Applying Selection Rules to Determine Non-Zero Ground State Perturbations

[Diracobama2181]

Homework Statement

Consider a rigid rotator (i.e. a bar shaped system of fixed separation) of moment of inertia I about an axis through its center perpendicular to the direction of the bar, with Hamiltonian $$H_0 = \frac{L_2}{2I}$$ and electric dipole moment d. Suppose that while it is in its ground state

it is subjected to a perturbation $$V (t) = −d · E(t)$$ due to a time-dependent external electric field

$$E(t) = zˆE0e^{t/τ}$$ which points in the z-direction and which is switched on at time t = 0.

Here E0 is a time-independent constant. Determine to which of its

excited states the rotator can make transitions in lowest order in V (t) ,

and calculate the transition probabilities for finding the rotator in each

of these states at time t → ∞.

Relevant Equations

$$d_f=\frac{i}{\hbar}\int_{0}^{T'} e^{iw_{fi}t}v_{ni} dx$$

Since E_i=0 for the ground state, and $$E_f=\frac{(\hbar)^2l(l+1)}{2I}$$, $$w_{fi}=\frac{E_f-E_i}{\hbar}=\frac{(\hbar)l(l+1)}{2I}$$.
So, $$d_f(\infty)=\frac{i}{\hbar}\int_{-\infty}^{\infty}<f|E_od_z|0>e^{\frac{i\hbar l(l+1)t}{2I}+\frac{t}{\tau}}dt$$

My question is in regards to $$<f|E_0d_z|0>$$. Does d_z have parity? Also, how can I apply the selection rules to determine which eigenstates (ie, spherical harmonics) will not give $$<f|E_0d_z|0>=0$$? Also, is my integral set up correctly, because it seems like it would diverge.

 

[DrClaude]

The trick is to convert $\hat{z}$ into a spherical harmonic, such that $\langle f | \hat{z} | i \rangle$ becomes the integral of a product of 3 spherical harmonics, for which there are closed formulas.

http://mathworld.wolfram.com/SphericalHarmonic.html

 

[DrClaude]

So, $$d_f(\infty)=\frac{i}{\hbar}\int_{-\infty}^{\infty}<f|E_od_z|0>e^{\frac{i\hbar l(l+1)t}{2I}+\frac{t}{\tau}}dt$$

Careful with the limits of integration here.

 

[Diracobama2181]

But wouldn't the dot product in V(t) get rid of the $$\hat Z$$?

 

[Diracobama2181]

Ok, made some changes. I first noted that $$\overrightarrow{d}=q\overrightarrow{d}$$.
So $$V(t)=E_0e^{\frac{t}{\tau}}qz$$. From here, I note $$z=dcos\theta$$, and hence $$Z=r2(\frac{\pi}{3})^(\frac{1}{2})Y_{l=1,m=0}$$. Therefore, $$ <l',m'|r2(\frac{\pi}{3})^(\frac{1}{2})Y_{l=1,m=0}|l=0.\,m=0>=0$$ unless l'=1 and m'=0. Seem right? The $$<l'=1,m'=0|r2(\frac{\pi}{3})^(\frac{1}{2})Y_{l=1,m=0}|l=0,m=0>=r2(\frac{\pi}{3})^(\frac{1}{2})$$ From there, I would integrate $$d(t)=\frac{i}{\hbar}\int_{0}^{t}qE_0r2(\frac{\pi}{2})^{\frac{1}{3}}e^{\frac{i\hbar t}{I}}e^{-\frac{t}{T}}dt$$