- #1
damarkk
- 7
- 2
- 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.
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.
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?
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?
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