Quantization of EM Field

Chasse_neige

EM Field in a Cavity

$$\begin{array}{c}\frac{1}{c^2}\frac{{\partial }^2}{\partial t^2}{E}_x(z,t)-\frac{{\partial }^2}{\partial z^2}{E}_x(z,t)=0\\ E_x(0,t)=E_x(L,t)=0\end{array}$$

Here we use the integral of vector potential to solve this problem

$$E_x(z,t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{d}}^{}k(A_k{e}^{-i\omega t+kz}+c.c.)$$

Adding the boundary conditions on $A_k$, we could know that

$$A_k=-A_{-k}$$

and the allowed wave number is

$$k_n=\frac{n\pi }{L}$$

Using the discrete values, we can write the EM field in the form of

$$E_x(z,t)=2i\sum _{n=1}^{\mathrm{\infty }}(A_{k_n}{e}^{-i{\omega }_{n}t}-c.c.)\sin (k_{n}z)$$

Quantization

Transfering the amplitude into operators

$$A_{k_n}\to {\hat{A}}_{k_n}{A}_{k_n}^{\ast }\to {\hat{A}}_{k_n}^{†}$$

We can define annihilation and creation operators

$${\hat{a}}_n=i\sqrt{\frac{2L{ϵ}_0}{\hslash {\omega }_n}}{\hat{A}}_{k_n}{\hat{a}}_n^{†}=-i\sqrt{\frac{2L{ϵ}_0}{\hslash {\omega }_n}}{\hat{A}}_{k_n}$$

So we can use the annihilation and creation operators to express the electromagnetic field

$$E_x(z,t)=\sum _{n=1}^{\mathrm{\infty }}\sqrt{\frac{2\hslash {\omega }_n}{L{ϵ}_0}}({\hat{a}}_n{e}^{-i{\omega }_{n}t}+{\hat{a}}_n^{†}{e}^{i{\omega }_{n}t})\sin (k_{n}z)$$$$B_y(z,t)=-i\sum _{n=1}^{\mathrm{\infty }}\sqrt{\frac{2\hslash {\omega }_n{\mu }_0}{L}}({\hat{a}}_n{e}^{-i{\omega }_{n}t}-{\hat{a}}_n^{†}{e}^{i{\omega }_{n}t})\cos (k_{n}z)$$

The total energy of the EM field is

$$\begin{array}{c}\hat{H}=\int \frac{1}{2}({ϵ}_0{E}^2+\frac{B^2}{{\mu }_0}){\mathrm{d}}^{}\tau \\ =\frac{\hslash }{2}\sum _{n=1}^{\mathrm{\infty }}{\omega }_n{({\hat{a}}_n{e}^{-i{\omega }_{n}t}+{\hat{a}}_n^{†}{e}^{i{\omega }_{n}t})}^2+\frac{\hslash }{2}\sum _{n=1}^{\mathrm{\infty }}{\omega }_n{({\hat{a}}_n{e}^{-i{\omega }_{n}t}-{\hat{a}}_n^{†}{e}^{i{\omega }_{n}t})}^2\\ =\sum _{n=1}^{\mathrm{\infty }}\frac{\hslash {\omega }_n}{2}\{{\hat{a}}_n,{\hat{a}}_n^{†}\}=\sum _{n=1}^{\mathrm{\infty }}\frac{\hslash {\omega }_n}{2}(2{\hat{a}}_n^{†}{\hat{a}}_n+[{\hat{a}}_n,{\hat{a}}_n^{†}])\end{array}$$

Considering the commutation rule for the annihilation and creation operators

$$[{\hat{a}}_n,{\hat{a}}_n^{†}]=1$$

So the total energy is

$$\hat{H}=\sum _{n=1}^{\mathrm{\infty }}\frac{\hslash {\omega }_n}{2}(2{\hat{a}}_n^{†}{\hat{a}}_n+1)$$

Each mode can be viewed as a quantum harmonic oscillator and the total energy is the sum of contributions from all of them.

Energy Eigenstate

We can express the total energy eigenstates in direct product of multiple-photon states, say

$$|N⟩=|N_{{\omega }_1}⟩\otimes |N_{{\omega }_2}⟩\otimes |N_{{\omega }_3}⟩\otimes \cdots$$

while the hamiltonian exerts on the eigenstate gives

$$\hat{H}|N⟩=\sum _n\hslash {\omega }_n(N_n+\frac{1}{2})$$

Quadratures

Using the linear combination of the annihilation and creation operators, we can define amplitude and phase quadratures

$$\begin{array}{c}{\hat{X}}_n=\frac{1}{\sqrt{2}}({\hat{a}}_n+{\hat{a}}_n^{†})(\text{amplitude quadrature})\\ {\hat{Y}}_n=\frac{1}{\sqrt{2}i}({\hat{a}}_n-{\hat{a}}_n^{†})(\text{phase quadrature})\end{array}$$

which analogous to position and momentum of a classical oscillator.

EM Field in the Free Space

Consider a ring cavity’s boundary condition, at this time

$$\begin{array}{c}{\omega }_n=\frac{2\pi cn}{L}\\ k_n=\frac{2\pi n}{L}\end{array}$$

Quantize the electric field

$${\hat{E}}_x(z,t)=\sum _{n=\pm 1}^{\mathrm{\infty }}\sqrt{\frac{\hslash {\omega }_n}{2L{ϵ}_0}}({\hat{a}}_n{e}^{-i{\omega }_{n}t+i{k}_{n}z}+{\hat{a}}_n^{†}{e}^{i{\omega }_{n}t-i{k}_{n}z})$$

And now take the continuous limit ($L\to \mathrm{\infty }$)

$$k_{n+1}-k_n\to \frac{2\pi }{L}{\mathrm{d}}^{}k$$$$\sum _{n=\pm 1}^{\mathrm{\infty }}\to {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{L}{2\pi }{\mathrm{d}}^{}k$$

We define that

$${\hat{a}}_k=\sqrt{\frac{2\pi }{{\mathrm{d}}^{}k}}{\hat{a}}_n$$

and the commutation rule for ${\hat{a}}_k$ is

$$[{\hat{a}}_k,{\hat{a}}_k^{†}]=\frac{2\pi }{{\mathrm{d}}^{}k}{\delta }_{n{n}^{\prime }}\approx 2\pi \delta (k-k^{\prime })$$

So the quantized electricmagnetic field is

$${\hat{E}}_x(z,t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}k}{2\pi }\sqrt{\frac{\hslash \omega }{2{ϵ}_0}}({\hat{a}}_n{e}^{-i\omega t+ikz}+{\hat{a}}_n^{†}{e}^{i\omega t-ikz})$$$${\hat{B}}_y(z,t)=\frac{1}{c}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}k}{2\pi }\sqrt{\frac{\hslash \omega }{2{ϵ}_0}}({\hat{a}}_n{e}^{-i\omega t+ikz}+{\hat{a}}_n^{†}{e}^{i\omega t-ikz})$$

If we ignore zero-point energy, we can write the hamiltonian as

$$\hat{H}={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}k}{2\pi }\hslash c|k|{\hat{a}}_n^{†}{\hat{a}}_n$$

Narrow Band Approximation

For a laser with central frequency ${\omega }_0$ and half band width $\mathrm{\Omega }$, where $\mathrm{\Omega }\ll {\omega }_0$, so the electric field can be written as

$$\begin{array}{c}{E}_x^{\to }(z,t)={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}k}{2\pi }\sqrt{\frac{\hslash \omega }{2{ϵ}_0}}({\hat{a}}_k{e}^{-i\omega t+ikz}+c.c.)\\ ={\int }_{-{\omega }_0}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}\mathrm{\Omega }}{2\pi }\sqrt{\frac{\hslash ({\omega }_0+\mathrm{\Omega })}{2{ϵ}_0}}({\hat{a}}_{\mathrm{\Omega }}{e}^{-i\mathrm{\Omega }(t-\frac{z}{c})}{e}^{-i{\omega }_{0}t+i{k}_{0}z}+c.c.)\\ \approx {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}\mathrm{\Omega }}{2\pi }\sqrt{\frac{\hslash {\omega }_0}{2{ϵ}_0}}({\hat{a}}_{\mathrm{\Omega }}{e}^{-i\mathrm{\Omega }(t-\frac{z}{c})}{e}^{-i{\omega }_{0}t+i{k}_{0}z}+c.c.)\end{array}$$

Define $\hat{a}(t)$ as

$$\hat{a}(t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{d}}^{}\mathrm{\Omega }}{2\pi }{\hat{a}}_{\mathrm{\Omega }}{e}^{-i\mathrm{\Omega }t}$$

So the electric field under approximation is

$$E_x^{\to }(z,t)\approx \sqrt{\frac{\hslash {\omega }_0}{2{ϵ}_0}}(\hat{a}(t-\frac{z}{c})e^{-i{\omega }_{0}t+i{k}_{0}z}+{\hat{a}}^{†}(t-\frac{z}{c})e^{i{\omega }_{0}t-i{k}_{0}z})$$

Using and the amplitude and phase quadratures

$$\begin{array}{c}\hat{X}(t)\equiv \frac{1}{\sqrt{2}}(\hat{a}(t)+{\hat{a}}^{†}(t))\\ \hat{Y}(t)\equiv \frac{1}{\sqrt{2}i}(\hat{a}(t)-{\hat{a}}^{†}(t))\end{array}$$

The field can be rewritten as

$$E_x(z,t)=\sqrt{\frac{\hslash {\omega }_0}{{ϵ}_0}}(\hat{X}(t-\frac{z}{c})\cos ({\omega }_{0}t-k_{0}z)+\hat{Y}(t-\frac{z}{c})\sin ({\omega }_{0}t-k_{0}z))$$