- #1
Sekonda
- 201
- 0
Hey,
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:
\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})
and so the expectation value becomes:
\sqrt{\frac{\hbar}{2m\omega}}<E_{1}|(a+a^{\dagger})|E_{0}>
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:
\sqrt{\frac{\hbar}{2m\omega}}=<E_{1}|\hat{x}|E_{0}>
Can someone confirm this?
Cheers,
SK
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:
\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dagger})
and so the expectation value becomes:
\sqrt{\frac{\hbar}{2m\omega}}<E_{1}|(a+a^{\dagger})|E_{0}>
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:
\sqrt{\frac{\hbar}{2m\omega}}=<E_{1}|\hat{x}|E_{0}>
Can someone confirm this?
Cheers,
SK