# Introduction to Mathematical Physics/Quantum mechanics/Linear response in quantum mechanics

Let $\mathrel{<} A\mathrel{>} (t)$ be the average of operator (observable) $A$. This average is accessible to the experimentator (see ([#References|references])). The case where $H(t)$ is proportional to $sin(\omega t)$ is treated in ([#References|references]) Case where $H(t)$ is proportional to $\delta(t)$ is treated here. Consider following problem:

Problem:

Find $\psi$ such that:

$i\hbar \frac{d\psi}{dt}=(H_0+W_i(t))\psi$

with

$W_i(t)=W_i^c.\delta(t)$

and evaluate:

$\mathrel{<} qZ\mathrel{>} = \mathrel{<} \psi|qZ|\psi\mathrel{>}$

Remark: Linear response can be described in the classical frame where Schr\"odinger equation is replaced by a classical mechanics evolution equation. Such models exist to describe for instance electric or magnetic susceptibility.

Using the interaction representation\footnote{ This change of representation is equivalent to a WKB method. Indeed, $\tilde{\psi(t)}$ becomes a slowly varying function of $t$ since temporal dependence is absorbed by operator $e^{\frac{iH_0t}{\hbar}}$}

$\tilde{\psi(t)}=e^{\frac{iH_0t}{\hbar}}\psi(t)$

and

$\tilde{W}_i(t)=e^{\frac{iH_0t}{\hbar}}W_ie^{\frac{-iH_0t}{\hbar}}$

Quantity $\mathrel{<} qZ\mathrel{>}$ to be evaluated is:

$\mathrel{<} qZ\mathrel{>} = \mathrel{<} \tilde{\psi}|q\tilde{Z}|\tilde{\psi}\mathrel{>}$

$i\hbar \frac{d\tilde{\psi}}{dt}=\tilde{W}_i(t)\tilde{\psi}$

At zeroth order:

$\frac{d\tilde{\psi}}{dt}=0$

Thus:

$\tilde{\psi}^0(t)=\tilde{\psi}^0(0)$

Now, $\tilde{\psi}$ has been prepared in the state $\psi_0$, so:

$\tilde{\psi}^0(t)=\psi_0(t) {{IMP/label|pert1}}$

At first order:

$\tilde{\psi}^1(t)=\tilde{\psi}^1(0)+\frac{1}{i\hbar}\int_0^t\tilde{W}_i(t\prime)\tilde{\psi}^0(t\prime)dt\prime$

thus, using properties of $\delta$ Dirac distribution:

$\tilde{\psi}^1(t)=\frac{1}{i\hbar}W^i_c\psi_0. {{IMP/label|pert2}}$

Let us now calculate the average: Up to first order,

$\begin{matrix} \mathrel{<} qZ\mathrel{>} &=& \mathrel{<} \tilde{\psi}^0+\tilde{\psi^1}|e^{\frac{iH_0t}{\hbar}} qZe^{\frac{iH_0t}{\hbar}}|\tilde{\psi}^0+\tilde{\psi^1}\mathrel{>} \\ &=& \mathrel{<} \tilde{\psi}^0| e^{\frac{iH_0t}{\hbar}}qZe^{\frac{iH_0t}{\hbar}}|\tilde{\psi}^1\mathrel{>} + \mathrel{<} \tilde{\psi}^1|e^{\frac{iH_0t}{\hbar}} qZe^{\frac{iH_0t}{\hbar}}|\tilde{\psi}^0\mathrel{>} \end{matrix}$

Indeed, $\mathrel{<} \tilde{\psi}^0|qZ|\tilde{\psi}^0\mathrel{>}$ is zero because $Z$ is an odd operator.

$matrix} \lefteqn{ \mathrel{<} \tilde{\psi}^0|e^{\frac{iH_0t}{\hbar}}qZ e^{\frac{iH_0t}{\hbar}}|\tilde{\psi}^1\mathrel{>} =}\\ &=& \mathrel{<} {\tilde{\psi}}^0| e^{\frac{iH_0t}{\hbar}}qZe^{\frac{-iH_0t}{\hbar}}|{\psi}_k\mathrel{>} \mathrel{<} {\psi}_k|{\tilde{\psi}}^1\mathrel{>} \end{matrix$

where, closure relation has been used. Using perturbation results given by equation ---pert1--- and equation ---pert2---:

$\mathrel{<} \tilde{\psi}^0|qZ|\tilde{\psi}^1\mathrel{>} =e^{i\omega_{0k}t} \mathrel{<} {\psi}^0|qZ|{\psi}^k\mathrel{>} \frac{1}{i\hbar} \mathrel{<} {\psi}^k|W^c_i|{\psi}^0\mathrel{>}$

We have thus:

$\begin{matrix} \mathrel{<} qZ\mathrel{>} (t)&=& 0 \mbox{ if } t < 0\\ \mathrel{<} qZ\mathrel{>} (t)&=& e^{i\omega_{0k}t} \mathrel{<} {\psi}^0|qZ|{\psi}^k\mathrel{>} \frac{1}{i\hbar} \mathrel{<} {\psi}^k|W^c_i|{\psi}^0\mathrel{>} +CC \mbox{ if not } \end{matrix}$

Using Fourier transform\footnote{ Fourier transform of:

$f(t)=e^{-i\omega_0t}$

and Fourier transform of:

$g(t)=H(t)e^{-i\omega_0t}$

are different: Fourier transform of $f(t)$ does not exist! (see ([#References|references])) }

$\mathrel{<} qZ\mathrel{>} (\omega)=2q^2E\sum_{k\neq 0}\omega_{0k}\frac{| \mathrel{<} \psi_0|Z|\psi_k\mathrel{>} |^2}{\omega_{k0}^2-\omega^2}$