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

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

Problem:

Find ${\displaystyle \psi }$ such that:

${\displaystyle i\hbar {\frac {d\psi }{dt}}=(H_{0}+W_{i}(t))\psi }$

with

${\displaystyle W_{i}(t)=W_{i}^{c}.\delta (t)}$

and evaluate:

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

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, ${\displaystyle {\tilde {\psi (t)}}}$ becomes a slowly varying function of ${\displaystyle t}$ since temporal dependence is absorbed by operator ${\displaystyle e^{\frac {iH_{0}t}{\hbar }}}$}

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

and

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

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

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

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

At zeroth order:

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

Thus:

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

Now, ${\displaystyle {\tilde {\psi }}}$ has been prepared in the state ${\displaystyle \psi _{0}}$, so:

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

At first order:

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

thus, using properties of ${\displaystyle \delta }$ Dirac distribution:

${\displaystyle {\tilde {\psi }}^{1}(t)={\frac {1}{i\hbar }}W_{c}^{i}\psi _{0}.{IMP/label|pert2}}$

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

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

Indeed, ${\displaystyle {\mathrel {<}}{\tilde {\psi }}^{0}|qZ|{\tilde {\psi }}^{0}{\mathrel {>}}}$ is zero because ${\displaystyle Z}$ 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---:

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

We have thus:

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

Using Fourier transform\footnote{ Fourier transform of:

${\displaystyle f(t)=e^{-i\omega _{0}t}}$

and Fourier transform of:

${\displaystyle g(t)=H(t)e^{-i\omega _{0}t}}$

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

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