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\heading{E6050: Pertrubation in a symmetric square box} 
\auname{Guy Nir \& Rotem Tzuk}


{\bf The problem:}

\Dn

A particle without spin, with mass $m$, is placed in a square box $x,y\in[-a,a]$. We then add a pertrubation of the form $V=u \delta (x)\delta (y)$
\Dn

(1) Write the Wavefunctions for the unpertrubed particle's ground state. 

(2) Also write the first three eigenstates that are coupled to the ground state. 

(3) Write the Hamiltonian as a sum of two $4\times 4$ matrices: $H=H_0+V$ 

(4) Write the eigenstates (as column vectors) and the eigen energies in the first order of $u$.

(5) Calculate the second order correction to the energy of the ground state. 


\Dn\Dn


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{\bf The solution:}

\Dn

 
The Hamiltonian without the pertrubation is:

\[
\displaystyle i\frac{\partial}{\partial t}\Psi=-\frac{\hbar^2}{2m}\nabla ^2\Psi+V\Psi \rightarrow V=\left \{ \amatrix { 0 & x,y\in[-a,a]\cr \infty & otherwise}\right
\]

\Dn

As usual the solution is time dependante according to $\Psi(x,y,t)=\psi (x,y) \eexp {i\frac E {\hbar} t}$ and we must now solve the eigenvector problem in the space coordinates only. We can also assume an independant solution for $x,y \rightarrow\psi(x,y)=\chi(x)\eta(y)$ when the obvious solutions are $\sin$ and $\cos$ functions. 

\Dn

So now we can calculate the solutions, taking into account the boundry conditions of the box: 

\[
\chi(x)=N_x \sin(k_x x)\mid_{x=\pm a}=0\hspace{30pt} \displaystyle \int_{-a}^a N_x \sin^2(k_x x)dx=1 \rightarrow N_x=\sqrt{\frac 1 a}
\]
The same goes for cosine solutions. The results for y are the same as for x (for obvious reasons). We now have a full discription of all the eigenfuntions of this Hamiltonian:

$$\displaystyle \psi(x,y) =\left\{ \begin{array}{lr} 
\frac 1 a sin(\frac{\pi n}{2a}x) sin (\frac{\pi m}{2a}y) & for\ even\ n,m \\
\frac 1 a sin(\frac{\pi n}{2a}x) cos (\frac{\pi m}{2a}y) & for\ even\ n,\ odd\ m \\
\frac 1 a cos(\frac{\pi n}{2a}x) sin (\frac{\pi m}{2a}y) & for\ odd\ n,\ even\ m \\
\frac 1 a cos(\frac{\pi n}{2a}x) cos (\frac{\pi m}{2a}y) & for\ odd\ n,m \\
\end{array}
\right
$$
\Dn
with the energy given by substitution into the Hamiltonian: $$E_{n,m}\psi(x,y)=-\frac{\hbar^2}{2m}\nabla^2 \psi(x,y)=\frac{\hbar^2 \pi^2}{8 m a^2}(n^2+m^2)$$
\Dn

(1) The ground state wavefunction is then $\frac 1 a \cos(\frac{\pi}{2 a}x)\cos(\frac {\pi}{2 a}y) $ for $n=m=1$ and the energy $E_{1,1}=\frac{\hbar^2 \pi^2}{4 m a^2} \equiv E_0$

\Dn
\newpage
(2) The only wavefunctions that are in any way affected by the $\delta(x)\delta(y)$ scatterer are those that are not zero at $x,y=0$, which means only the cosine functions. so we get $n=1,3$ and $m=1,3$ with all combinations allowed (notice that $n=m=1$ is the ground state). we can define:
\Dn
$$
|1>=\frac 1 a \cos(\frac{\pi }{2a}x) cos (\frac{\pi }{2a}y)
$$
$$|2>=\frac 1 a cos(\frac{3\pi}{2a}x) cos (\frac{\pi}{2a}y)$$

$$|3>=\frac 1 a cos(\frac{\pi}{2a}x) cos (\frac{3 \pi}{2a}y)$$

$$|4>=\frac 1 a cos(\frac{3\pi}{2a}x) cos (\frac{3\pi}{2a}y)$$

\Dn
The energies involved are $E_1=\frac{\hbar^2 \pi^2}{4 m a^2}\ ,\ E_2=E_3=\frac{\hbar^2 \pi^2}{8 m a^2}10\ ,\ E_4=\frac{\hbar^2 \pi^2}{8 m a^2}18$

(3) In the given basis we get the unpertrubed Hamiltonian to be:
$$
H_0=\frac{\hbar^2 \pi^2}{8 m a^2} \left( \begin{array}{cccc}
2 & 0 & 0 & 0 \\
0 & 10 & 0 &0 \\
0 & 0 & 10 &0\\
0 & 0 & 0 & 18\\
 \end{array} \right)$$
 Now we must find the pertrubing potential matrix by sandwiching the different wavefunctions of our base, according to the Hamiltonian with the $\delta (x)\delta(y) $ function:
 $$
 \displaystyle E_{n,m}=<n|E|m>=<n|H|M>=
 \int_{-a}^a \int_{-a}^a \frac 1 {a^2} \cos(\frac {\pi n}{2a}x)\cos(\frac {\pi n}{2a}y)(\frac{\partial^2}{\partial x^2}\cos(\frac {\pi m}{2a}x)\cos(\frac {\pi m}{2a}y)dxdy+
 $$ 
 $$\int_{-a}^a \int_{-a}^a \frac 1 {a^2} \cos(\frac {\pi n}{2a}x)\cos(\frac {\pi n}{2a}y)\left[ u\delta (x,y)\right] \cos(\frac {\pi m}{2a}x)\cos(\frac {\pi m}{2a}y)dxdy
 $$
 The first integral gives the energies of the unpertrubed Hamiltonian, which is the diagonal matrix $H_0$. 
 The second integral simply gives $ < n | V | m > = \frac {u} {a^2} $ for any n,m. 
 \Dn
  $$
  H=H_0+V=\frac{\hbar^2 \pi^2}{8 m a^2} \left( \begin{array}{cccc}
2 & 0 & 0 & 0 \\
0 & 10 & 0 &0 \\
0 & 0 & 10 &0\\
0 & 0 & 0 & 18\\
 \end{array} \right)
 +
 \frac{u}{a^2} \left( \begin{array}{cccc}
1 & 1 & 1 & 1 \\
1 & 1 & 1 &1 \\
1 & 1 & 1 &1\\
1 & 1 & 1 & 1\\
 \end{array} \right)$$
 
 we can also define the following constants to shorten writing: $E_0=\frac{\hbar^2 \pi^2}{4 m a^2}\hspace{15} \lambda =\frac u {a^2}$, giving the following matrix:
 \Dn
 


 $$H=H_0+V=E_0 \left( \begin{array}{cccc}
1&0&0&0\\ 
0&5&0&0\\
0&0&5&0\\
0&0&0&9\end{array} \right)
+\lambda \left( \begin {array}{cccc}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1\end{array}\right) $$
\Dn
\newpage
Notice that we have a degeneracy in the energies $E_2,E_3$. we will use the symmetric and anti symmetric solutions to avoid this problem. 
$$
|S>=\frac 1 {\sqrt 2}(|2>+|3>)\hspace{20},\hspace{20}|A>=\frac 1 {\sqrt 2} (|2>-|3>)
$$
We may find the matrix $V$ in the new basis by operating the old $V$ on the $|A>$ and $|S>$ states in the standard base. We also add the diagonal elements of $V$ to the already diagonal matrix $H_0$.We end up with the following states: $|1>,|A>,|S>,|4>$.\newline In the new base our matrices take the following form:

$$
H=H_0+V=E_0 \left( \begin {array}{cccc}
1+\frac{\lambda}{E_0} & 0 &0 &0 \\
0 & 5 & 0 &0 \\
0 & 0 & 5+2\frac{\lambda}{E_0} & 2\\
0 & 0 & 0 & 9+\frac{\lambda}{E_0}
\end{array} \right) 
+\lambda \left( \begin{array}{cccc}
0 & 0 & \sqrt2 & 1\\
0 & 0 & 0 & 0 \\
\sqrt2 & 0 & 0 & \sqrt2 \\
1 & 0 & \sqrt2 & 0 \end{array} \right)
$$
\Dn
So we can now easily find the eigenstates and energies of the first order in $\lambda$: 

$$
\Psi_n^{[0,1]}=\Psi_n ^{[0]}+\Psi_n ^{[1]}=\Psi_n ^{[0]}+\frac{V_{n,n_0}}{E_{n_0}^{[0]}-E_n^{[0]}}
$$
With the (first order) energy corrections given directly from the addition of the diagonal elements of V into $H_0$ i.e:
$$
E_1^{[0,1]}=E_0+\lambda\hspace{15},\hspace{15}E_A^{[0,1]}=5E_0 \hspace{15},\hspace{15}E_S^{[0,1]}=5E_0+2\lambda \hspace{15},\hspace{15}E_4^{[0,1]}=9E_0+\lambda$$
The eigenstates given in coloumn form in the previous base ($|1>,|A>,|S>,|4>$):
$$
|\Psi_1^{[0,1]}>=\left( \begin {array}{c} 1 \\ - \frac{\sqrt2\lambda}{4E_0-\lambda} \\ 0 \\ -\frac{\lambda}{8E_0} \end{array}\right)\hspace{15},\hspace{15}
|\Psi_A^{[0,1]}>=\left( \begin {array}{c} 0 \\ 1 \\ 0 \\ 0  \end{array}\right)

$$
|\Psi_S^{[0,1]}>=\left( \begin {array}{c}  \frac{\sqrt2\lambda}{4E_0+\lambda}\\ 0 \\ 1 \\- \frac{\sqrt2\lambda}{4E_0+\lambda} \end{array}\right)\hspace{15},\hspace{15}
|\Psi_4^{[0,1]}>=\left( \begin {array}{c} \frac{\lambda}{8E_0} \\  \frac{\sqrt2\lambda}{4E_0+\lambda} \\ 0 \\ 1 \end{array}\right)$$
\Dn 
We can see the Anti Symmetric solution is unaffected by the pertrubation. That is because the wavefunction $\frac 1{\sqrt2 a}(\cos(\frac{3\pi}{2a}x)\cos(\frac{\pi}{2a}y)- \cos(\frac{\pi}{2a}x)\cos(\frac{3\pi}{2a}y))$ is zero at $(0,0)$ and so does not feel the pertrubation. 
\newline
\Dn
(5) We can calculate the second order correction to the energy using:$\displaystyle \sum_{m(\!=n_0)}\frac{|V{n_0,m}|^2}{E_{n_0}^{[0]}-E_n ^{[0]}}$

\Dn
$$
\displaystyle E_1^{[2]}=\frac {V_{1,A}^2}{E_1^{[0]}-E_A^{[0]}}+\frac {V_{1,S}^2}{E_1^{[0]}-E_S^{[0]}}+\frac {V_{1,4}^2}{E_1^{[0]}-E_4^{[0]}}=-\frac {2\lambda^2}{4E_0+\lambda}-\frac{\lambda^2}{8E_0}$$
The first term is zero (anti symmetric solution). We can see both states $|S>$ and $|4>$ 'push' down the ground state energy (the term is always negative for ground state energies). 
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\Dn


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