Showing energy is expectation of the Hamiltonian

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Homework Help Overview

The discussion revolves around demonstrating that the energy of a quantum harmonic oscillator can be expressed as the expectation values of its momentum and position operators. The original poster presents a normalized eigenvector and its corresponding energy level, seeking to establish a relationship involving the Hamiltonian of the system.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the validity of taking the expectation of both sides of the Hamiltonian equation. Some express uncertainty about whether this approach leads to the correct formulation. Others suggest starting from the expectation of the Hamiltonian to derive the desired result.

Discussion Status

The conversation is ongoing, with participants exploring different interpretations and approaches. Some have suggested the use of ladder operators, indicating a potential direction for solving the problem, while others express confusion about how to proceed without that knowledge.

Contextual Notes

There is mention of the need for additional information to fully address the problem, as well as references to university notes that may include relevant concepts not yet covered in the discussion.

Unredeemed
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Homework Statement



The vector \psi =\psi_{n} is a normalized eigenvector for the energy level E=E_{n}=(n+\frac{1}{2})\hbar\omega of the harmonic oscillator with Hamiltonian H=\frac{P^{2}}{2m}+\frac{1}{2}m\omega^{2}X^{2}. Show that:

E=\frac{\mathbb{E}_{\psi}[P^{2}]}{2m}+\frac{1}{2}m\omega^{2}\mathbb{E}_{\psi}[X^{2}]

Homework Equations



Time independent Schrödinger: H\psi=E\psi

The Attempt at a Solution



Am I wrong in thinking it's as simple as taking the expectation of both sides? I feel like I must be as that gives \mathbb{E}_{\psi}[E]=\frac{\mathbb{E}_{\psi}[P^{2}]}{2m}+\frac{1}{2}m\omega^{2}\mathbb{E}_{\psi}[X^{2}] which isn't quite right.

But I can't see how else I'd do it?

Thanks for your help in advance!

EDIT: I believe the extra information is needed to answer the rest of the question (this is just part 1).
 
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Unredeemed said:
\mathbb{E}_{\psi}[E]=\frac{\mathbb{E}_{\psi}[P^{2}]}{2m}+\frac{1}{2}m\omega^{2}\mathbb{E}_{\psi}[X^{2}]
I guess you mean (using your notation)
$$
\mathbb{E}_{\psi}[H]=\frac{\mathbb{E}_{\psi}[P^{2}]}{2m}+\frac{1}{2}m\omega^{2}\mathbb{E}_{\psi}[X^{2}]
$$

I would start from ##E = \mathbb{E}_{\psi}[H]## and develop the right-hand-side until the desired result (the equation just above) is obtained.
 
Indeed, taking the EV of both sides is the solution.
Disturbingly simple ;>) .
 
We're then asked to show by considering <\psi\mid (P\pm im\omega X)^{k}\psi> for k=1,2 and using orthogonality properties of eigenvectors that:

\mathbb{E}_{\psi}(P)=0=\mathbb{E}_{\psi}(X) and \mathbb{E}_{\psi}(P^{2})=m^{2}\omega^{2}\mathbb{E}_{\psi}(X^{2})=mE

I've shown that (P-im\omega X)*=P+im\omega X and I can see that <\psi\mid (P- im\omega X)\psi>=\mathbb{E}_{\psi}(P)-im\omega\mathbb{E}_{\psi}(X) etc.

But I really can't see how to get to the answer?

Am I supposed to show that <\psi\mid (P- im\omega X)\psi>=0=<\psi\mid (P+ im\omega X)\psi>? Because I can't seem to do that...
 
Have you seen ladder operators?
 
Not yet. But they do come in the next chapter of the notes?
 
Then I don't get how you're supposed to solve the problem. Which textbook are you using?
 
It's my university notes - we could well have to use ladder operators to solve this. In the past we've occasionally had questions that can only be covered with material from the next chapter. What in specific should I be looking for with regard to ladder operators?
 
You should be able to express ##P \pm i m \omega X## in terms of the ladder operators. Once this is done, you can figure out what ##(P \pm i m \omega X) \left| \psi \right\rangle## does, keeping in mind that ##\psi## is actually ##\psi_n##.
 
  • #10
I've got it! Thanks very much.
 

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