Phy5645/HO problem1: Difference between revisions
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First, we rewrite the position operator in terms of the raising and lowering operators: | |||
<math>\ | <math>\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^{\dagger})</math> | ||
<math>=\sqrt{\frac{\hbar}{2m\omega}}(\langle | Now we determine the expectation value of the position for an arbitrary harmonic oscillator eigenstate <math>|n\rangle:</math> | ||
<math>\langle n|\hat{x}|n\rangle=\sqrt{\frac{\hbar}{2m\omega}}\langle n|(\hat{a}+\hat{a}^{\dagger})|n\rangle</math> | |||
<math>=\sqrt{\frac{\hbar}{2m\omega}}(\langle n|\hat{a}|n\rangle+\langle n|\hat{a}^{\dagger}|n\rangle)</math> | |||
<math>=\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n}\langle | <math>=\sqrt{\frac{\hbar}{2m\omega}}(\sqrt{n}\langle n|n-1\rangle+\sqrt{n+1}\langle n|n+1\rangle)</math> | ||
<math>=0</math> | <math>=0\!</math> | ||
We | We may also see this intuitively; we know that, because the potential is an even function of <math>x,\!</math> the eigenfunctions must be either even or odd functions of <math>x.\!</math> This means that the probability density is always an even function, meaning that the particle is just as likely to be found at <math>-x\!</math> as it is to be found at <math>x.\!</math> | ||
Back to [[Harmonic Oscillator Spectrum and Eigenstates]] | Back to [[Harmonic Oscillator Spectrum and Eigenstates]] | ||
Revision as of 16:41, 8 August 2013
First, we rewrite the position operator in terms of the raising and lowering operators:
Now we determine the expectation value of the position for an arbitrary harmonic oscillator eigenstate
We may also see this intuitively; we know that, because the potential is an even function of the eigenfunctions must be either even or odd functions of This means that the probability density is always an even function, meaning that the particle is just as likely to be found at as it is to be found at
