Phy5645/HO problem1: Difference between revisions
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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 | ||
Revision as of 01:20, 6 December 2009
Calculate the expectation value of x in eigenstate.
Solution:
We can compute the expectation value of x simply.
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)} =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