A hydrogen atom in a weak time-dependent perturbation

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Urvabara
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Homework Statement



A hydrogen atom, which is in its ground state 1s (i.e. [tex]\|1,0,0\rangle[/tex]), is put into a weak time-dependent external electric field, which points into the z direction:
[tex]\boldsymbol{E}(t,\boldsymbol{r}) = \frac{C\hat{\text{\boldsymbol{e}}}_{z}}{t^{2}+\tau^{2}}[/tex], where C and [tex]\tau > 0[/tex] are constants. This gives rise to a perturbation potential [tex]V(t) = C\frac{e\hat{z}}{t^{2}+\tau^{2}}[/tex], where e denotes the electron charge.

(a) Using lowest-order time-dependent perturbation theory, find the selection rules for transitions from the ground state, i.e. find out which final state values for the quantum numbers n, l and m are possible in transitions from the ground state.

(b) Calculate the probability of transition from the ground state 1s to to the state 2p during an infinitely long period of time, setting [tex]t_{0}\rightarrow -\infty[/tex] and [tex]t\rightarrow\infty[/tex].

Homework Equations


This is useful, isn't it?
[tex]P_{fi}(t,t_{0})\equiv |\langle\phi_{f}|\psi(t)\rangle|^{2}\approx \frac{1}{\hbar^{2}}\left|\int_{t_{0}}^{t}\text{d}t_{1}\langle\phi_{f}|V_{S}(t_{1})|\phi_{i}\rangle \text{e}^{\text{i}(E_{f}-E_{i})t_{1}/\hbar}\right|^{2}.[/tex]


The Attempt at a Solution



(b) I probably should use the equation given in the section "2. Homework Equations " and "sandwich" perturbation potential between the states [tex]\langle 2,1,m|[/tex] and [tex]|1,0,0\rangle[/tex], shouldn't I? What m-value should I use?

Any hints to (a)?
 
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Urvabara said:

Homework Statement




(b) I probably should use the equation given in the section "2. Homework Equations " and "sandwich" perturbation potential between the states [tex]\langle 2,1,m|[/tex] and [tex]|1,0,0\rangle[/tex], shouldn't I? What m-value should I use?

Any hints to (a)?

what kind of spherical tensor is z? What is its rank?
Then you will get your selection rules, so you know what m values you should use for the matrix elements (i.e the set if final states that can be reached)

You might want to look in the chapter of tensor operators in your textbook.


Or you can write z as a spherical harmnonics and use the wave functions for the hydrogen-states (that is a radial part times a spherical harmonic) and use some well known spherical harmonics identities.

EDIT: You might also use parity relations for the Spherical harmonics, if you are not allowed to use collection of formulas for spherical harmonics.
 
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malawi_glenn said:
How's it going?

Thanks for your reply!

I think that [tex]z = r_{0} = r\cos (\theta)[/tex]. Right?

I also think that [tex]\langle 2,1,m|r_{0}|1,0,0\rangle = \sqrt{\frac{4\pi}{3}}\langle 2,1|r_{0}|1,0\rangle\langle 1,m|Y_{10}(\theta ,\phi)|0,0\rangle[/tex] Right?

The radial part is easy to calculate: [tex]\langle 2,1|r_{0}|1,0\rangle = \int_{0}^{\infty}r^{3}R^{*}_{21}(r)R_{10}(r)\,\text{d}r = ...[/tex] Right?

The angular part is: [tex]\langle 1,m|Y_{10}(\theta ,\phi)|0,0\rangle = \sqrt{\frac{3(2*0+1)}{4\pi (2*1+1)}}\langle 0,1;0,0|1,0\rangle\langle 0,1;0,0|1,m\rangle[/tex] Right?

The C-G coefficients [tex]\langle 0,1;0,0|1,m\rangle[/tex] vanish unless [tex]m=0+0=0[/tex] and [tex]0-1\leq 0 \leq 0+1[/tex] or [tex]\Delta m = m-0 = 1,0,-1[/tex] and [tex]\Delta l = 1-0=1[/tex] Hmm. Right?

PS. Sorry, because my reply is so late. I was somewhat busy.
 
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Urvabara said:
Thanks for your reply!

I think that [tex]z = r_{0} = \cos (\theta)[/tex]. Right?

I also think that [tex]\langle 2,1,m|r_{0}|1,0,0\rangle = \sqrt{\frac{4\pi}{3}}\langle 2,1|r_{0}|1,0\rangle\langle 1,m|Y_{10}(\theta ,\phi)|0,0\rangle[/tex] Right?

The radial part is easy to calculate: [tex]\langle 2,1|r_{0}|1,0\rangle = \int_{0}^{\infty}r^{3}R^{*}_{21}(r)R_{10}(r)\,\text{d}r = ...[/tex] Right?

The angular part is: [tex]\langle 1,m|Y_{10}(\theta ,\phi)|0,0\rangle = \sqrt{\frac{3(2*0+1)}{4\pi (2*1+1)}}\langle 0,1;0,0|1,0\rangle\langle 0,1;0,0|1,m\rangle[/tex] Right?

The C-G coefficients [tex]\langle 0,1;0,0|1,m\rangle[/tex] vanish unless [tex]m=0+0=0[/tex] and [tex]0-1\leq 0 \leq 0+1[/tex] or [tex]\Delta m = m-0 = 1,0,-1[/tex] and [tex]\Delta l = 1-0=1[/tex] Hmm. Right?

PS. Sorry, because my reply is so late. I was somewhat busy.

You have some errors that you need to fix.

[tex]z = r\cdot \cos \theta[/tex]

and the "selection" rules from CG are:
[tex]m = m_1 + m_2[/tex]
and
[tex]|j_1 - j_2| \leq j \leq j_1 + j_2[/tex]

so z is essential [tex]r \cdot Y_{l=1}^{m=0}[/tex], which is a spherical tensor of rank (k) = 1, with q = 0
The so called m-selection rule is:
[tex]<\alpha , j_1,m_1 |T^{(k)}_{m}|\beta , j_2, m_2> = 0[/tex] ; unless m_1 = q + m_2

So now you know which final state your pertubation can take the initial state (the ground state in this case).

Then you just evaluate the integrals, you can look up explicit wave functions for the hydrogen atoms in your textbook and just do the math, straighforward but lenghty. Remember to use spherical coordinates when you integrate.
 
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