Timeline for Is the zero point energy of this system zero?
Current License: CC BY-SA 4.0
29 events
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| Aug 16, 2021 at 8:58 | history | edited | DanielC | CC BY-SA 4.0 |
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| Aug 16, 2021 at 8:23 | history | edited | Qmechanic♦ |
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| Aug 8, 2021 at 9:49 | vote | accept | Lost | ||
| S Aug 8, 2021 at 9:46 | history | bounty ended | Lost | ||
| S Aug 8, 2021 at 9:46 | history | notice removed | Lost | ||
| Aug 7, 2021 at 22:43 | answer | added | Leviathan | timeline score: 0 | |
| Aug 7, 2021 at 20:15 | answer | added | AccidentalFourierTransform | timeline score: 5 | |
| Aug 7, 2021 at 19:23 | answer | added | Chiral Anomaly | timeline score: 5 | |
| Aug 7, 2021 at 14:16 | comment | added | my2cts | What is your time dependent hamitonian? It seems that you shift energy reference depending on the solution of your unshifted problem. | |
| Aug 7, 2021 at 12:35 | history | edited | Lost | CC BY-SA 4.0 |
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| Aug 7, 2021 at 12:27 | history | edited | Lost | CC BY-SA 4.0 |
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| Aug 7, 2021 at 11:21 | comment | added | ACuriousMind♦ | Please edit information that clarifies your question into the question instead of just leaving it in comments | |
| Aug 7, 2021 at 11:15 | comment | added | Lost | The above indicates that a constant offset has no consequences in the dynamics. That I know. But how come it can change the real fluctuations of the ground state? | |
| Aug 7, 2021 at 11:14 | comment | added | Lost | @Maurice Also, "I think what is typically meant by "zero-point energy" in this context is actually zero-point kinetic energy. Exactly as you say, a constant offset E0 in the Hamiltonian H=H0+E0 has no consequences on the dynamics and classically speaking, it corresponds to picking a different reference point for your potential energy function." | |
| Aug 7, 2021 at 11:13 | comment | added | Lost | @Maurice All the related answers seem to indicate that the zero-point energy is just a matter of convention ("can be set to zero with a suitable shift"). My doubt is how could that be true? Doesn't the zero point energy represent real fluctuations? | |
| Aug 7, 2021 at 11:10 | comment | added | Mauricio | Related questions here seem to answer the question already. | |
| Aug 7, 2021 at 10:54 | comment | added | Lost | It could be a trouble of terminology. But I think the replies I gave to your question in the comments above clarify where my confusion lies. | |
| Aug 7, 2021 at 10:52 | comment | added | Lost | @ACuriousMind♦ Consider the comment I gave above of removing energy.... Is the zero point energy not the kinetic energy left at 0K. If it is then how does it not physically matter? | |
| Aug 7, 2021 at 9:55 | comment | added | ACuriousMind♦ | I still don't understand how this is not already answered by the answers to the question you link that point out that it is differences in zero-point energy that matter, not the absolute value of zero-point energy. | |
| S Aug 7, 2021 at 8:12 | history | bounty started | Lost | ||
| S Aug 7, 2021 at 8:12 | history | notice added | Lost | Draw attention | |
| Aug 5, 2021 at 9:28 | comment | added | Lost | @BySymmetry I have read that. When we measure the excitations, it doesn't matter what reference we take and we can simply set ground state energy to be zero for convenience. But I don't think zero-point energy is just a matter of convention or arbitrary setting of reference level. | |
| Aug 5, 2021 at 9:24 | comment | added | Lost | @ACuriousMind Since I used to think that "zero-point energy" corresponds to real physical fluctuations I don't think it should get changed simply by adding a constant term in the hamiltonian. So, if I keep on removing energy from a quantum mechanical oscillator I can at minimum reach its zero-point energy. Now if I apply a shift in the potential (add a constant factor to the Hamiltonian ) and do the same thing I can get to zero energy. Isn't this wrong somehow? | |
| Aug 5, 2021 at 9:19 | comment | added | Lost | "I think what is typically meant by "zero-point energy" in this context is actually zero-point kinetic energy. Exactly as you say, a constant offset $E_0$ in the Hamiltonian $H=H_0+E_{0}$ has no consequences on the dynamics and classically speaking, it corresponds to picking a different reference point for your potential energy function." | |
| Aug 5, 2021 at 9:18 | comment | added | Lost | @ACuriousMind To put my question in a better context let me quote the part from an answer given here by Sahand: physics.stackexchange.com/q/614186 | |
| Aug 4, 2021 at 18:32 | comment | added | By Symmetry | You might find this question, and the associated answers interesting | |
| Aug 4, 2021 at 17:17 | comment | added | ACuriousMind♦ | "since I have just added a constant factor to the Hamiltonian of a Harmonic oscillator which is nothing but shifting the potential level and I don't think that should change the zero point energy" - what do you think the "potential level" is if not the energy? | |
| Aug 4, 2021 at 17:11 | history | edited | Lost | CC BY-SA 4.0 |
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| Aug 4, 2021 at 16:51 | history | asked | Lost | CC BY-SA 4.0 |