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\heading{E263: Second order perturbation of an atom in electric field } 
\auname{Etai Rosenkrantz}


{\bf The problem:}

\Dn

(1) Explain why, in a generic way, an electron eiganfunction, in an atom, has a defined parity.

 How does the above explains that in an electric field the first order correction to the energy is zero.


An atom can be described as a two level system with level spacing $ \Delta $.

(2) Write the Hamiltonian of an atom in electric field.

 Use $ \epsilon $ for the perturbation strenght.
 What is the connection of $ \epsilon $ with the strenght of the electric field ?
 		
(3) Using the perturbation theory, find the second order correction to the energy of the atom above.

 Compare the solution to the exact solution from the diagonalization of the Hamiltonian.
 
Be sure you understand the following:
An electric field strenght can change in different places in space. That is why an atom will feel an effective potential 
$ V(R)\propto |\epsilon(R)|^2 $
This potential explains the energy drop caused by the atom's polarization in electric field.

(4) Define and calculate the polarization using the perturbation theory. 

\Dn\Dn


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{\bf The solution:}

\Dn

(1) Elecron in an atom has $r \longrightarrow -r$ symetry and therefore has definite parity.

Consequently $ \langle x \rangle = \langle y \rangle = \langle z \rangle $, since $  V(x) = -e \mathcal{E} \hat{x}$,
it follows that $ \langle V(x) \rangle  = 0 $,  that is why the first order correction is zero.


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\Dn

(2)
The hamiltonian is:
\[
H = H_0 + V 
\]
Where: $ V = - e \mathcal{E} \hat{x} $  and  $ \hat{x} = \frac{a}{2} \sigma_{z} $ .
\[
H = \left( \amatrix { \frac{ea}{2} \mathcal {E} & c \cr c & -\frac{ea}{2} \mathcal {E} } \right) 
\]
Hence we see that
\[ \Delta = 2c \]
\[ \epsilon = e \mathcal{E}a \]
In the $H_0$ Basis we have:
\[
\mathcal {H} = \frac {\Delta}{2}\left( \amatrix { -1 & 0 \cr 0 & 1 } \right ) + \frac {ea}{2} \mathcal{E} \left( \amatrix { 0 & 1 \cr 1 & 0} \right ) = -\frac{\Delta}{2} \sigma_z + \frac{ea \mathcal{E}}{2} \sigma_x
\]
(3)
By using:
\[ E_0 = E^{[0]}_0 + V_{nn} + \sum_{p\neq n} \frac{|V_{np}|^2}{E^{[0]}_n - E^{[0]}_p}
\]
We obtain the atom's energy, in the ground state, up to the second order:
\[
E^{[0]}_{0}= -\frac {\Delta}{2} \] \[ E^{[1]}_{0} = 0 \] \[ E^{[2]}_{0} = -\frac {|ea \mathcal{E}|^2}{4 \Delta} \]
So we can write:
\[
E_{0} = E^{[0]}_{0} + E^{[1]}_{0} + E^{[2]}_{0}= -\frac{\Delta}{2} - \frac {|ea \mathcal{E}|^2}{4 \Delta}
\]

To compare the result with the exact solution, we first diagonalize the Hamiltonian: 
\[
\mathcal{H}_{diag} = \left( \amatrix { -\sqrt{\left(\frac{\Delta}{2}\right)^2+\left(\frac{ea\mathcal{E}}{2}\right)^2} & 0 \cr 0 & \sqrt{\left(\frac{\Delta}{2}\right)^2+\left(\frac{ea\mathcal{E}}{2}\right)^2} } \right )
\]
From here the exact solution for the ground state:
\[ E_{exact} = -\sqrt{\left(\frac{\Delta}{2}\right)^2+\left(\frac{ea\mathcal{E}}{2}\right)^2}
\]
By Tylor expending $E_{exact}$ we obtain the same result as the perturbe energy:
\[ E_{exact} = -\frac{\Delta}{2} - \frac {|ea \mathcal{E}|^2}{4 \Delta}
\]
(4)
Let us assume that the electron is in the $ |1 \rangle $ state, and now calculate the wave function corrections using:
\[
|\psi_n \rangle = |\phi_n \rangle + \sum_{p\neq n} \frac{\langle \phi_p|V|\phi_n \rangle}{E^{[0]}_n - E^{[0]}_p} |\phi_p \rangle
\]
We obtain:
\[
|\psi \rangle = |1^{[0]} \rangle + \frac {ea \mathcal {E}}{2 \Delta} |2^{[0]}\rangle
\]
The polarization is given by the expectation value of $ \hat{x} $ (in the $ H_{0} $ basis):
\[
\langle \hat{x} \rangle = \langle \psi| \hat{x} | \psi \rangle = \frac{ea \mathcal{E}}{2 \Delta} 
\]
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