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\begin{document}
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\heading{E621: Stark effect in Hydrogen atom} 
\auname{Naama Shukrun, Ella Yonai}

{\bf The problem:}

The electron in the Hydrogen atom is a particle of mass M and charge
e with

potential of V (r) = -$\frac{r}{\alpha}$.

Assume that we can neglect the probability to find

the particle in an energy level n>2 so we have a five dimensional
space for

the particle's states, two spherical and three polaric orbitals so
the basis for

the representation is : $\left|s1\right\rangle $,$\left|s2\right\rangle $,$\left|px\right\rangle $,$\left|py\right\rangle $,$\left|pz\right\rangle $
. 

We also neglect spin orbit interaction so we can ignore the spin.

The atom is placed in constant electric field in the z direction ,

this causes the atom to polarize. 
\begin{quote}
(1) Write the Hamiltonian in the standard form using the dynamic variables:\ \ \ $\mathcal{H}\left(p_{r},L,r,\theta,\mathcal{E}\right)$

(2) Write the two matrices that represent the unperturbed 

Hamiltonian and the interaction with the electric field.

(3) Find the expression for the ground state energy $E_{gr}$$\left(\mathcal{E}\right)$
in the second order of perturbation theory.

(4) Find the expression for the first excited state energy $E_{ex}$$\left(\mathcal{E}\right)$
in the second order of perturbation theory.

(5) Write the expression for the polarization $\widetilde{P}=e\left\langle z\right\rangle $
of the atom that was prepared in the ground state.

(6) Write the expression for the polarization of the atom that was
prepared in the excited state.
\end{quote}

\textbf{$\quad$Guidance}:

$\quad$(a) The solutions should be expressed with ($M$,$a$,$e$,$c_{1}$,$c_{2}$)

$\quad$

$\quad$(b) $a=\frac{1}{\alpha M}$ the Bohr atom radius, $c_{1}=\frac{8\sqrt{2}}{9},$$c_{2}=3\sqrt{3}$
constants

$\quad$that represent the solution of the radial integral. 

$\quad$

$\quad$(c) The radial functions $\frac{1}{a^{3/2}}R^{l\nu}$are:
$R^{01}=2e^{-x}$, $R^{02}$ $=\sqrt{\frac{1}{6}}(\frac{x}{2})e^{-\frac{x}{2}}$,

$\quad$$R^{11}=\sqrt{\frac{1}{2}}(1-\frac{x}{2})e^{-\frac{x}{2}}$
. 

$\quad$

$\quad$(d) Solution will include c1, c2 and other factors. 

$\quad$

$\quad$(e) Solution to clauses 5-6 may be derived without 

$\quad$complicated calculations. 

$\quad$




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

$\quad$

(1) In the $\left|\upsilon lm\right\rangle $basis the Hamiltonian
is :

\[
\mathcal{H}=\frac{P_{r}^{2}}{2M}+\frac{L^{2}}{2mr^{2}}-\frac{\alpha}{r}-e\mathcal{E}rcos\theta\]


when the last term stands for the perturbation caused by the electric
field

in the z direction.

$\quad$

(2) The Eigenvalues for the unperturbed Hydrogen atom are:

\[
E_{\upsilon lm}=\frac{1}{2Ma^{2}\left(l+\upsilon\right)^{2}}\]


\[
\mathcal{H}=\mathcal{H}_{0}+V\]


The unperturbed Hamiltonian:

\[
\mathcal{H}_{0}=-\frac{1}{2Ma^{2}}\left(\begin{array}{ccccc}
1\\
 & \frac{1}{4}\\
 &  & \frac{1}{4}\\
 &  &  & \frac{1}{4}\\
 &  &  &  & \frac{1}{4}\end{array}\right)\]


the interaction with the electric field:

\[
V=-e\mathcal{E}\left\langle \nu^{'}l^{'}m^{'}\mid cos\theta\mid\upsilon lm\right\rangle =-e\mathcal{E}\intop_{0}^{\infty}R^{l'\nu'*}r^{3}R^{l\nu}dr\intop_{0}^{4\pi}Y_{l'm'}^{*}Y_{lm}cos\theta d\Omega\]


\[
cos\theta=\sqrt{\frac{4\pi}{3}}Y_{10}\]


\[
V=-\sqrt{\frac{4\pi}{3}}e\mathcal{E}\intop_{0}^{\infty}R^{l'\nu'*}r^{3}R^{l\nu}dr\intop_{0}^{4\pi}Y_{l'm'}^{*}Y_{lm}Y_{10}d\Omega\]


from parity considerations we see that the only case in which the
expression

is nonzero is when the function is odd, meaning $l'\neq l$ and $m'=m=0$.

\[
-a\left\langle s1\mid\mathcal{E}e\frac{r}{a}cos\theta\mid pz\right\rangle =-\sqrt{\frac{4\pi}{3}}\mathcal{E}ea\int R^{01*}x^{3}R^{11}dx\intop_{0}^{4\pi}Y_{00}^{*}Y_{10}Y_{10}d\Omega=\frac{-\mathcal{E}ea}{\sqrt{3}}c_{1}\]


\[
-a\left\langle s2\mid\mathcal{E}e\frac{r}{a}cos\theta\mid pz\right\rangle =\frac{-\mathcal{E}ea}{\sqrt{3}}c_{2}\]


\[
V=-\frac{\mathcal{E}ea}{\sqrt{3}}\left(\begin{array}{ccccc}
0 & 0 & 0 & 0 & c_{1}\\
0 & 0 & 0 & 0 & c_{2}\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
c_{1} & c_{2} & 0 & 0 & 0\end{array}\right)\]


$\quad$

(3)\,The expression for the ground state energy with second order
correction:

\[
E_{gr}\left(\mathcal{E}\right)=E_{10}-\frac{\left|\left\langle s1\mid V\mid pz\right\rangle \right|^{2}}{\left|E_{10}-E_{11}\right|}=-\frac{1}{2Ma^{2}}-\frac{8c_{1}^{2}e^{2}\mathcal{E^{\textrm{2}}}Ma^{4}}{9}\]


$\quad$

(4)\,The expression for the lowest excited state energy will be obtained 

after using degenerate perturbation theory:

\[
\left(\begin{array}{cc}
-\frac{1}{8Ma^{2}} & -\frac{\mathcal{E}ea}{\sqrt{3}}c_{2}\\
-\frac{\mathcal{E}ea}{\sqrt{3}}c_{2} & -\frac{1}{8Ma^{2}}\end{array}\right)\]


\[
E_{ex}\left(\mathcal{\mathcal{E}}\right)=-\frac{1}{8Ma^{2}}-\frac{\mathcal{E}ea}{\sqrt{3}}c_{2}\]


$\quad$

(5)\,The polarization can be simply obtain from the concept of 

generalized forces: $\mathcal{F=\textrm{-}}\frac{\partial\mathcal{\mathbb{\mathcal{H}}}}{\partial X}$

\[
\mathcal{\left\langle \widetilde{P_{\textrm{gr}}}\right\rangle =\textrm{-}}\frac{\partial E_{gr}}{\partial\mathcal{E}}=\frac{16c_{1}^{2}e^{2}\mathcal{E}Ma^{4}}{9}\]


$\quad$

(6)\ For the first excited state due to degenerate perturbation theory
the

energy has different dependence on the electric field:

\[
\mathcal{\left\langle \widetilde{P_{\textrm{ex}}}\right\rangle =\textrm{-}}\frac{\partial E_{ex}}{\partial\mathcal{E}}=\frac{ea}{\sqrt{3}}c_{2}\]


Thus the polarization for this state will be constant.
\end{document}
