772 reputation
157
bio website
location London, United Kingdom
age
visits member for 3 years, 6 months
seen Jun 29 at 14:48

University College London


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 suggested 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 suggested 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
Feb
17
comment Momentum representation of a state
@yankeefaan Sorry, now I see that the condition actually is only $\Delta x \Delta p= \hbar/2$ (like for squeezed coherent states). So you are right, the factor of two actually doesn't matter. It is a different state, but it still satisfies your conditions. In fact, any state of the form $A e^{-a(q+q_0)^2/2}e^{i p_0x/\hbar}$ will satisfy your condition! (But don't forget we are still in coordinate space representation.)
Feb
17
revised Momentum representation of a state
corrected a factor of 2 + some additional small corrections.
Feb
17
suggested suggested edit on Momentum representation of a state
Feb
16
comment Momentum representation of a state
@yankeefan11 Comment after the edit: You are almost done! You found the correct wave function (apart from a small factor 2 typo, which I will correct in your answer now), it is $A e^{-(q+q_0)^2/2}e^{i p_0x/\hbar}$. However, this is the correct wave function in coordinate space representation. To transform this wave function into momentum space representation, you still have to Fourier transform it (see en.wikipedia.org/wiki/Position_and_momentum_space), and then you are really done.