# 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 $e^{\frac {iH_{0}t}{\hbar }}$ }

${\tilde {\psi (t)}}=e^{\frac {iH_{0}t}{\hbar }}\psi (t)$ and

${\tilde {W}}_{i}(t)=e^{\frac {iH_{0}t}{\hbar }}W_{i}e^{\frac {-iH_{0}t}{\hbar }}$ 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:

${\tilde {\psi }}^{1}(t)={\frac {1}{i\hbar }}W_{c}^{i}\psi _{0}.{IMP/label|pert2}$ Let us now calculate the average: Up to first order,

${\begin{matrix}{\mathrel {<}}qZ{\mathrel {>}}&=&{\mathrel {<}}{\tilde {\psi }}^{0}+{\tilde {\psi ^{1}}}|e^{\frac {iH_{0}t}{\hbar }}qZe^{\frac {iH_{0}t}{\hbar }}|{\tilde {\psi }}^{0}+{\tilde {\psi ^{1}}}{\mathrel {>}}\\&=&{\mathrel {<}}{\tilde {\psi }}^{0}|e^{\frac {iH_{0}t}{\hbar }}qZe^{\frac {iH_{0}t}{\hbar }}|{\tilde {\psi }}^{1}{\mathrel {>}}+{\mathrel {<}}{\tilde {\psi }}^{1}|e^{\frac {iH_{0}t}{\hbar }}qZe^{\frac {iH_{0}t}{\hbar }}|{\tilde {\psi }}^{0}{\mathrel {>}}\end{matrix}}$ Indeed, ${\mathrel {<}}{\tilde {\psi }}^{0}|qZ|{\tilde {\psi }}^{0}{\mathrel {>}}$ is zero because $Z$ is an odd operator.

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

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

${\mathrel {<}}{\tilde {\psi }}^{0}|qZ|{\tilde {\psi }}^{1}{\mathrel {>}}=e^{i\omega _{0k}t}{\mathrel {<}}{\psi }^{0}|qZ|{\psi }^{k}{\mathrel {>}}{\frac {1}{i\hbar }}{\mathrel {<}}{\psi }^{k}|W_{i}^{c}|{\psi }^{0}{\mathrel {>}}$ We have thus:

${\begin{matrix}{\mathrel {<}}qZ{\mathrel {>}}(t)&=&0{\mbox{ if }}t<0\\{\mathrel {<}}qZ{\mathrel {>}}(t)&=&e^{i\omega _{0k}t}{\mathrel {<}}{\psi }^{0}|qZ|{\psi }^{k}{\mathrel {>}}{\frac {1}{i\hbar }}{\mathrel {<}}{\psi }^{k}|W_{i}^{c}|{\psi }^{0}{\mathrel {>}}+CC{\mbox{ if not }}\end{matrix}}$ Using Fourier transform\footnote{ Fourier transform of:

$f(t)=e^{-i\omega _{0}t}$ and Fourier transform of:

$g(t)=H(t)e^{-i\omega _{0}t}$ are different: Fourier transform of $f(t)$ does not exist! (see ([#References|references])) }

${\mathrel {<}}qZ{\mathrel {>}}(\omega )=2q^{2}E\sum _{k\neq 0}\omega _{0k}{\frac {|{\mathrel {<}}\psi _{0}|Z|\psi _{k}{\mathrel {>}}|^{2}}{\omega _{k0}^{2}-\omega ^{2}}}$ 