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\exnumber{7122}
\heading{Transition between energy levels due to wall displacement} 
\auname{Michael Pukshanski, Eyal Shoham, Lior Weitzhandler}


{\bf The problem:}

\Dn

A particle with mass M is set in one dimensional box $x \in [0,a]$ with length $a$. At time $t=-\infty$ we prepare the particle at ground state $n=1$ of the box. Displacement of the left wall is described by the step function $x=\epsilon (t)$.\\
The velocity in which we move the wall is $\dot{\epsilon}=\frac{\epsilon_0}{\sqrt{2\pi}\tau_0} exp[-\frac{1}{2}(\frac{t}{\tau_0})^2]$.\\
We measure the energy level of the particle at time $t=\infty$.

\Dn

(1) Calculate the perturbation element $\dot{\epsilon}W_{n.1}$ for transitions of adiabatic states.

(2) Write the condition for the adiabatic approximation validity.

(3) Assume a small displacement $\epsilon_0<<a$ of the wall. Calculate the first order, using the perturbation theory (in the adiabatic basis), the probability $P(n)$ to find the particle in the exited level $n$.

\Dn

We change the protocol as follows:
\\
- First we move the left wall small displacement $\epsilon_0<<a$.
\\
- We move the right wall in perfectly adiabatically formation to the point $x=b$.
\\
- The displacement of the right wall is described by $x=a+v_0t$ so the total time for the process is $t_0=(b-a)/v_0$.
\\
- Finally we return the left wall to it's initial place.

\Dn

(4) Find the probability $P(n)$ for the new protocol.

\Dn
Guidance: the perturbation element for infinitesimal displacement of the wall is 
\\
$<\varphi|V|\psi>=-(\frac{\epsilon}{2M})[\frac{\partial\varphi}{\partial x}][\frac{\partial\psi}{\partial x}]$.
\\
The partial derivative are calculated in the point the wall is placed.\\
Express the results with the given parameters $(M,a,b,v_0,\epsilon_0,\tau_0)$.\\
Write the result of section 3 in the form $P(n)=|A(a)|^2$.\\
Express the result of section 4 with the function $A(\cdot)$ and the given parameters. 


\Dn\Dn


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{\bf The solution:}

\Dn

(1) According to time dependent perturbation theory we can write, in the adiabatic basis, the perturbation element as follows: 
\[
W_{nm}=i<n|\frac{\partial m}{\partial x}>=\frac{-iV_{nm}}{E_n-E_m}
\]
We can use the equation in the guidance:
\[
V_{nm}=\frac{1}{2M}(\frac{\partial \varphi_n}{\partial x})|_{x=0}(\frac{\partial\psi_m}{\partial x})|_{x=0}
\]
And for a particle in a one-dimensional box:
\[
\varphi_n=\sqrt{\frac{2}{a}}\cdot sin(\frac{\pi n x}{a})
\]
With the energies:
\[
E_n=\frac{\pi^2 n^2}{2 M a^2}
\]
Hence:
\[
(\frac{\partial \varphi}{\partial x})|_{x=0}=\sqrt{\frac{2}{a}}\cdot \frac{\pi\cdot n}{a}
\]
Therefore one gets:
\[
V_{nm}=\frac{ \pi^2\cdot n m}{Ma^3}
\]
\[
W_{nm}=\frac{-i \pi^2 m n}{Ma^3 \cdot  (E_n-E_m)}
\]
And we get the result:
\[
W_{n1}= -\frac{2i \cdot n}{a(n^2-1)}
\]

%%%%%%%%%%%%%%%
\Dn

(2) The adiabatic condition is given by:
\[
\dot X<<\frac{\Delta^2}{|V|}
\]
One can observe that the energy level spacing in one dimensional box is given by:
\[
\Delta = \frac{\pi \cdot  v_E}{L}
\]
For a wall displacement we have:
\[
|V|=\frac{M v_E^2}{L}
\]
Then:
\[
\dot{X}<<\frac{\pi^2}{ML}
\]


%%%%%%%%%%%%%%
\Dn

(3) The leading order of the transition probability from the ground state to state $n$ is given by:
\[
P_t(n|1)=|W_{n1}|^2 \times |FT[\dot{\epsilon}]|^2
\]
Which results in:
\[
P(n)=(\frac{2n}{a(n^2-1)})^2\epsilon_0^2e^{-(w_n\tau_0)^2}
\]
Defining:
\[
w_n \equiv E_n-E_1
\]
%%%%%%%%%%%%%
\Dn

(4) The leading order of the non-adiabatic transition probability is:
\[
P_t (n|1)=|\int_{0}^{t} \frac{V_{n1}}{E_n-E_1}  e^{-i(\Phi_n-\Phi_1)} \dot{X}dt|^2
\]
Where:
\[
\Phi_n(t)=\int_{0}^{t}E_ndt'
\]
We notice that $\dot{X}$ isn\textsc{\char13}t zero only when the right wall moves, so we can divide the integral into two integrals as follows:
\[
P_t (n|1)=|\int_{part 1} W_{n1}  e^{-i(\Phi_{n}^{(1)})} \dot{X} dt' - \int_{part 2} W_{n1}  e^{-i(\Phi_{n}^{(2)})} \dot{X}dt'|^2
\]
The velocity in the second part is in the opposite direction which results in a minus sign. Also, $\omega_n$ depends on $X$ so the phase will be different for each integral. 
The phase integral limits will be from $0$ to the current time and will be divided into 3 intervals:
left wall movement, adiabatic right wall movement and left wall movement.\\
The expression for the frequency is: 
\[
\omega_{n}(X)=\frac{\pi^2 (n^2-1)}{2 M X^2}
\]
We can see that for the left integral the phase will be:
\[
\Phi_{n}^{(1)}=\int_{0}^{t}\omega_{n}(a)dt=\omega_{n}(a)t
\]
The second phase includes the phase accumulated during the adiabatic process.
\[
\Phi_{n}^{(2)}=\int_{0}^{t}\omega_{n}(a)dt+\int_{adiabatic}\omega_{n}(X)dt
\]
Calculating the adiabatic part as follows:
\[
\Phi_{adiabatic}=\int_{a}^{b}\frac{\pi^2 (n^2-1)}{2 M X^2} \frac{dt}{dX} dX=\frac{1}{v_0} \frac{\pi^2 (n^2-1)}{2M} (\frac{1}{a}-\frac{1}{b})
\] 
Finally, the probability is:
\[
P_t(n|1)=|A(a)-e^{-i\Phi_{adiabatic}} A(b)|^2
\]
Defining:
\[
A(x)\equiv2 \frac{\epsilon_0}{x} (\frac{n}{n^2-1}) e^{-\frac{1}{2} (\tau_0\omega_n(x))^2}
\]
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