Applying Selection Rules to Determine Non-Zero Ground State Perturbations

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In summary, the ground state has zero energy and the final energy can be calculated using the formula E_f = (h-bar)^2l(l+1)/2I. The transition probability w_fi is then determined using the formula (E_f-E_i)/h-bar. The integral for d_f at infinity involves the dot product of the operator dz with the final and initial states. To determine which eigenstates will result in a non-zero value for this dot product, the operator dz is converted into a spherical harmonic function. The limits of integration should be carefully considered, and the final integral should involve the product of three spherical harmonics.
  • #1
Diracobama2181
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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.
 
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  • #2
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
 
  • #3
Diracobama2181 said:
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.
 
  • #4
But wouldn't the dot product in V(t) get rid of the $$\hat Z$$?
 
  • #5
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 $$d(t)=\frac{i}{\hbar}\int_{0}^{t}qE_0r2(\frac{\pi}{2})^{\frac{1}{3}}e^{\frac{i\hbar t}{I}}e^{-\frac{t}{T}}dt$$
 

1. What is a rigid rotator perturbation?

A rigid rotator perturbation is a mathematical model used to describe the behavior of a rotating body that is subject to external forces. It is commonly used in fields such as quantum mechanics and molecular spectroscopy.

2. How does a rigid rotator perturbation differ from a simple rigid rotator?

A simple rigid rotator assumes that the body is rotating without any external forces acting on it, whereas a rigid rotator perturbation takes into account the effects of external forces on the rotation of the body.

3. What types of external forces can be included in a rigid rotator perturbation?

The external forces considered in a rigid rotator perturbation can include electric and magnetic fields, as well as mechanical forces and torques.

4. How is a rigid rotator perturbation solved?

A rigid rotator perturbation is typically solved using mathematical techniques such as perturbation theory, which involves breaking down a complex problem into simpler, solvable parts.

5. What are some real-world applications of rigid rotator perturbation?

Rigid rotator perturbation is commonly used in fields such as molecular spectroscopy to study the rotational behavior of molecules. It is also used in quantum mechanics to solve problems related to rotating systems, such as the behavior of spinning particles in a magnetic field.

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