Rates and Energy Shift (RWA)

Let us have a closer look at the expressions

$$\hat{C}_{12}(z)=\int_0^{\infty} d\omega {\rho(\omega) [1+n_B(\omega)}] \int_{0}^{\infty}dt e^{-(z+i\omega)t}.$$ (41)

The Laplace transform exists for Im$(z)>0$ to ensure convergence of the integral, but in the expressions above we need $\hat{C}_{12}(z=-i\Omega)$ etc., i.e. purely imaginary arguments! The limit $t\to \infty$, if explicitely written, reads

$$\hat{C}_{12}(z=-i\Omega)=\lim_{t\to \infty}\int_0^{\infty} d\omega {\rho(\omega) [1+n_B(\omega)}] \int_{0}^{t}dt' e^{i(\Omega-\omega)t'}.$$ (42)

Now,

$$\lim_{t\to \infty} \int_{0}^{t}dt' e^{i x t'}=\lim_{t\to\infty} \... ...t}{x} + i \frac{1-\cos xt}{x}\right]=\pi \delta(x) +iP\left(\frac{1}{x}\right),$$ (43)

where $P$ denotes the principal value.

For the first term, we used a very useful

Theorem:

For any integrable, normalised function $f(x)$ with $\int_{-\infty}^{\infty}dx f(x)=1$,

$$\lim_{\varepsilon\to 0 } \frac{1}{\varepsilon}f\left( \frac{x}{\varepsilon}\right)=\delta(x).$$ (44)

Since $\int_{-\infty}^{\infty}dx \sin(x)/x=\pi$, this yields the Delta function above.

We split the two bath correlation functions into real and imaginary parts,

$$\hat{C}_{12}(-i\Omega) \equiv \frac{1}{2}\gamma_+ + i \Delta_+,\quad \hat{C}_{21}( i\Omega)\equiv\frac{1}{2}\gamma + i \Delta$$

$$\gamma_+ \equiv \gamma_+(\Omega)\equiv 2\pi \rho(\Omega)[1+n_B(\Omega)],\quad \gamma\equiv \gamma(\Omega) \equiv 2\pi \rho(\Omega) n_B(\Omega)$$

$$\Delta_+ \equiv P\int_{0}^{\infty}\frac{d\omega}{2\pi}\frac{\gamma_+(\omega)}{\Om... ... - P\int_{0}^{\infty}\frac{d\omega}{2\pi}\frac{\gamma (\omega)}{\Omega-\omega}.$$ (45)

Remarks:

• Real and imaginary parts of the correlation functions are related to each other: Kramers-Kronig relations.
• Note the minus-sign in the definition of $\Delta$.