First-order time-dependent perturbation theory on a Hydrogen atom

[boyu]

Homework Statement

A Hydrogen atom is initially in its ground state and then subject to a pulsed electric field [tex]E(t)=E_{0}\delta(t)[/tex] along the [tex]z[/tex] direction. We neglect all fine-structure and hyperfine-structure corrections.

Homework Equations

1. It is important to use selection rules to avoid unnecessary calculations of many zero transition matrix elements. According to the selection rule associated with electric dipole transitions, what is the final angular momentum quantum number [tex]l[/tex] in order to have a nonzero transition amplitude?

2. For an excited state [tex]|nlm>[/tex], its angular part [tex]Y^{m}_{l}(\theta,\phi)[/tex] will contain a term [tex]exp(im\phi)[/tex]. Based on this observation, show that the transition probability from the ground state to a final state would be zero if the quantum number [tex]m[/tex] of the final state is not zero.

3. Calculate the transition probability to an arbitrary excited state using the first-order time-dependent perturbation theory. You don't need to evaluate the matrix elements of the dipole operator explicitly.

4. Can you calculate the total transition probability (that is, the sum of transition proba-
bilities to all excited states)? Note that here we cannot use Fermi's golden rule because
the final states are a collection of discrete states.

The Attempt at a Solution

1. Selection rules: [tex]\Delta l=1[/tex] ---> [tex]l=1[/tex]

2. [tex]m=0[/tex]

3. Transition probability from first order time-dependent perturbation theory:
[tex]P_{n<-m}=|<\psi^{0}_{n}|\hat{U}(t,t_{0})|\psi^{0}_{m}>|^{2}=|<\psi^{0}_{n}|\hat{U}_{I}(t,t_{0})|\psi^{0}_{m}>|^{2}=(\frac{1}{hbar})^{2}|\int^{t}_{0}<\psi^{0}_{n}|\hat{V}(t_{1})|\psi^{0}_{m}>e^{i(E^{0}_{n}-E^{0}_{m})(t_{1}-t_{0})/hbar}dt_{1}|^{2}[/tex]

where [tex]\psi_{m}=\psi_{1}=\psi_{100}[/tex]
[tex]\psi_{n}=\psi_{n}=\psi_{n10}[/tex]
[tex]\hat{V}(t)=-q\overline{r}\cdot E_{0}\delta(t)\widehat{z}[/tex]
[tex]t_{0}=0[/tex]
[tex]\omega_{nm}=\frac{(E^{0}_{n}-E^{0}_{m})}{hbar}[/tex]

Then I have

[tex]P_{n1}=(\frac{|\overline{d}_{n1}\cdot\widehat{z}|E_{0}}{hbar})^{2}|\int^{t}_{0}\delta(t_{1})e^{i\omega_{n1}t_{1}}dt_{1}|^{2}=(\frac{|\overline{d}_{n1}\cdot\widehat{z}|E_{0}}{hbar})^{2}|e^{i\omega_{n1}t}|^{2}=(\frac{|\overline{d}_{n1}\cdot\widehat{z}|E_{0}}{hbar})^{2}[/tex]

4. Total transition probability
[tex]P_{total}=\sum^{\infty}_{n=2}P_{n<-1}=(\frac{E_{0}}{hbar})^{2}\sum^{\infty}_{n=2}|\vec{d_{n1}}\cdot \hat{z}|^{2}[/tex]

where [tex]\vec{d_{n1}}\cdot \hat{z}=q<n10|z|100>[/tex]

After this, I have to use the formula of radial wavefunction for Hydrogen atom
[tex]R_{nl}=\sqrt{(\frac{2}{na})^{3}\frac{(n-l-1)!}{2n[(n+1)!]^{3}}}e^{-r/na}(2r/na)^{l}[L^{2l+1}_{n-l-1}(2r/na)][/tex]
set [tex]l=1[/tex], and get the function of n, then plug into the formula for total transition probability which is a summation of n.

The problem is, that I can't find the formula of [tex]R_{nl}[/tex] for n=n and [tex]l=1[/tex].
Can anyone show me how to continue my derivation?

Many thanks!

 

[fzero]

In your expression

[tex]
R_{nl}=\sqrt{(\frac{2}{na})^{3}\frac{(n-l-1)!}{2n[(n+1)!]^{3}}}e^{-r/na}(2r/na)^{l}[L^{2l+1}_{n-l-1}(2r/na)]
[/tex]

you have the generalized Laguerre polynomials, which can be computed from

[tex] L^p_q(x) = \frac{ x^{-p} e^x}{q! } \frac{d^q}{dx^q}(e^{-x} x^{p+q}).[/tex]

There are more relations that you can probably find in your textbook or at http://en.wikipedia.org/wiki/Laguerre_polynomial#Generalized_Laguerre_polynomials