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\heading{E143: 2 sites Bose-Einstein Condensation experiment} 
\auname{Guy Bareli 021356613 ; Ben Harel 031199292}


%{\bf The Problem:}

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{\bf The solution:}

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(1)  The wave function that describes the particle is the symetric eigenstate:
\[
\psi(x)=e^{(-\frac{(x+a)^2}{2\sigma^2})}+e^{(-\frac{(x-a)^2}{2\sigma^2})}
\]
In order to find the momentum distribution function, we will have to perform a Fourier Transform.\\
We know that for a Gaussian function, the F.T. is:
\[
F.T.\left[e^{-\frac{x^2}{2\sigma^2}}\right]\propto e^{-\frac{a^2k^2}{2}}
\]
Therefore:
\[
\tilde{\psi}(k)=F.T[\psi (x)] = e^{-ika}\cdot e^{-\frac{\sigma^2 k^2}{2}} + e^{+ika}\cdot e^{-\frac{\sigma^2 k^2}{2}}= 
e^{-\frac{\sigma^2 k^2}{2}}\cdot 2 cos(ka) 
\]
For the symetric wave function, the distribution function will be:
\[
P(k)=\left|\tilde{\psi}(k)\right|^2 = e^{-\sigma^2 k^2}\cdot 4cos^2(ka)
\]
(2)  The wave function describes the summation of the \\RIGHT location $|R>=e^{(-\frac{(x-a)^2}{2\sigma^2})}$  and the LEFT location  $|L>=e^{(-\frac{(x+a)^2}{2\sigma^2})}$ , \\therefore the symetric wave function can be written as $\psi(x)\equiv|S>=|L>+|R>$. \\ It is easy to see that the anti-symetric wave function, $|A>$ will be described as \\ $|A>=|L>-|R>=e^{(-\frac{(x+a)^2}{2\sigma^2})}-e^{(-\frac{(x-a)^2}{2\sigma^2})}$ \\ Hance, after a Fourier Transform, we will get:
\[
|\tilde{A}(k)>=F.T[|A>]= e^{-\frac{\sigma^2 k^2}{2}}\cdot 2 i \cdot sin(ka) 
\]
And for the anti-symetric wave function, the distribution function will be:
\[
P(k)= e^{-\sigma^2 k^2}\cdot 4sin^2(ka)
\]
(3)  For a 2-sites system with a potential $u$ and transmission amplitude $c$ we will get the Hemiltonian:
\[
\mathcal{H} = \left( \amatrix { u/2 & c    \cr
            						        c   & -u/2 \cr} \right) 
\]
The first scenario describes "sudden change" where a potential $u$ is turned on. The condition $u>>c$ allows us to rewrite the Hemiltonian as :
\[
\mathcal{H} = \left( \amatrix { u/2 & 0    \cr
            						        0   & -u/2 \cr} \right) 
\]
The energies are $\pm u/2$ and therefore we can write the time dependant wave function:
\[
\psi_t(x)=e^{-i\frac{u}{2}t}|L>+e^{+i\frac{u}{2}t}|R>
\]
After a Fourier Transform we will get:
\[
\tilde{\psi_t}(k)=[e^{-i(ka+\frac{u}{2}t)}+e^{+i(ka+\frac{u}{2}t)}]\cdot e^{-\frac{\sigma^2 k^2}{2}}=
cos(ka+\frac{u}{2}t)\cdot e^{-\frac{\sigma^2 k^2}{2}}
\]
Therefore the momentum distribution function will be:
\[
P(k)=cos^2(ka+\frac{u}{2}t)\cdot e^{-\sigma^2 k^2}
\]
(4)  In order to find the time $t$ where minimum amount of particles will stay in the system (where most of the particles will be in the anti-symetric system) we should remember that the anti-symetric momentum distribution function is proportional to $sin^2(ka)$ and the time $t$ will be the time where the function $P(k)$ changes from $cos^2(ka)$ to $sin^2(ka)$.\\
The condition is:
\[
\frac{ut}{2}=\frac{\pi}{2}+2\pi n \ \ \ \longmapsto \ \ \  t=\frac{\pi}{u} + \frac{4\pi n}{u}
\]
(5)  Scenario 2 describes an Adiabatic change where the potential $u$ is slowly turned on, and when it is much larger than the transmission amplitude $c$, $u>>c$, it is suddenly turned off. At that point $u=0$ and after time $t$ the momentum distribution is examined. \\
We will assume that the potential is set in a way where the left state $|L>$ has higher energy level than the right state $|R>$. We Know that in 2 state system, the symetric state is the Base state, therefore the lowest state. \\
During an adiabatic change, the wavefunction might change, but the energy levels don't! So in time $t=0$ (the first moment when $u=0$) the wave function is $\psi=|R>$. \\
The Hemiltonian based on the standard eigen states will be:
\[
\mathcal{H} = \left( \amatrix { 0 & c    \cr
            						        c & 0    \cr} \right) 
\]
The Hemiltonian based on the symetric and anti-symetric states will be:
\[
\mathcal{H} = \left( \amatrix { c & 0     \cr
            						        0 & -c    \cr} \right) 
\]
Now we have to take the wavefunction and find it's time evolution:
\[
\psi_t(x)=\hat{U}(t)|R> =|S>\cdot e^{-ict} -|A>\cdot e^{+ict}
\]
\[
\tilde{\psi}(k) = e^{-ict} \cdot e^{-\frac{\sigma^2 k^2}{2}}\cdot 2 cos(ka) - e^{+ict} \cdot e^{-\frac{\sigma^2 k^2}{2}}\cdot 2i \cdot sin(ka)
\]
\[
P(k)=\left|\tilde{\psi}(k)\right|^2 = 4e^{-\sigma^2 k^2}\cdot\left[ 1+sin(2ka)\cdot sin(2ct) \right]
\]
(6)  The wave function is occilating between the two optional states $|L>$, $|R>$.\\
The time $t$ when the wave function is a perfect Gaussian (and does not modulate) is the time when the wave function is exactly in one of the two states. \\
After a full time period the wavefunction returns to it's original location ($|R>$). after half a period the function will be in the other state ($|L>$).
therefore......
\[
t=\frac{\pi}{\omega}\cdot integer \ \ \rightarrow \ \ \omega = 2c \ \ \rightarrow \ \ t=\frac{\pi}{2c}\cdot integer
\]





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The problem???
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