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\heading{E6050: Perturbed particle in a square box} 
\auname{Elperin Ariel}


{\bf The problem:}

\Dn

A particle with no spin, of mass m, in placed in a square box x,y$\in{[-a,a]}$.
Later the particle is presented with the perturbation V=u$\delta(x)\delta(y)$.

\Dn

(1) Write the wavefunction $\psi(x,y)$ of the unperturbed ground state.

(2) Write the wavefunction of the 3 lowest states which are coupled by the perturbation to the ground state.

(3) Write the hamiltonian $H=H_0+V$ as a sum of two 4x4 matrices.

(4) Write the eigenstates and the first order eigenenergies in u.

(5) Calculate the second order energy shift for the ground state energy.



\Dn\Dn


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{\bf The solution:}

\Dn

(1) 

we have a particle in the potential:   

 \[ V(x.y)= \left\{ \begin{array}{ll}
   0 & \mbox{$x,y\in[-a,a]$}\\
 \infty & \mbox{$x,y\not\in[-a,a]$}\end{array} \right. \] 

In order to find the eigenstates, we need to solve:
\[H\psi=-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi\]

After losing the time dependence, since $\psi=\psi(x,y)e^{-i\frac{E}{\hbar}t}$ , and using separation of variables \\ $\psi(x,y)=X(x)Y(y)$ we get an infinite well in each axis.
After adding the  boundary conditions: $\psi(x,-a)=\psi(x,a)=\psi(y,-a)=\psi(y,a)=0$ we have:
 \[ X(x)= \left\{ \begin{array}{ll}
  A_n\sin(\frac{\pi n}{2a}x) & \mbox{n is even}\\
  A_n\cos(\frac{\pi n}{2a}x) & \mbox{n is odd}\end{array} \right. \] 

Since we have the same solution for Y(y), after normalizing we get:
 \[ \psi(x,y)= \left\{ \begin{array}{ll}
  \frac{1}{a}\sin(\frac{\pi n}{2a}x)\sin(\frac{\pi m}{2a}y) & \mbox{n,m=even}\\
  \frac{1}{a}\sin(\frac{\pi n}{2a}x)\cos(\frac{\pi m}{2a}y) & \mbox{n=even , m=odd}\\
  \frac{1}{a}\cos(\frac{\pi n}{2a}x)\sin(\frac{\pi m}{2a}y) & \mbox{n=odd, m=even} \\
  \frac{1}{a}\cos(\frac{\pi n}{2a}x)\cos(\frac{\pi m}{2a}y) & \mbox{n,m=odd}
\end{array} \right. \] 

with the corresponding eigenenergies:
\[ E_{n,m}=\frac{\hbar^2\pi^2}{8a^2}(n^2+m^2) \]

The ground state is $n=m=1$ , and has the wavefunction: 
\[ \psi(x,y)=\frac{1}{a}\cos(\frac{\pi x}{2a})\cos(\frac{\pi y}{2a}) \]
And energy: 
\[  E_0=\frac{\hbar^2\pi^2}{4a^2}  \]

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\Dn

(2)

The perturbation $V(x)=u\delta(x)\delta(y)$ only affects states with a wavefunction $\psi(x,y)$ \\ such that $\psi(0,0)\not=0$, otherwise
the particle already cannot be found in (0,0) and the perturbation has no effect.
This happens only for states $\psi_{nm}$ in which n,m are odd.
\\
\\
Therefore, the lowest states which are coupled by the perturbation to the ground state are $\psi_{13}$, $\psi_{31}$, $\psi_{33}$.
\\
\[ |1,3 \rangle \equiv|1 \rangle =  \frac{1}{a}\cos(\frac{\pi}{2a}x)\cos(\frac{3\pi}{2a}y) \]
\[ |3,1 \rangle \equiv|2 \rangle =  \frac{1}{a}\cos(\frac{3\pi}{2a}x)\cos(\frac{\pi}{2a}y) \]						
\[ |3,3 \rangle \equiv|3 \rangle =  \frac{1}{a}\cos(\frac{3\pi}{2a}x)\cos(\frac{3\pi}{2a}y) \]


(3)
We shall now find the perturbation matrix $V_{ij}$ in the basis $|0 \rangle,|1 \rangle,|2 \rangle,|3 \rangle$ \\
when $|0 \rangle=\psi_{11}$

\[ V_{00}=\langle 0|V|0\rangle=\frac{1}{a^2}\int_{-a}^{a} \! \int_{-a}^{a} \cos(\frac{\pi}{2a}x)\cos(\frac{\pi}{2a}y) u\delta(x)\delta(y) \cos(\frac{\pi}{2a}x)\cos(\frac{\pi}{2a}y)\,dx\,dy=\frac{u}{a^2} \]

It is obvious that  $\forall i,j \hspace{2 mm}  \langle i|V|j\rangle=\langle 0|V|0\rangle$, Therefore the representation of $V$ in our basis is as follows:

\[ V=\frac{u}{a^2}\begin{pmatrix}
  1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 \\
 1 & 1 & 1 &1
 \end{pmatrix}\] 

The unperturbed hamiltonian is diagonal, with the eigenenergies: \\
\[ H_0|0 \rangle =E_0|0 \rangle ,  H_0|1 \rangle =5E_0|1 \rangle ,  H_0|2 \rangle =5E_0|2 \rangle ,  H_0|3 \rangle =9E_0|3 \rangle \] where $E_0=\frac{\hbar^2\pi^2}{16a^2}.$ \\ \\
Hence, the perturbed hamiltonian is:

\[ H=E_0\begin{pmatrix}   1 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\0 & 0 & 5 & 0 \\0 & 0 & 0 & 9\end{pmatrix}+\frac{u}{a^2}\begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\1 & 1 & 1 &1 \end{pmatrix}\]

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

(4)
We notice that there is degeneracy in the eigenenergy $E=5E_0$ which corresponds to the two \\
different eigenstates : $|1 \rangle,|2 \rangle.$ In order to use the perturbation theory, we must first get rid of this degeneracy. We shall do this by 
the following transformation of basis: \\
\[|0 \rangle \rightarrow |0 \rangle \hspace{2 mm}
|1 \rangle \rightarrow |S \rangle \hspace{2 mm}
|2 \rangle \rightarrow |A \rangle \hspace{2 mm}
|3 \rangle \rightarrow |3 \rangle \]

where
\[
|S \rangle=\frac{1}{\sqrt{2}}(|1 \rangle+|2 \rangle) \hspace{2 mm}  and \hspace{2 mm}
|A \rangle=\frac{1}{\sqrt{2}}(|1 \rangle-|2 \rangle).
\]

The transformation matrix is:
\[ T=\begin{pmatrix}
  1 & 0 & 0 & 0 \\
  0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
  0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
  0 & 0 & 0 &1
 \end{pmatrix}\] 
The unperturbed hamiltonian $H_0$ has the same representation in the new basis, hence, in the new basis we have:
\[H_{new}=H_0+T^{-1}VT=E_0\begin{pmatrix}   1 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\0 & 0 & 5 & 0 \\0 & 0 & 0 & 9\end{pmatrix}+\frac{u}{a^2}\begin{pmatrix}   1 & \sqrt{2} & 0 & 1 \\ \sqrt{2} & 2 & 0 & \sqrt{2} \\0 & 0 & 0 & 0 \\1 & \sqrt{2} & 0 & 1\end{pmatrix} \]

We notice that the perturbation $V$ does not couple eigenstates with the same eigenenergey anymore, therefore we can now use the approximation:
\[E_n=E_n^{(0)}+\lambda V_{nn}+\lambda^2\sum_{k\not=n}\frac{|V_{nk}|^2}{E_n^{(0)}-E_k^{(0)}} \]
\[ |n \rangle=|n^{(0)} \rangle+\lambda \sum_{k\not=n}\frac{V_{kn}}{E_n^{(0)}-E_k^{(0)}}|k^{(0)} \rangle \]
where
\[H_0|n^{(0)} \rangle=E_n^{(0)}|n^{(0)} \rangle \]
Using the approximation, we have the first order correction in $\frac{u}{a^2}$ to the eigenenergies:
\[ E_0=E_0^{(0)}+\frac{u}{a^2} ,\hspace{2 mm} E_S=E_S^{(0)}+\frac{2u}{a^2} ,\hspace{2 mm} E_A=E_A^{(0)},\hspace{2 mm} E_3=E_3^{(0)}+\frac{u}{a^2} \]
And the first order correction to the eigenstates:
\[|0 \rangle=|0^{(0)} \rangle+\frac{u}{a^2} \left (\frac{\sqrt{2}}{-4E_0}|S^{(0)} \rangle+\frac{1}{-8E_0}|3^{(0)}\rangle \right) =\begin{pmatrix}  1  \\  -\frac{1}{2\sqrt{2}E_0}\frac{u}{a^2} \\ 0 \\ -\frac{1}{8E_0}\frac{u}{a^2} \end{pmatrix} \]
\[|S \rangle=|S^{(0)} \rangle+\frac{u}{a^2} \left (\frac{\sqrt{2}}{4E_0}|0^{(0)} \rangle+\frac{\sqrt{2}}{-4E_0}|3^{(0)}\rangle \right) =\begin{pmatrix}   \frac{1}{2\sqrt{2}E_0}\frac{u}{a^2}  \\ 1 \\ 0 \\ -\frac{1}{2\sqrt{2}E_0}\frac{u}{a^2} \end{pmatrix} \]
\[|A \rangle=|A^{(0)} \rangle=\begin{pmatrix}  0  \\ 0 \\ 1 \\ 0 \end{pmatrix} \]
\[|3 \rangle=|3^{(0)} \rangle+\frac{u}{a^2} \left (\frac{1}{8E_0}|0^{(0)} \rangle+\frac{\sqrt{2}}{4E_0}|S^{(0)}\rangle \right) =\begin{pmatrix}  \frac{1}{8E_0}\frac{u}{a^2}  \\ \frac{1}{2\sqrt{2}E_0}\frac{u}{a^2} \\ 0 \\ 1 \end{pmatrix} \]

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\Dn

(5)
The correction of the ground state energy up to the second order is:
\[ E_0=E_0^{(0)}+\frac{u}{a^2}+\frac{u^2}{a^4} \left( \frac{2}{-4E_0}+\frac{1}{-8E_0}   \right) \]


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