Phy5645/HO problem1: Difference between revisions
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Calculate the expectation value of x in eigenstate. | Problem 1: Calculate the expectation value of x in eigenstate. | ||
Solution: | Solution: | ||
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<math>\langle\Psi_n|x|\Psi_n\rangle=\sqrt{\frac{\hbar}{2m\omega}}\langle\Psi_n|\hat{a}+\hat{a}^{\dagger}|\Psi_n\rangle=\sqrt{\frac{\hbar}{2m\omega}}(\langle\Psi_n|\hat{a}\Psi_n\rangle+\langle\Psi_n|\hat{a}^{\dagger}\Psi_n\rangle)=\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n}\langle\Psi_n|\Psi_n-1\rangle+\sqrt{n+1}\langle\Psi_n|\Psi_n+1\rangle)</math>=0 | <math>\langle\Psi_n|x|\Psi_n\rangle=\sqrt{\frac{\hbar}{2m\omega}}\langle\Psi_n|\hat{a}+\hat{a}^{\dagger}|\Psi_n\rangle=\sqrt{\frac{\hbar}{2m\omega}}(\langle\Psi_n|\hat{a}\Psi_n\rangle+\langle\Psi_n|\hat{a}^{\dagger}\Psi_n\rangle)=\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n}\langle\Psi_n|\Psi_n-1\rangle+\sqrt{n+1}\langle\Psi_n|\Psi_n+1\rangle)</math>=0 | ||
We should have seen that coming. Since each term in the operator changes the eigenstate, the dot product with the original (orthogonal) state must give zero | We should have seen that coming. Since each term in the operator changes the eigenstate, the dot product with the original (orthogonal) state must give zero. | ||
(Submitted by Team 4-Hang Chen) |
Revision as of 01:26, 6 December 2009
Problem 1: Calculate the expectation value of x in eigenstate.
Solution:
We can compute the expectation value of x simply.
=0
We should have seen that coming. Since each term in the operator changes the eigenstate, the dot product with the original (orthogonal) state must give zero.
(Submitted by Team 4-Hang Chen)