# Rigid Rotator in an Electric Field

• Diracobama2181
In summary, the selection rules for angular momentum and parity dictate which eigenstates will give a non-zero matrix element for $<f|E_0d_z|0>$, and the integral should converge as long as $\tau$ is finite.

#### 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.

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.

## 1. What is a rigid rotator in an electric field?

A rigid rotator in an electric field is a physical system in which a particle is confined to a specific region and is subject to both rotational and electric forces. It is used to study the behavior of molecules in an electric field, such as diatomic molecules.

## 2. How is the motion of a rigid rotator affected by an electric field?

The motion of a rigid rotator is affected by an electric field through the torque exerted on the particle. This torque causes the particle to rotate and align with the direction of the electric field.

## 3. What is the equation for the energy of a rigid rotator in an electric field?

The energy of a rigid rotator in an electric field can be calculated using the equation E = (J(J+1)ℏ²/2I) - μE, where J is the quantum number, ℏ is the reduced Planck's constant, I is the moment of inertia, μ is the dipole moment, and E is the electric field strength.

## 4. How does the energy of a rigid rotator change with increasing electric field strength?

The energy of a rigid rotator increases with increasing electric field strength. This is because the electric field creates a torque on the particle, causing it to rotate and align with the field. As the field strength increases, the torque and thus the rotational energy also increase.

## 5. What is the significance of studying a rigid rotator in an electric field?

Studying a rigid rotator in an electric field allows us to understand the behavior of molecules in an electric field, which has important applications in fields such as chemistry and materials science. It also helps us to better understand the fundamental principles of rotational motion and the interaction between electric fields and particles.