Timeline for Energy eigenvalues of a Q.H.Oscillator with $[\hat{H},\hat{a}] = -\hbar \omega \hat{a}$ and $[\hat{H},\hat{a}^\dagger] = \hbar \omega \hat{a}^\dagger$
Current License: CC BY-SA 3.0
9 events
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| May 7, 2013 at 9:54 | comment | added | 71GA | Thank you! the most helpfull was the fact that $\hat{H} \neq W_n$ but $\hat{H} \psi_n = W_n \psi_n$. It looks a bit WIERD though... | |
| May 7, 2013 at 9:53 | vote | accept | 71GA | ||
| May 7, 2013 at 9:29 | comment | added | Ondřej Černotík | You can't say $\hat{H} = W_n$ but you can say $\hat{H}\psi_n = W_n\psi_n$ if $\psi_n$ is an eigenstate of $\hat{H}$ with eigenvalue $W_n$. And that is your starting point, together with the commutator, to find what $\hat{H}\hat{a}\psi_n$ is. | |
| May 7, 2013 at 9:28 | comment | added | 71GA | Does this mean that eigenvalue in Hilbert space is equivalent of an expectation value??? | |
| May 7, 2013 at 9:26 | comment | added | 71GA | Ok i understand now that you swapped opperator $\hat{H}$ with its eigenvalue $W_n$. All i know from this is that i get an expectation value for energy $\langle W \rangle$ like this: $$\langle W \rangle = \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x$$ and $$\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$$ But i am weak on eigenvalues, eigenvectors and Hilbert space... So how could i connect what i already know to confirm that $\hat{H} = W_n$? | |
| May 7, 2013 at 8:52 | comment | added | Ondřej Černotík | I edited the answer to give more detail on this step. Is it understandable now? | |
| May 7, 2013 at 8:51 | history | edited | Ondřej Černotík | CC BY-SA 3.0 |
extended calculation
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| May 7, 2013 at 8:44 | comment | added | 71GA | I don't quite understand this statement: ˝Because we have $\hat{H}\psi_n = W_n\psi_n$ we get $(W_n-\hbar\omega)\hat{a}\psi_n$˝. | |
| May 7, 2013 at 7:43 | history | answered | Ondřej Černotík | CC BY-SA 3.0 |