A quantum mechanical rigid roator

In summary, the conversation discusses the calculation of the eigenfunctions of the Hamiltonian for a rigid rotator with an external electric field applied. The eigenfunctions are found to be double degenerate and the degeneracy is not removed even in high orders of perturbation. The calculation of the perturbation matrix elements between the degenerate kets is also discussed.
  • #1
hokhani
483
8

Homework Statement



Consider the Hamiltonian for a rigid rotator, constrained to rotate in xy plane and with moment of inertia I and electric dipole moment[itex]\mu[/itex] in the plane, as [itex]H_0=\frac{L_z^2}{2I}[/itex]. Then suppose that a constant and weak external electric field,[itex]E[/itex], in the direction of x is applied so the perturbation is [itex]H'=-\mu E cos (\phi)[/itex].

Homework Equations



The eigenfunctions of [itex]H_0[/itex] are double degenerate and the degeneracy is not removed even in high orders of perturbation. How should I solve it? could anyone please help me?

The Attempt at a Solution

 
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  • #2
The problem statement doesn't include a question.

Have you calculated ##\langle m' | H' | m \rangle##? And found it to be always zero for ##m'\neq m##?
 
  • #3
The degenerate eigenfunctions are [itex]|m\rangle=\frac{1}{\sqrt {2\pi}} e^(im\phi)[/itex] and [itex]|-m\rangle=\frac{1}{\sqrt {2\pi}} e^(-im\phi)[/itex] and [itex]\langle -m|cos(\phi)|m\rangle =0[/itex] (ms are integer). Hence all the elements of perturbation matrix between degenerate kets would be zero.
 
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  • #4
I made a quick calculation, and it seems indeed that the degeneracy is not lifted. How is that a problem? The field still shifts the levels.
 
  • #5
Could you please give me your solution? Or at least, how much does the electric field move the levels?
 
  • #6
You should calculate ##\langle m' | H' | m \rangle## yourself. Rewriting ##\cos \phi## in terms of exponentials helps in evaluating the integrals.
 
  • #7
DrClaude said:
You should calculate ##\langle m' | H' | m \rangle## yourself. Rewriting ##\cos \phi## in terms of exponentials helps in evaluating the integrals.
Ok, it is very easy to calculate it:
[itex]\langle m' | H' | m \rangle=-\frac{\mu E}{2}(\delta(m',m+1)+\delta(m',m-1))[/itex]
Making the [itex]H'[/itex] matrix on the two degenerate kets, m and -m, we observe that all the four matrix elements are zero. Could you please tell me what is my mistake?
 
  • #8
I'm sorry, but the I don't understand what question you are trying to answer.
 

Related to A quantum mechanical rigid roator

1. What is a quantum mechanical rigid rotor?

A quantum mechanical rigid rotor is a model used in quantum mechanics to describe the rotational motion of a molecule or atom. It is a simplified model that assumes the molecule or atom is rigid and does not vibrate or bend.

2. How does a quantum mechanical rigid rotor differ from a classical rigid rotor?

A quantum mechanical rigid rotor takes into account the quantization of energy levels and the uncertainty principle, whereas a classical rigid rotor does not. This means that the energy levels and angular momentum of a quantum mechanical rigid rotor are discrete, rather than continuous as in a classical rigid rotor.

3. What is the significance of the quantization of energy levels in a quantum mechanical rigid rotor?

The quantization of energy levels in a quantum mechanical rigid rotor is significant because it allows for the prediction of the rotational energy levels and transitions of a molecule or atom. This is important in understanding the behavior and properties of these particles.

4. How is the quantum mechanical rigid rotor model used in spectroscopy?

The quantum mechanical rigid rotor model is used in spectroscopy to analyze the rotational transitions of molecules or atoms. By measuring the wavelengths of light absorbed or emitted by the particles, information about their rotational energy levels and structure can be obtained.

5. Can the quantum mechanical rigid rotor model be applied to any molecule or atom?

No, the quantum mechanical rigid rotor model is only applicable to small, rigid molecules or atoms. It does not take into account the effects of molecular vibrations or bending, so it is not suitable for larger or more complex molecules.

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