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

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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.

Failed to parse (unknown function "\lefteqn"): \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^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}