Cavity quantum electrodynamics (cavity QED) is the study of the interaction between light confined in a reflective cavity and atoms or other particles, under conditions where the quantum nature of photons is significant.
For a free scalar field, the Lagrangian is
L = 1 2 ( ∂ μ ϕ ) ∂ μ ϕ − 1 2 m 2 ϕ 2 \mathcal{L}=\frac{1}{2}\left(\partial^\mu \phi\right) \partial_\mu \phi-\frac{1}{2} m^2 \phi^2
L = 2 1 ( ∂ μ ϕ ) ∂ μ ϕ − 2 1 m 2 ϕ 2
The conjugate momentum π \pi π
π ( x ) = ∂ L ∂ ( ∂ 0 ϕ ) = ∂ 0 ϕ ( x ) \pi(x)=\frac{\partial \mathcal{L}}{\partial\left(\partial_0 \phi\right)}=\partial_0 \phi(x)
π ( x ) = ∂ ( ∂ 0 ϕ ) ∂ L = ∂ 0 ϕ ( x )
Stipulate equal time commutation relations
[ ϕ ( x , t ) , π ( y , t ) ] = i δ ( 3 ) ( x − y ) , [ ϕ ( x , t ) , ϕ ( y , t ) ] = 0 , [ π ( x , t ) , π ( y , t ) ] = 0 [\phi(\boldsymbol{x}, t), \pi(\boldsymbol{y}, t)]=\mathrm{i} \delta^{(3)}(\boldsymbol{x}-\boldsymbol{y}), \quad[\phi(\boldsymbol{x}, t), \phi(\boldsymbol{y}, t)]=0, \quad[\pi(\boldsymbol{x}, t), \pi(\boldsymbol{y}, t)]=0
[ ϕ ( x , t ) , π ( y , t ) ] = i δ ( 3 ) ( x − y ) , [ ϕ ( x , t ) , ϕ ( y , t ) ] = 0 , [ π ( x , t ) , π ( y , t ) ] = 0
by which we quantize the coefficients a p , a p † a_\boldsymbol{p}, a_\boldsymbol{p}^\dagger a p , a p †
ϕ ( x , t ) = ∫ d 3 p ( 2 π ) 3 1 2 E p ( a p e − i p ⋅ x + a p † e i p ⋅ x ) \phi(\boldsymbol{x}, t)=\int \frac{\mathrm{d}^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\boldsymbol{p}}}}\left(a_{\boldsymbol{p}} \mathrm{e}^{-\mathrm{i} p \cdot x}+a_{\boldsymbol{p}}^{\dagger} \mathrm{e}^{\mathrm{i} p \cdot x}\right)
ϕ ( x , t ) = ∫ ( 2 π ) 3 d 3 p 2 E p 1 ( a p e − i p ⋅ x + a p † e i p ⋅ x )
where p ⋅ x = p 0 t − p ⋅ x p \cdot x=p^0 t-\boldsymbol{p} \cdot \boldsymbol{x} p ⋅ x = p 0 t − p ⋅ x p 0 > 0 p^0>0 p 0 > 0 on shell
p 0 = E p ≡ ∣ p ∣ 2 + m 2 p^0=E_{\boldsymbol{p}} \equiv \sqrt{|\boldsymbol{p}|^2+m^2}
p 0 = E p ≡ ∣ p ∣ 2 + m 2
And it is easy to check a p , a p † a_\boldsymbol{p}, a_\boldsymbol{p}^\dagger a p , a p †
[ a p , a q † ] = ( 2 π ) 3 δ ( 3 ) ( p − q ) , [ a p , a q ] = [ a p † , a q † ] = 0 \left[a_{\boldsymbol{p}}, a_{\boldsymbol{q}}^{\dagger}\right]=(2 \pi)^3 \delta^{(3)}(\boldsymbol{p}-\boldsymbol{q}), \quad\left[a_{\boldsymbol{p}}, a_{\boldsymbol{q}}\right]=\left[a_{\boldsymbol{p}}^{\dagger}, a_{\boldsymbol{q}}^{\dagger}\right]=0
[ a p , a q † ] = ( 2 π ) 3 δ ( 3 ) ( p − q ) , [ a p , a q ] = [ a p † , a q † ] = 0
It is important to emphasize that, the quantization is performed in the Heisenberg picture. What we have now is actually
ϕ H ( x , t ) = ∫ d 3 p ( 2 π ) 3 1 2 E p ( a p e − i p ⋅ x + a p † e i p ⋅ x ) \phi^\mathrm{H}(\boldsymbol{x}, t)=\int \frac{\mathrm{d}^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\boldsymbol{p}}}}\left(a_{\boldsymbol{p}} \mathrm{e}^{-\mathrm{i} p \cdot x}+a_{\boldsymbol{p}}^{\dagger} \mathrm{e}^{\mathrm{i} p \cdot x}\right)
ϕ H ( x , t ) = ∫ ( 2 π ) 3 d 3 p 2 E p 1 ( a p e − i p ⋅ x + a p † e i p ⋅ x )
Therefore, In Heisenberg picture, the field operator is time dependent. Moreover, a p , a p † a_\boldsymbol{p}, a_\boldsymbol{p}^\dagger a p , a p †
a p H ( t ) = e i H t a p e − i H t = e − i E p t a p , a p † H ( t ) = e i E p t a p † a_\boldsymbol{p}^\mathrm{H}(t) = \mathrm{e}^{\mathrm{i}Ht}a_\boldsymbol{p}\mathrm{e}^{-\mathrm{i}Ht} = \mathrm{e}^{-\mathrm{i} E_\boldsymbol{p} t} a_\boldsymbol{p}, \quad a_\boldsymbol{p}^{\dagger\mathrm{H}}(t) = \mathrm{e}^{\mathrm{i} E_\boldsymbol{p} t} a^\dagger_\boldsymbol{p}
a p H ( t ) = e i H t a p e − i H t = e − i E p t a p , a p † H ( t ) = e i E p t a p †
If switching to the Schrodinger picture, one will find the field operator is time independent
ϕ S ( x , t ) = ∫ d 3 p ( 2 π ) 3 1 2 E p ( a p e i p ⋅ x + a p † e − i p ⋅ x ) \phi^\mathrm{S}(\boldsymbol{x}, t)=\int \frac{\mathrm{d}^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\boldsymbol{p}}}}\left(a_{\boldsymbol{p}} \mathrm{e}^{\mathrm{i} \boldsymbol{p} \cdot \boldsymbol{x}}+a_{\boldsymbol{p}}^{\dagger} \mathrm{e}^{-\mathrm{i}\boldsymbol{p} \cdot \boldsymbol{x}}\right)
ϕ S ( x , t ) = ∫ ( 2 π ) 3 d 3 p 2 E p 1 ( a p e i p ⋅ x + a p † e − i p ⋅ x )
The Hamiltonian of the system, however, is always time independent
H = ∫ d 3 p ( 2 π ) 3 E p a p † a p H = \int \frac{\mathrm{d}^3 p}{(2 \pi)^3} E_\boldsymbol{p} a_\boldsymbol{p}^\dagger a_\boldsymbol{p}
H = ∫ ( 2 π ) 3 d 3 p E p a p † a p
in whichever picture.
For a free massless vector field, the Lagrangian is
L = − 1 4 F μ ν F μ ν \mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}
L = − 4 1 F μ ν F μ ν
where F μ ν ≡ ∂ μ A ν − ∂ ν A μ F^{\mu \nu} \equiv \partial^\mu A^\nu-\partial^\nu A^\mu F μ ν ≡ ∂ μ A ν − ∂ ν A μ
[ A μ ( x , t ) , π ν ( y , t ) ] = i δ μ ν δ ( 3 ) ( x − y ) , [ A μ ( x , t ) , A ν ( y , t ) ] = [ π μ ( x , t ) , π ν ( y , t ) ] = 0 \left[A^\mu(\boldsymbol{x}, t), \pi_\nu(\boldsymbol{y}, t)\right]=\mathrm{i} \delta^\mu{ }_\nu \delta^{(3)}(\boldsymbol{x}-\boldsymbol{y}), \quad\left[A^\mu(\boldsymbol{x}, t), A^\nu(\boldsymbol{y}, t)\right]=\left[\pi_\mu(\boldsymbol{x}, t), \pi_\nu(\boldsymbol{y}, t)\right]=0
[ A μ ( x , t ) , π ν ( y , t ) ] = i δ μ ν δ ( 3 ) ( x − y ) , [ A μ ( x , t ) , A ν ( y , t ) ] = [ π μ ( x , t ) , π ν ( y , t ) ] = 0
thus we obtain
A μ ( x , t ) = ∫ d 3 p ( 2 π ) 3 1 2 E p ∑ σ = 0 3 e μ ( p , σ ) ( b p , σ e − i p ⋅ x + b p , σ † e i p ⋅ x ) A^\mu(\boldsymbol{x}, t)=\int \frac{\mathrm{d}^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_{\boldsymbol{p}}}} \sum_{\sigma=0}^3 e^\mu(\boldsymbol{p}, \sigma)\left(b_{\boldsymbol{p}, \sigma} \mathrm{e}^{-\mathrm{i} p \cdot x}+b_{\boldsymbol{p}, \sigma}^{\dagger} \mathrm{e}^{\mathrm{i} p \cdot x}\right)
A μ ( x , t ) = ∫ ( 2 π ) 3 d 3 p 2 E p 1 σ = 0 ∑ 3 e μ ( p , σ ) ( b p , σ e − i p ⋅ x + b p , σ † e i p ⋅ x )
in which e μ ( p , σ ) e^\mu(\boldsymbol{p}, \sigma) e μ ( p , σ ) polarization vectors , depending on the 3-momentum p \boldsymbol{p} p σ \sigma σ b p , σ , b p , σ † b_{\boldsymbol{p},\sigma}, b_{\boldsymbol{p}, \sigma}^{\dagger} b p , σ , b p , σ † p \boldsymbol{p} p σ \sigma σ
In cavity quantum electrodynamics (cavity QED), quantization is performed in an optical cavity with certain quantization volume V V V
For polarized mode in a specific direction e j \boldsymbol{e}_j e j
A ( r , t ) = ∑ j A j ( r , t ) = ∑ j e j ℏ 2 V ε 0 ω j { a j e − i ( ω j t − k j ⋅ r ) + a j † e i ( ω j t − k j ⋅ r ) } \boldsymbol{A}(\boldsymbol{r}, t)=\sum_j \boldsymbol{A}_j(\boldsymbol{r}, t)=\sum_j \boldsymbol{e}_j \sqrt{\frac{\hbar}{2 V \varepsilon_0 \omega_j}}\left\{a_j \mathrm{e}^{-\mathrm{i}\left(\omega_j t-\boldsymbol{k}_j \cdot \boldsymbol{r}\right)}+a_j^\dagger \mathrm{e}^{\mathrm{i}\left(\omega_j t-\boldsymbol{k}_j \cdot \boldsymbol{r}\right)}\right\}
A ( r , t ) = j ∑ A j ( r , t ) = j ∑ e j 2 V ε 0 ω j ℏ { a j e − i ( ω j t − k j ⋅ r ) + a j † e i ( ω j t − k j ⋅ r ) }
The EM fields are
E ( r , t ) = ∑ j E j ( r , t ) = i ∑ j e j E j ( r ) { a j e − i ( ω j t − k j ⋅ r ) − a j † e i ( ω j t − k j ⋅ r ) } \boldsymbol{E}(\boldsymbol{r}, t)=\sum_j \boldsymbol{E}_j(\boldsymbol{r}, t)=\mathrm{i} \sum_j \boldsymbol{e}_j E_j^{(r)}\left\{a_j \mathrm{e}^{-\mathrm{i}\left(\omega_j t-\boldsymbol{k}_j \cdot \boldsymbol{r}\right)}-a_j^\dagger \mathrm{e}^{\mathrm{i}\left(\omega_j t-\boldsymbol{k}_j \cdot \boldsymbol{r}\right)}\right\}
E ( r , t ) = j ∑ E j ( r , t ) = i j ∑ e j E j ( r ) { a j e − i ( ω j t − k j ⋅ r ) − a j † e i ( ω j t − k j ⋅ r ) }
B ( r , t ) = ∑ j B j ( r , t ) = i ∑ j ( k j k j × e j ) B j ( r ) { a j e − i ( ω j t − k j ⋅ r ) − a j † e i ( ω j t − k j ⋅ r ) } \boldsymbol{B}(\boldsymbol{r}, t)=\sum_j \boldsymbol{B}_j(\boldsymbol{r}, t)=\mathrm{i} \sum_j\left(\frac{\boldsymbol{k}_j}{k_j} \times \boldsymbol{e}_j\right) B_j^{(r)} \left\{a_j \mathrm{e}^{-\mathrm{i}\left(\omega_j t-\boldsymbol{k}_j \cdot \boldsymbol{r}\right)}-a_j^\dagger \mathrm{e}^{\mathrm{i}\left(\omega_j t-\boldsymbol{k}_j \cdot \boldsymbol{r}\right)}\right\}
B ( r , t ) = j ∑ B j ( r , t ) = i j ∑ ( k j k j × e j ) B j ( r ) { a j e − i ( ω j t − k j ⋅ r ) − a j † e i ( ω j t − k j ⋅ r ) }
where E j ( r ) = ℏ ω j 2 V ε 0 E_j^{(r)} = \sqrt{\frac{\hbar \omega_j}{2V \varepsilon_0}} E j ( r ) = 2 V ε 0 ℏ ω j
H = ∑ j H j = ∑ j ℏ ω j ( a j † a j + 1 2 ) H = \sum_j H_j = \sum_j \hbar \omega_j \left(a_j^\dagger a_j +\frac{1}{2} \right)
H = j ∑ H j = j ∑ ℏ ω j ( a j † a j + 2 1 )
In Schrodinger picture, we have time independent field operators
E S ( r ) = ∑ j E j ( r ) = i ∑ j e j E j ( r ) { a j e i k j ⋅ r − a j † e − i k j ⋅ r } \boldsymbol{E}^\mathrm{S}(\boldsymbol{r})=\sum_j \boldsymbol{E}_j(\boldsymbol{r})=\mathrm{i} \sum_j \boldsymbol{e}_j E_j^{(r)}\left\{a_j \mathrm{e}^{\mathrm{i} \boldsymbol{k}_j \cdot \boldsymbol{r}}-a_j^\dagger \mathrm{e}^{-\mathrm{i} \boldsymbol{k}_j \cdot \boldsymbol{r}}\right\}
E S ( r ) = j ∑ E j ( r ) = i j ∑ e j E j ( r ) { a j e i k j ⋅ r − a j † e − i k j ⋅ r }
