Timeline for In quantum mechanics, why is $\langle T\rangle=\frac{\langle p^2 \rangle}{2m}$ rather than $\langle T\rangle=\frac{\langle p \rangle^2}{2m}$?
Current License: CC BY-SA 4.0
16 events
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| Nov 5, 2018 at 9:08 | history | edited | knzhou | CC BY-SA 4.0 |
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| Nov 2, 2018 at 0:00 | history | tweeted | twitter.com/StackPhysics/status/1058146875226886144 | ||
| Nov 1, 2018 at 15:05 | history | edited | jjepsuomi | CC BY-SA 4.0 |
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| Nov 1, 2018 at 14:20 | vote | accept | jjepsuomi | ||
| Nov 1, 2018 at 14:11 | answer | added | Alfred Centauri | timeline score: 8 | |
| Nov 1, 2018 at 14:08 | comment | added | knzhou | If $A = B$, then $\langle A \rangle = \langle B \rangle$, because we can do the same thing to both sides of an equation. That's absolutely all there is to it. If you think $H = p^2/2m$, then $\langle H \rangle = \langle p^2 / 2m \rangle$. | |
| Nov 1, 2018 at 13:48 | comment | added | BioPhysicist | @ZeroTheHero I don't think the OP is asking about the difference between $\langle p\rangle^2$ and $\langle p^2\rangle$. I think the OP is just wondering why one appears in the average kinetic energy instead of the other. | |
| Nov 1, 2018 at 13:47 | comment | added | ZeroTheHero | A simpler example of the distinction between $\langle p\rangle^2$ and $\langle p^2\rangle$ would be $\langle x\rangle^2\ne \langle x^2\rangle$; you would need $\langle x^2\rangle$ to obtain the average potential energy of a harmonic oscillator for instance. | |
| Nov 1, 2018 at 13:43 | history | edited | Qmechanic♦ |
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| Nov 1, 2018 at 13:42 | answer | added | BioPhysicist | timeline score: 8 | |
| Nov 1, 2018 at 13:19 | comment | added | jjepsuomi | @AaronStevens thank you, the intuition already helped :) Of course the details are still in the mist for me. Any other book recommendation where this might be explicitly derived? | |
| Nov 1, 2018 at 13:16 | comment | added | jjepsuomi | @MartinC. Thank you, I actually checked that answer before but it didn't really open up to me unfortunately :/ | |
| Nov 1, 2018 at 13:13 | comment | added | BioPhysicist | In the mean time, I will explain with intuition rather than math. Momentum has a direction, kinetic energy does not. You can have a mean momentum of $0$ but a non-zero mean kinetic energy. Performing the average of the square of momentum fixes this. | |
| Nov 1, 2018 at 13:11 | comment | added | Martin C. | The first answer here might help you: physics.stackexchange.com/questions/424800/… | |
| Nov 1, 2018 at 13:08 | history | edited | jjepsuomi | CC BY-SA 4.0 |
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| Nov 1, 2018 at 13:03 | history | asked | jjepsuomi | CC BY-SA 4.0 |