Zoltan Zimboras
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 Dec 17 awarded Nice Answer Jul 7 awarded Yearling Sep 30 awarded Explainer Sep 24 awarded Autobiographer May 4 awarded Fanatic May 3 comment Calculating phase diagrams (Calphad) @Qmechanic: Good idea, thanks for the permalink! Apr 16 comment Software for calculating Feynman Diagrams I just clicked on your link, nice description of your problem and software (with link to the original homework problem). This is great!! Thanks for sharing with us. Mar 8 comment Is the concept of tensor rank useful in physics? Yes, it was the OP who used the term "rank". The best term would be "generalized Schmidt rank", as for a tensor product of two vector spaces this notion is called "Schmidt rank". (And the linked paper in the question simply generalizes this concept to tensor products of $d$ number of vector spaces.) Feb 23 awarded Enthusiast Feb 17 comment Translation Operator for Position on Momentum @DavidZ Thanks for letting me know. Next time I mention it in the comment, and wait for the OP to edit it. Feb 17 revised Translation Operator for Position on Momentum I corrected the definition of the translation operator (see comments) + did some additional small cahnges. Feb 17 suggested approved edit on Translation Operator for Position on Momentum Feb 17 comment Translation Operator for Position on Momentum @Slaviks Nothing changed. Apparently they didn't accept this change. But it should be there, I'll try again. Feb 17 comment Translation Operator for Position on Momentum @yankeefan11 Exactly! $T^{\dagger}qpT=T^{\dagger}qTp=(q+c)p$. Feb 17 comment Translation Operator for Position on Momentum @Slaviks I think you are correct. I already changed that (and I used that definition in my answer), but my change awaits confirmation. Cheers, ZZ Feb 17 comment Translation Operator for Position on Momentum @yankeefan11 I hope this explains it. Btw, in the question you defined $T$ a bit ambiguously. I will exchange there $x$ with $c$ in the definition to make it compatible with the formula below (and not confuse anybody with the fact that $x$ is not an operator there). I hope these changes are okay with you. Feb 17 revised Translation Operator for Position on Momentum added 423 characters in body Feb 17 suggested rejected edit on Translation Operator for Position on Momentum Feb 17 comment Translation Operator for Position on Momentum Yes, let $\hbar=1$. Since $\hat{T}=\exp(-ic\hat{p})=\sum_n (-ic\hat{p})^n/n!$, we have that $\hat{p}$ commutes with all terms (as $[\hat{p}^n,\hat{p}]=0$ holds, of course). Hence $[\hat{T}, \hat{p}]=0$. Feb 17 answered Translation Operator for Position on Momentum