3.1.5 Electromagnetic field

As example for the response function of the bath, we give the result for the blackbody radiation (Maxwell field)

$$\Gamma\_{\vec{\eta}}(\boldsymbol{\alpha}) = \int\_0^\infty d\tau e^{\mathbf{i}(\boldsymbol{\alpha} + \boldsymbol{i}\cdot\mathbf{r})\cdot\mathbf{r}} \langle E\_i(\boldsymbol{\tau}) E\_j(\mathbf{0}) \rangle\_\mathbf{B} = \delta\_{\vec{\eta}} \left( \frac{\mathbf{1}}{2} \boldsymbol{\gamma}(\boldsymbol{\alpha}) + \mathbf{i} \mathbf{S}(\boldsymbol{\alpha}) \right) \tag{68}$$

with

Nonequilibrium Statistical Operator DOI: http://dx.doi.org/10.5772/intechopen.84707

∂ ∂t

ρSðÞ�t

is easily proven,

d dt <sup>a</sup>† � �<sup>t</sup>

h i <sup>n</sup> <sup>t</sup> <sup>¼</sup> <sup>a</sup>† h i <sup>a</sup> <sup>t</sup> <sup>¼</sup> pnð Þ<sup>t</sup>

with the solution

<sup>¼</sup> <sup>a</sup>† a � �t<sup>0</sup> e

the initial distribution <sup>a</sup>† h i <sup>a</sup> <sup>t</sup><sup>0</sup>

blackbody radiation (Maxwell field)

ð<sup>∞</sup> 0 dτe

3.1.5 Electromagnetic field

Γijð Þ¼ ω

with

18

a† a � �<sup>t</sup> 1

Non-Equilibrium Particle Dynamics

<sup>¼</sup> γ ωð Þ<sup>0</sup> <sup>a</sup>ρSð Þ<sup>t</sup> <sup>a</sup>† � <sup>1</sup>

are immediately calculated as

¼ TrS

with the renormalized frequency ω<sup>0</sup>

∂ ∂t

a† � �<sup>t</sup>

Similar expressions are obtained for h i <sup>a</sup> <sup>t</sup>

d dt <sup>a</sup>† a � �<sup>t</sup>

<sup>ρ</sup>Sð Þ<sup>t</sup> <sup>a</sup>† � �

<sup>¼</sup> <sup>a</sup>† � �<sup>t</sup><sup>0</sup>

�½ � γ ωð Þ�<sup>0</sup> <sup>γ</sup>ð Þ �ω<sup>0</sup> ð Þ <sup>t</sup>�t<sup>0</sup> <sup>þ</sup> <sup>γ</sup>ð Þ �ω<sup>0</sup>

γð Þ �ω<sup>0</sup> γ ωð Þ� <sup>0</sup> γ ωð Þ<sup>0</sup>

.

<sup>i</sup><sup>ℏ</sup> <sup>H</sup>S; <sup>ρ</sup><sup>S</sup> ½ �� ð Þ<sup>t</sup>

1

<sup>2</sup> <sup>a</sup>† <sup>a</sup>; <sup>ρ</sup>Sð Þ<sup>t</sup> � � � �

<sup>i</sup><sup>ℏ</sup> ½ðSð Þ <sup>ω</sup><sup>0</sup> <sup>a</sup>†

<sup>a</sup> <sup>þ</sup> <sup>S</sup>ð Þ �ω<sup>0</sup> aa†

The curly brackets in the dissipator denote the anticommutator. There are eight additional terms containing aa or a†a†. In interaction picture, they are

γð Þ¼ �ω<sup>0</sup> γ ωð Þ<sup>0</sup> e

¼ iω<sup>0</sup>

e

proportional to e�2iω0ð Þ <sup>t</sup>�t<sup>0</sup> and are dropped within the rotating wave approximation. For a bath in thermal equilibrium, using eigenstates the detailed balance relation

The evolution equations for the averages <sup>a</sup>† h i<sup>t</sup> <sup>¼</sup> TrS <sup>ρ</sup>Sa† f g, a† h i <sup>a</sup> <sup>t</sup> <sup>¼</sup> TrS <sup>ρ</sup>Sa† f g<sup>a</sup>

γ ωð Þ� <sup>0</sup> γð Þ �ω<sup>0</sup>

The asymptotic behavior t � t<sup>0</sup> ! ∞ is determined by the properties of the bath,

the system relaxes to the thermal equilibrium distribution that is independent on

<sup>¼</sup> <sup>1</sup>

As example for the response function of the bath, we give the result for the

<sup>B</sup> ¼ δij

1 2

<sup>i</sup>ð Þ <sup>ω</sup>þi<sup>ϵ</sup> <sup>τ</sup> Eið Þ<sup>τ</sup> Ejð Þ <sup>0</sup> � �

<sup>þ</sup> <sup>γ</sup>ð Þ �ω<sup>0</sup> <sup>a</sup>†ρSð Þ<sup>t</sup> <sup>a</sup> � <sup>1</sup>

, ρSð Þ� t

½ � γ ωð Þ� <sup>0</sup> γð Þ �ω<sup>0</sup> � �

<sup>¼</sup> <sup>γ</sup>ð Þ� �ω<sup>0</sup> ½ � γ ωð Þ� <sup>0</sup> <sup>γ</sup>ð Þ �ω<sup>0</sup> <sup>a</sup>†<sup>a</sup> � �<sup>t</sup> (65)

1 � e

<sup>e</sup>�ℏω0=kB<sup>T</sup> � <sup>1</sup> <sup>¼</sup> <sup>n</sup>Bð Þ <sup>ω</sup><sup>0</sup> , (67)

γ ωð Þþ iSð Þ ω � �

<sup>2</sup> aa† ; <sup>ρ</sup>Sð Þ<sup>t</sup> � � � �

�ℏω0=kBT: (62)

<sup>0</sup> ¼ ω<sup>0</sup> þ ½ � Sð Þþ ω<sup>0</sup> Sð Þ �ω<sup>0</sup> =ℏ. The solution is

. We find for the occupation number

�½ � γ ωð Þ�<sup>0</sup> γð Þ �ω<sup>0</sup> ð Þ t�t<sup>0</sup> h i

: (66)

(68)

½ � <sup>i</sup>ω00�γ ωð Þ <sup>0</sup> <sup>=</sup>2þγð Þ �ω<sup>0</sup> <sup>=</sup><sup>2</sup> ð Þ <sup>t</sup>�t<sup>0</sup> : (64)

:

a† � �<sup>t</sup> (63)

(61)

$$\gamma(\boldsymbol{\omega}) = \frac{4\alpha^3}{3\hbar c^3} [\mathbf{1} + n\_\mathbf{B}(\boldsymbol{\alpha})], \quad \mathbf{S}(\boldsymbol{\alpha}) = \frac{2}{3\pi\hbar c^3} \boldsymbol{P} \int\_0^\infty d\boldsymbol{\alpha}\_k \boldsymbol{\alpha}\_k^3 \left[ \frac{\mathbf{1} + n\_\mathbf{B}(\boldsymbol{\alpha}\_k)}{\boldsymbol{\alpha} - \boldsymbol{\alpha}\_k} + \frac{n\_\mathbf{B}(\boldsymbol{\alpha}\_k)}{\boldsymbol{\alpha} + \boldsymbol{\alpha}\_k} \right]. \tag{69}$$

Note that the Planck distribution satisfies nBð Þ¼� �ω ½ � 1 þ nBð Þ ω such that γ ωð Þ¼ <sup>4</sup>ω3½ � <sup>1</sup> <sup>þ</sup> <sup>n</sup>Bð Þ <sup>ω</sup> <sup>=</sup> <sup>3</sup>ℏc<sup>3</sup> ð Þ for <sup>ω</sup>>0 and γ ωð Þ¼ <sup>4</sup>j j <sup>ω</sup> <sup>3</sup> <sup>n</sup>Bð Þ <sup>j</sup>ω<sup>j</sup> <sup>=</sup> <sup>3</sup>ℏc<sup>3</sup> ð Þ for <sup>ω</sup>< 0.

The resulting quantum optical master equation which, e.g., describes the coupling of atoms to the radiation field Hint ¼ �D � E in dipole approximation,

$$\frac{\partial}{\partial t}\rho\_{\mathcal{S}}(t) - \frac{1}{\mathbf{i}\hbar}[H\_{\mathcal{S}},\rho\_{\mathcal{S}}(t)] - \frac{1}{\mathbf{i}\hbar}[H\_{\text{inf}},\rho\_{\mathcal{S}}(t)] = \mathcal{D}'[\rho\_{\mathcal{S}}(t)],\tag{70}$$

has the Lindblad form. The influence Hamiltonian

<sup>H</sup>infl <sup>¼</sup> <sup>Ð</sup> <sup>d</sup>ωℏSð Þ <sup>ω</sup> <sup>D</sup>† ð Þ� ω Dð Þ ω leads to a renormalization of the system Hamiltonian H<sup>S</sup> that is induced by the vacuum fluctuations of the radiation field (Lamb shift) and by the thermally induced processes (Stark shift). The dissipator of the quantum master equation reads

$$\begin{split} \mathcal{D}'[\rho\_{\mathcal{S}}(t)] &= \int\_{0}^{\infty} d\boldsymbol{\alpha} \frac{4\boldsymbol{\alpha}^{3}}{3\hbar c^{3}} [\boldsymbol{\mathbbm{1}} + \boldsymbol{n}\_{\mathcal{B}}(\boldsymbol{\omega})] \left[ \mathbf{D}(\boldsymbol{\alpha}) \rho\_{\mathcal{S}}(t) \mathbf{D}^{\dagger}(\boldsymbol{\omega}) - \frac{1}{2} \left\{ \mathbf{D}^{\dagger}(\boldsymbol{\alpha}) \mathbf{D}(\boldsymbol{\omega}), \rho\_{\mathcal{S}}(t) \right\} \right] \\ &+ \int\_{0}^{\infty} d\boldsymbol{\alpha} \frac{4\boldsymbol{\alpha}^{3}}{3\hbar c^{3}} \boldsymbol{n}\_{\mathcal{B}}(\boldsymbol{\omega}) \left[ \mathbf{D}^{\dagger}(\boldsymbol{\alpha}) \rho\_{\mathcal{S}}(t) \mathbf{D}(\boldsymbol{\omega}) - \frac{1}{2} \left\{ \mathbf{D}(\boldsymbol{\omega}) \mathbf{D}^{\dagger}(\boldsymbol{\omega}), \rho\_{\mathcal{S}}(t) \right\} \right], \end{split} \tag{71}$$

where the integral over the negative frequencies has been transformed into positive frequencies. This result can be interpreted in a simple way. The application of the destruction operator Dð Þ ω on a state of the system lowers its energy by the amount ℏω and describes the emission of a photon. The transition rate 4ω<sup>3</sup> <sup>3</sup>ℏc<sup>3</sup> ½ � 1 þ nBð Þ ω contains the spontaneous emission as well as the thermal emission of photons. The term D† ð Þ ω gives the creation of excitations with transition rate 4ω<sup>3</sup> <sup>3</sup>ℏc<sup>3</sup> nBð Þ ω describing the absorption of photons.
