Hydrogen Atom in an electric field along ##z##

[damarkk]

Homework Statement

Hydrogen Atom, expectation value of $L^2$, $L_z$ Perturbation Theory.

Relevant Equations

$\langle |L^2|\rangle$, $\langle |L_z| \rangle$

There is an hydrogen atom on a electric field along $z$ $E_z= E_{0z}$ .

Consider only the states for $n=2$. Solving the Saecular matrix for find the correction to first order for the energy and the correction to zero order for the states, we have:

$| \Psi_{211} \rangle$, $| \Psi_{21-1} \rangle$ for $E^1_{2} = 0$ that has degeneration equal to two, and

$\frac{1}{\sqrt{2}}(| \Psi_{200} \rangle ) +|\Psi_{210} \rangle)$ for $E^1_{2} = W$

$\frac{1}{\sqrt{2}}(| \Psi_{200} \rangle )- |\Psi_{210} \rangle)$ for $E^1_{2} = -W$

Where $W$ is the value of $\langle \Psi_{200}| H'|\Psi_{210} \rangle$, $H'=-eEz$.

Now the statement is: consider the state of the system at $t=0$ equal to $|\Psi (0) \rangle = |\Psi_{210} \rangle$. Find the state at $t>0$ and calculate $\langle L^2 \rangle$, $\langle L_z \rangle$.


MY ATTEMPT


I find that $|\Psi(t) \rangle = e^{-iE_2t/\hbar}(\cos{\omega t} |\Psi_{200} \rangle -i\sin{\omega t}|\Psi_{210} \rangle)$.

Now I compute $\langle L_z \rangle$ like that. The state of the physical system is described by eigenstates of $L_z$ and $L^2$ because I know that $L_z |\Psi_{nlm} \rangle = m\hbar |\Psi_{nlm} \rangle$ and $L^2 |\Psi_{nlm} \rangle = \hbar^2 l(l+1) |\Psi_{nlm} \rangle$.

If this is correct, then $\langle | L_z |\rangle = 0$ and $\langle | L^2 |\rangle = 2\hbar^2 \sin{\omega t}$ where $\omega = \frac{W}{\hbar}$.

Is this a correct answer and method?


I have also another question.

Spoiler: Spoiler

Why $\langle L_z \rangle = 0$? Because the electric field is along $z$, and the electron move along z axis: it doens't have any angular momentum component along z.

Why $\langle L^2 \rangle \neq 0$? On the other hand, electron oscillates perpendicular to x and y axis and there is a component of angular momentum along it. But my question is: what is the variance of $L_z$? In fact, $\sigma^2 = \langle L_z^2 \rangle - \langle L_z \rangle ^2$, but if $L_z |\Psi_{nlm} \rangle= m\hbar |\Psi_{nlm} \rangle$, then $L_z^2 |\Psi_{nlm} \rangle= m^2\hbar^2 |\Psi_{nlm} \rangle$ and is equal to zero because all the states before examinated ($|\Psi_{210} \rangle, |\Psi_{200} \rangle$) has $m=0$. But if this is true, I can say correctly that $\langle | L^2 |\rangle = \langle | L^2_x |\rangle+\langle | L^2_y |\rangle$ and this last two are equal?