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\heading{E732: Potential well with delta function} 
\auname{Yonatan Gilon}


{\bf The problem:}

\Dn

A particle is placed in a one dimensional potential well with boundary conditions equal to zero at $x=\pm\frac{L}{2}$.

The potential is $ V(x,t)= \epsilon e^{-\frac{1}{2} (\frac{t}{\tau})^2}\delta(x) $ 

At a time $t\ll -\tau$ the particle was at the lowest energy state.
\Dn
 
 1. Using the perturbation theory (up to the first order) calculate the probability of finding the particle at energy state ${n}$ at a time $\tau \ll t$
 
 2. Define the condition where the calculation is valid in the last problem (careful).
 
 3. Define the condition where the effect of the pulse will be adiabatic.
 
\Dn\Dn


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{\bf The solution:}

\Dn

1. First, we separate the potential to $W=\delta(x)\ $ and  $f(t) =\epsilon e^{-\frac{1}{2} (\frac{t}{\tau})^2}\ $.


We will use the formula: 

\[
P_t(n|m) \approx  |W_{nm}|^2|FT(f(t))|^2
\]


We are given that the particle starts at energy level ${n=1}$ so we need to find $W_{n,1}$

\[
W_{n,1} = \langle n | W | 1 \rangle = \int^{\frac{L}{2}}_{\frac{-L}{2}} \Psi  ^{n*}(x) W \Psi ^1 (x)dx
\]

We remember the wave function of a particle inside a potential well
  
\[
\Psi ^n(x) = \sqrt{\frac{2}{L}} \cos(\frac{n\pi x}{L}) \\ $for n=1,3,5,...$
\]

\[
\Psi ^n(x) = \sqrt{\frac{2}{L}} \sin (\frac{n\pi x}{L}) \\ $for n=2,4,6,...$
\]

Therefore the ground state is
\[
\Psi ^1(x) = \sqrt{\frac{2}{L}} \cos(\frac{\pi x}{L})
\]
And for $W_{n,1}$
\[
\langle \Psi ^n | W | \Psi ^1 \rangle = \int^{\frac{L}{2}}_{\frac{-L}{2}} \Psi  ^{n*}(x) W \Psi ^1 (x)dx 
\]
Due to the delta function all the Sin states equal zero.

And all the Cos state will equal $\frac{2}{L}$.

\[
W_{n,1}=\frac{2}{L} \\ $for n=1,3,5,...$
\]
\[
W_{n,1}=0 \\ $    for n=2,4,6,...$
\]

And for $f(t)$
\[
FT[\epsilon e^{-\frac{1}{2} (\frac{t}{\tau})^2}]=\sqrt{2\pi} \tau \epsilon e^{-2\pi ^2 \tau^2 \omega^2}
\]
We define that
\[
\omega = E_{n} - E_{1}
\]
\[
E_{n} = \pi ^2 n^2/2mL^2 
\]
\[
E_{1}=\pi^2 \ / 2mL^2
\] 
\[ 
P_t(n,1) = (\frac{2\epsilon}{L})^2 2\pi \tau ^2 Exp \left( -\frac{4\pi ^4 \tau ^2 (n^2-1)}{2mL^2} \right)  \ \ \ \ \ \ $ for n=1,3,5,...$
\]
\[  
P_t(n,1) = 0 \ \ \ \ \  \ \  \ \ \ \ \ \  \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \  \  \ \ \ \ \  \ \  \   \ \  \    $for n=2,4,6,...$
\] 
 

2. The condition is $ |W|\ll \Delta $ ,and from that we derive $ \epsilon \ll \frac{1}{ML} $ (it might be that the actual condition, which is $p\ll 1$ where $p$ is the total transition probability, is weaker).\\ \\

3. The condition is:

\[
\left|\frac{df}{dt}\right|\ll \frac{\Delta^2}{|W|}
\]

and we define that:

\[
\left|\frac{df}{dt}\right| \approx \frac{\epsilon}{\tau}
\]

Therefore:

\[ 
\frac{1}{L}\frac{\epsilon}{\tau}\ll {\left(\frac{1}{mL^2}\right)}^2
\]
  

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