Timeline for Expectation value of momentum
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 13, 2013 at 21:25 | vote | accept | Denver Dang | ||
| Jun 10, 2013 at 13:07 | comment | added | Will | As requested, my solution has been posted below. | |
| Jun 10, 2013 at 13:07 | answer | added | Will | timeline score: 3 | |
| Jun 10, 2013 at 7:51 | comment | added | Neuneck | @Will could you write this as an answer please? | |
| Jun 10, 2013 at 0:28 | comment | added | Will | *product rule. You're welcome. | |
| Jun 10, 2013 at 0:21 | comment | added | Denver Dang | Ahhh, didn't think of it as a product. Thank you Will. | |
| Jun 10, 2013 at 0:13 | comment | added | Will | The momentum operator in position space is a derivative. So you're getting the chain rule. e.g. $\frac{\partial}{\partial x}\left(e^{ikx}\phi(x)\right) = \left(\frac{\partial}{\partial x}e^{ikx}\right)\phi(x) + e^{ikx}\left(\frac{\partial}{\partial x}\phi(x)\right)$. Hence addition. | |
| Jun 10, 2013 at 0:09 | comment | added | Denver Dang | Hmmm, I see what you mean. Then I get the $\hbar \vec{k}$ actually. But why is that not multiplied with $\vec{p}_{0}$ insead of adding it ? When the operator works on the exponential function, I just get the $\hbar \vec{k}$, but where does the plus sign come from ? :/ | |
| Jun 9, 2013 at 23:51 | comment | added | Will | Write the momentum operator in position space and think how this acts on the wavefunction. | |
| Jun 9, 2013 at 23:48 | history | asked | Denver Dang | CC BY-SA 3.0 |