Timeline for answer to Deriving a QM expectation value for a square of momentum $\langle p^2 \rangle$ by nervxxx
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| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2013 at 0:01 | vote | accept | 71GA | ||
| Mar 24, 2013 at 0:01 | comment | added | 71GA | Thanks for pointing me into the future :) I hope i will find my wavefunction hehehe | |
| Mar 23, 2013 at 21:22 | comment | added | nervxxx | @71GA if you mean my note... well. Griffiths, from the outset, uses the position basis to teach QM. There's nothing wrong with that, but it obscures the fact that the fundamental object is a ket vector $| \psi \rangle>$ in an abstract Hilbert space, and the wavefunction $\psi(x)$ is just a representation of that ket vector in real space: $\langle x | \psi \rangle$. After you're done with Griffiths, Shankar will be a good 2nd read and Sakurai a good 3rd read. | |
| Mar 23, 2013 at 21:12 | comment | added | nervxxx | Any introductory text to QM will have this. Even Griffiths has this - there are a few sentences after he 'derives' $\hat{p}$ where he talks about constructing the quantum operator of any classical operator. Any classical operator can be constructed in terms of $p$ and $x$, and the rule he gives to 'quantize' a classical theory is exactly what I have described - change $x$ to $\hat{x}$ and $p$ to $\hat{p}$. Try reading the chapter again :) | |
| Mar 23, 2013 at 16:04 | comment | added | 71GA | Where can i read more about this? Is this somehow connected to operator theory??? | |
| Mar 22, 2013 at 23:29 | comment | added | David Z | @71GA that's the definition of what it means to square an operator. | |
| Mar 22, 2013 at 20:35 | comment | added | nervxxx | One way to look at it is to start from the classical theory and then quantize it - anytime you see a $x$ you change it to $\hat{x}$ and anything you see a $p$ you write a $\hat{p}$. Then you impose the commutation relation $[x,p] = i\hbar$. There are some ordering issues so you need to choose an ordering too. Now to quantize the classical $p^2$ term, $p^2 = p p \to \hat{p} \hat{p} = \hat{p}^2$. So the quantum version of $p^2$, which we call $\widehat{p^2}$, is equal to $\hat{p}^2$. The expectation values not being equal as you have mentioned follow from this. | |
| Mar 22, 2013 at 20:14 | comment | added | 71GA | Are you sure i can do this $\widehat{p^2} = \hat{p}^2= \hat{p} \hat{p}$? Afterall i am not alowed to do it with expectation values as $\langle p^2 \rangle \neq {\langle p \rangle} ^2$. Could you point me to any proof? | |
| Mar 22, 2013 at 20:03 | history | answered | nervxxx | CC BY-SA 3.0 |