf{k},\lambda}\,e^{-i\mathbf{k}\cdot\mathbf{r}} \right)$$

Switching to spherical coordinates in $\mathbf{k}$-space with $k = |\mathbf{k}|$, $\omega_k = ck$, and $\mathrm{d}^3k = k^2\,\mathrm{d}k\,\mathrm{d}\Omega$:

$$\hat{\mathbf{E}}(\mathbf{r}) = \sum_\lambda \int_0^\infty \frac{\mathrm{d}\omega}{(2\pi)^3 c^3} \int \mathrm{d}\Omega\, \omega^2 \sqrt{\frac{\hbar\omega}{2\epsilon_0}} \left( \hat{a}(\omega,\hat{k},\lambda)\,\hat{\mathbf{e}}_{\hat{k},\lambda}\,e^{i\omega\hat{k}\cdot\mathbf{r}/c} + \text{h.c.} \right)$$

For an atom at the origin (dipole approximation, $e^{i\mathbf{k}\cdot\mathbf{r}_\text{atom}} \approx 1$) or when only the frequency content matters, one typically integrates out the angular and polarization degrees of freedom. This yields the spectral density of the field, which is the physically relevant quantity for decay rates and is manifestly $V$-independent.

Positive- and negative-frequency parts in 3D§

As in 1D, we split

$$\hat{\mathbf{E}}(\mathbf{r}) = \hat{\mathbf{E}}^{(+)}(\mathbf{r}) + \hat{\mathbf{E}}^{(-)}(\mathbf{r})$$

where

$$\hat{\mathbf{E}}^{(+)}(\mathbf{r}) = \sum_{\mathbf{k},\lambda} \mathcal{E}_k\, \hat{a}_{\mathbf{k},\lambda}\,\hat{\mathbf{e}}_{\mathbf{k},\lambda}\,e^{i\mathbf{k}\cdot\mathbf{r}}$$

contains only annihilation operators (positive frequency, annihilates vacuum), and $\hat{\mathbf{E}}^{(-)} = \left[\hat{\mathbf{E}}^{(+)}\right]^\dagger$ contains only creation operators. This decomposition is used throughout quantum optics: photodetection theory, the optical Bloch equations, and the input-output formalism all rely on it.