- #1

Sekonda

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My question is on determing the expectation value of position of the Harmonic Oscillator using raising and lowering operators, the question is part d) below:

I have determined the position operator to be:

[tex]\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})[/tex]

and so the expectation value becomes:

[tex]\sqrt{\frac{\hbar}{2m\omega}}<E_{1}|(a+a^{\dagger})|E_{0}>[/tex]

Though this is the bit I'm not sure about, I think only the a-dagger operator acts to increase the E-zero ket and the 'a' operator just causes the state to go to zero thus:

[tex]\sqrt{\frac{\hbar}{2m\omega}}=<E_{1}|\hat{x}|E_{0}>[/tex]

Can someone confirm this?

Cheers,

SK