# Relationship between noise spectral density and quantum absorption/emission: I don't understand the derivation

In this paper, on page 57 he performs a calculation that I don't understand. Maybe there is an implicit physical assumption behind so this is why I am asking it here.

## The context (not necessary to understand the question)

I give a little bit of context: the goal is to find how noise in a driving field can affect a qubit evolution. I assume the noise being stationary.

Assuming a Hamiltonian:

$$H=-\frac{\hbar \omega_0}{2}\sigma_z+A \sigma_x F(t)$$

With $$F(t)$$ a classical field, the goal is to find the probability to find the qubit in the excited state after a time $$t$$, assuming it was initially in the ground state.

Going in the interacting picture, we have at first order perturbative expansion.

$$|\psi(t)\rangle \approx |\psi(0)\rangle-\frac{1}{i \hbar} \int_0^t dt' \ V_I(t') |\psi(0)\rangle$$

With $$V_I(t')=e^{\frac{i}{\hbar}H_0 t} \sigma_x e^{-\frac{i}{\hbar}H_0 t}$$, $$H_0 \equiv -\frac{\hbar \omega_0}{2}\sigma_z$$

We find that for a state initially in $$|\psi(0)\rangle=|g\rangle$$ (the ground state), the probability to find it excited is $$p_e(t)=|c_e(t)|^2$$, where:

$$c_e(t)=\frac{A}{i \hbar} \int_0^t dt' e^{-i \omega_0 t'} F(t')$$

It gives:

$$p_e(t)=\frac{A^2}{\hbar^2} \int_0^t dt_1 \int_0^t dt_2 e^{-i \omega_0(t_1-t_2)} F(t_1) F(t_2)$$

## My question

Calling $$S_{FF}$$ the noise spectral density in the fluctuating variable $$F$$:

$$S_{FF}(\omega) = \int_{-\infty}^{+\infty} d \tau e^{i \omega \tau} \langle F(\tau) F(0) \rangle$$

In the document, he manages to prove that:

$$\langle p_e(t) \rangle=t \frac{A^2}{\hbar^2} S_{FF}(-\omega_0)$$

I don't understand how he arrives at this formula.

## My attempt:

I start from $$p_e(t)=\frac{A^2}{\hbar^2} \int_0^t dt_1 \int_0^t dt_2 e^{-i \omega_0(t_1-t_2)} F(t_1) F(t_2)$$. I perform a change of variable: $$t_2 \to u \equiv t_1-t_2$$:

$$\langle p_e(t) \rangle=\frac{A^2}{\hbar^2} \int_0^t dt_1 \int_0^t dt_2 e^{-i \omega_0(t_1-t_2)} \langle F(t_1) F(t_2) \rangle =\frac{A^2}{\hbar^2} \int_0^{t_1} dt_1 \int_{t_1-t}^t du e^{-i \omega_0 u} \langle F(u) F(0) \rangle$$

I used the stationnary property of the noise ($$\langle F(a)F(b) \rangle = \langle F(a-b)F(0)\rangle$$)

I can use the inverse Fourier transform:

$$\langle F(u) F(0) \rangle=\int_{-\infty}^{+\infty} d \omega e^{-i \omega u} S_{FF}(\omega)$$

But replacing it, it makes appear some $$\sin$$ function when integrating and I think it is not the good way to proceed (I get stuck). Indeed:

$$\langle p_e(t) \rangle=\frac{A^2}{\hbar^2} \int_0^{t_1} dt_1 \int_{t_1-t}^t du e^{-i \omega_0 u} \langle F(u) F(0) \rangle=\frac{A^2}{\hbar^2} \int_0^{t_1} dt_1 \int_{t_1-t}^t du \int_{-\infty}^{+\infty} \frac{d \omega}{2 \pi} e^{-i (\omega_0+\omega) u} S_{FF}(\omega)$$

If I perform the integration on $$u$$, $$\sin$$ function appears from $$e^{-i (\omega_0+\omega) u}$$.

How is it possible to show his result? Are there some implicit assumption he uses ?