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\heading{E621: The linear Stark effect.} 
\auname{Ilya Bukchin}


{\bf The problem:}

\Dn

A Hydrogen atom is positioned in a constant electric field $\mathcal{E}_{0}\hat{z}$.

As a result , the degeneracy at the first exited energy level (four dimentional sub-space) is removed.

What are the Energy levels and the Eigenvalues if we ignore the other energy levels?

Write an Expression for the first order of the perturbing Hamiltonian.

\Dn\Dn


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{\bf The solution:}

\Dn
First we define the Hamiltonian for the Hydrogen atom,
A Hydrogen atom is a particle of charge $e$ and reduced mass $M$ that is positioned in a potential: $V_0=-\frac{\alpha}{r} $.
After solving for each $l$ get :
  $E_{\nu ,l,m}=- \frac{\alpha^2 M}{2(l+\nu )^2}$.
  
The Hamiltonain that describes the first exited level $(n = l+\nu =2)$ 
In the basis : 
\[
  |n,l,m \rangle \rightarrow |200\rangle , |210\rangle , |211\rangle, |21-1\rangle
\]
Is:
\[
  \mathcal{H}_0^{(n=2)}=-\frac{\alpha^2 M}{8}\left( \begin{array}{cccc}
                                         1 & 0 & 0 & 0 \\
                                         0 & 1 & 0 & 0 \\
                                         0 & 0 & 1 & 0 \\
                                         0 & 0 & 0 & 1 \end{array} \right)
\]



The perturbing potential for this problem is : $ V=-e\mathcal{E}_{0}z=-e\mathcal{E}_{0}r\cos{\theta} $

Therefore, the new Hamiltonian is :
\begin{equation}
  \mathcal{H}=\mathcal{H}_{0}-e\mathcal{E}_{0}r\cos{\theta}
\end{equation}

Now we will calculate the perturping potential:
\begin{equation}
  V_{n',l',m',n,l,m}= -e\mathcal{E}_{0}\langle n'l'm'|r\cos{\theta}|nlm\rangle=-e\mathcal{E}_{0}\int_{0}^{\infty}R^{*}_{n',l'}r^3 R_{n,l}dr \int^{4 \pi }_0 Y^{*}_{l'm'}Y_{lm}  cos( \theta ) d\Omega
\end{equation}

Where:
\begin{equation}
  \langle r \theta \phi | n l m \rangle = R_{n,l} Y_{l,m} 
\end{equation}
And in paticular :
\begin{equation}
  n=2
\end{equation}
\begin{equation}
  R_{20}=\left( \frac{M\alpha}{2}\right)^{\frac{3}{2}} \frac{rM \alpha }{\sqrt{3}} e^{-\frac{rM \alpha }{2}}\\
\end{equation}
\begin{equation}
  R_{21}=\left( \frac{M\alpha}{2}\right)^{\frac{3}{2}} \left( 2 - rM \alpha \right) e^{-\frac{rM \alpha }{2}}\\
\end{equation}
\begin{equation}
  Y_{00}=\sqrt{\frac{1}{4\pi}}\\
\end{equation}
\begin{equation}
  Y_{10}=\sqrt{\frac{3}{4\pi}}\cos{\theta}
\end{equation}

From now on We will omit $n$.
\vspace{10 mm}

Now we can write the perturbing electric field as follows : $ V=-\sqrt{\frac{4 \pi }{3}} e\mathcal{E}_0 Y_{10} $ .

(substituting $cos(\theta)$ for $\sqrt{\frac{4 \pi }{3}}Y_{10} $).

Then , equation (2) becomes :  
\begin{equation}
  V_{l',m',l,m}=-\sqrt{\frac{4 \pi }{3}}e\mathcal{E}_{0}\int_{0}^{\infty}R^{*}_{2,l'}r^3 R_{2,l}dr \int^{4 \pi }_0 Y^{*}_{l'm'}Y_{lm} Y_{10} d\Omega
\end{equation}

And then , from parity considerations we see that
the only case in which the expression is nonzero is when the function is odd,

meaning :  $l'\not= l$  and $m'=m=0$

and because for $m=0 $ , $ Y_{lm}$ and $R_{nl}$ are Real , 
the expression is identical for $l=1$ and $l=0$

Then :
subsituting $Y_{00}$ for $\frac{1}{\sqrt{4\pi }}$
and recalling the normlization of $Y_{10}$
we get right away the radial integral: 
\begin{equation}
  V_{1,0,0,0}=V_{0,0,1,0}=-\frac{e\mathcal{E}_0 (M \alpha )^4}{24} \int^{\infty}_0 r^4 \left( 2-rM \alpha \right) e^{-rM \alpha}
\end{equation}
\begin{equation}
  \left\{  I_n = \int^{\infty}_0 r^ne^{-r} : I_n=-nI_{n-1} : I_0=1 \right\}
\end{equation}
\begin{equation}
=...=\frac{3e\mathcal{E}_0}{M \alpha }
\end{equation}


\[
  V=\frac{3e\mathcal{E}_0}{M \alpha } \left( \begin{array}{cccc}
                   0 & 1 & 0 & 0 \\
                   1 & 0 & 0 & 0 \\
                   0 & 0 & 0 & 0 \\
                   0 & 0 & 0 & 0 \end{array} \right)
\]

Now we Diagonalize $\mathcal{V}$ in order to find the first order energy correction.
\[
  V=\frac{3e\mathcal{E}_0}{M \alpha } \left( \begin{array}{cccc}
                   1 & 0 & 0 & 0 \\
                   0 & -1 & 0 & 0 \\
                   0 & 0 & 0 & 0 \\
                   0 & 0 & 0 & 0 \end{array} \right)
\]

And the new energy levels are : 
\[
 E_1 = -\frac{\alpha^2 M}{8} + \frac{3e\mathcal{E}_0}{M \alpha }
\]
\[
 E_2 = -\frac{\alpha^2 M}{8} - \frac{3e\mathcal{E}_0}{M \alpha }
\]
\[
 E_3 = -\frac{\alpha^2 M}{8}
\]
\[
 E_4 = -\frac{\alpha^2 M}{8}
\]
With the corresponding eigenstates
\begin{equation}
  |E_1 \rangle = \frac{1}{\sqrt{2}}\left( |210 \rangle + |200 \rangle \right)
\end{equation}
\begin{equation}
  |E_2 \rangle = \frac{1}{\sqrt{2}}\left( |210 \rangle - |200 \rangle \right)
\end{equation}
\begin{equation}
  |E_3 \rangle = |211 \rangle
\end{equation}
\begin{equation}
  |E_4 \rangle = |21-1 \rangle
\end{equation}

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\end{document} 
