Rigid Rotator in an Electric Field

  • #1
Diracobama2181
74
2
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.
 

Answers and Replies

Suggested for: Rigid Rotator in an Electric Field

  • Last Post
Replies
4
Views
905
Replies
17
Views
884
Replies
5
Views
629
Replies
5
Views
853
  • Last Post
Replies
1
Views
661
Replies
4
Views
134
Replies
8
Views
1K
  • Last Post
Replies
9
Views
733
Top