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visits member for 1 year, 10 months
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Currently doing a PhD in Controlled Quantum Dynamics at Imperial College and the University of Oxford.


Aug
23
awarded  Revival
Aug
22
comment Stabilization of von Neumann equation
@XingdongZuo You're welcome. Yes, a probability distribution that does not evolve in time can be called a stationary state or steady state in physics.
Aug
22
comment Stabilization of von Neumann equation
@XingdongZuo This expansion holds whether or not $[H,\rho(0)] = 0$. However, $[H,\rho(0)] = 0$ implies that $\rho_{mn} = 0$ for $E_m\neq E_n$, and therefore there are no oscillations. That is, if $[H,\rho(0)]=0$, it follows that $\rho_{mn}(t) = \rho_{mn}(0)e^{i(E_m-E_n)t} = \rho_{mn}(0)$, since either $e^{i(E_m-E_n)t}=1$ or $\rho_{mn}(0) = 0$.
Aug
22
comment Stabilization of von Neumann equation
@XingdongZuo But the state of a system does not reach thermal equilibrium as $t\to 0$ under unitary evolution. If $[\rho,H]\neq 0$ you will just see coherent oscillations forever. There is no damping of fluctuations without some external influence.
Aug
22
comment Stabilization of von Neumann equation
Hi @XingdongZuo, I'm not sure I understand what you mean by "stabilised". Do you mean reaching thermal equilibrium?
Aug
16
comment Unknown letter ℑ used in an equation
This question is about typesetting and so would be appropriate for our sister site LaTeX StackExchange.
Aug
15
comment What is the physical meaning of commutators in quantum mechanics?
@AlexeyBobrick Cool, I think we both agree, just wanted to be clear. Thanks :)
Aug
15
comment What is the physical meaning of commutators in quantum mechanics?
@AlexeyBobrick But this is just restating the uncertainty principle. The commutator contains more information than just the statement of the uncertainty principle, see my answer for example. I also don't see how the stuff to do with symmetry transformations can be "put simply" as a statement about measurement of $A$ and $B$. This is why I made my original comment.
Aug
14
comment What is the physical meaning of commutators in quantum mechanics?
@AlexeyBobrick No, I just mean that a measurement of $B$ (or $A$) is not formally represented by the action of the operator $B$ ($A$). Therefore it does not automatically follow that the formal statement $[A,B]\neq 0$ tells you something about "measuring B then A". Of course the two are connected but the relationship seems to be non-trivial.
Aug
14
comment What is the physical meaning of commutators in quantum mechanics?
@AlexeyBobrick Except that the action of measuring $B$ does not correspond to application of the operator $B$ to the state. I do agree with your heuristic statement, but making it precise is not so straightforward.
Aug
14
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14
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14
revised What is the physical meaning of commutators in quantum mechanics?
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14
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Aug
14
comment What is the physical meaning of commutators in quantum mechanics?
@ValterMoretti Yes. I feel that this sentence is the essence of what's difficult (for me) about Emilio's question.
Aug
14
comment What is the physical meaning of commutators in quantum mechanics?
@ChrisWhite One difference is that in GR the Riemann tensor quantifies the difference between two physically understandable and visualisable processes, namely parallel transport of a vector in different directions. In QM it is not so clear what "applying A or B" corresponds to physically.
Aug
14
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Aug
14
answered What is the physical meaning of commutators in quantum mechanics?
Aug
13
comment How many different types of (hypothetical) multiverse theory exist?
Yep. Sorry I couldn't be of more help in refining this into an acceptable question for the community.