Timeline for Is the eigenvalue of Hamiltonian invariant under linear transformation of momentum operator?
Current License: CC BY-SA 3.0
9 events
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| Apr 23, 2016 at 15:26 | comment | added | Valter Moretti | OK, I have re-corrected the sign. Your proof is formally correct (actually the true proof is different because you cannot expand that way exponentials of unbounded operators like $X$ and $P$, but this is another class of problems, do not worry at this stage of your knowledge...) | |
| Apr 23, 2016 at 15:24 | comment | added | VladeKR | Oh I didn't even realize that... I was just working out the transformation, because the transform sequence wasn't immediately obvious to me. | |
| Apr 23, 2016 at 15:24 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Apr 23, 2016 at 15:22 | comment | added | Valter Moretti | Are you saying that the initial sign was correct? (In the exponent I mean) | |
| Apr 23, 2016 at 15:13 | comment | added | VladeKR | It is given $e^{A} B e^{-A} = B + [A,B] + \frac{1}{2!}[A,[A,B]] + ...$ where, $B= P $ and $A =i\lambda X/ \hbar $ Therefore, the transformation of $P$ is $P + [i\lambda X/\hbar,P] +\frac{1}{2!}[i\lambda X/\hbar,[i\lambda X/\hbar,P]] ...$ but because $ [i\lambda X/\hbar,P] = i \lambda / \bar [x,p] = -\lambda$ thus, further commutation relations will yield zero. | |
| Apr 23, 2016 at 14:44 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Apr 23, 2016 at 14:43 | comment | added | Valter Moretti | @VladeKR done! (I also corrected the sign in the exponent) | |
| Apr 23, 2016 at 14:16 | comment | added | VladeKR | I have carried out the transformation calculation given $U = e^{i\lambda X/\hbar}$ but failed to get the transformed $H'$, could you elaborate? | |
| Apr 23, 2016 at 11:33 | history | answered | Valter Moretti | CC BY-SA 3.0 |