Does the Rigid Rotator in an Electric Field Obey Parity Selection Rules?

[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.

 

[member 553083]

Yes, the operator $d_z$ has parity. The selection rules that apply are those of angular momentum and parity. In particular, since $d_z$ is an odd parity operator, only transitions between states with different parities will give a non-zero matrix element. This means that for $<f|E_0d_z|0>$, the final state must have different parity from the initial state. The integral is set up correctly, however it is important to note that for the exponential term in the integral, $l(l+1)$ must be an integer, as it is the angular momentum quantum number. Thus, the terms that will be non-zero are those with $l=0$ or $l=2$. The integral itself should converge as long as $\tau$ is finite.