Timeline for A limit of a particular Quantum Fidelity
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2023 at 9:04 | history | bounty ended | CommunityBot | ||
| Sep 28, 2023 at 14:18 | vote | accept | Hldngpk | ||
| Sep 25, 2023 at 13:07 | comment | added | Quantum Mechanic | @Hldngpk I'm not yet sure about normalization. Maybe the definition needs to be $|x;t\rangle=\int d(pt) exp(-iptx)|pt\rangle$, and these states are delta-orthogonal in the large-$t$ limit. This is just some intution for starting the calculation and seeing what happens | |
| Sep 25, 2023 at 2:05 | comment | added | Hldngpk | by utilizing the parameter $t$ that you have introduce . Do we not have $\langle x;t\big|x;t\rangle = \frac{1}{t}\delta(0) $ ( in distribution). Which means that we have given up normalization? | |
| Sep 25, 2023 at 1:59 | comment | added | Hldngpk | Also, I made a typo in my second to last message. I meant to write " by assuming that $e^{-ity\mathbf{\hat{B}}}\big|x\big\rangle = \big|x+ty\big\rangle$". | |
| Sep 25, 2023 at 1:53 | comment | added | Hldngpk | very interesting response. I have to process it. By the way, you wrote $tB|pt\rangle=tB|pt\rangle$ above? Is this a typo? Did you mean to write $ tB|pt\rangle=tp|pt\rangle$ ? | |
| Sep 24, 2023 at 13:46 | comment | added | Quantum Mechanic | @Hldngpk say we want $|x;t\rangle$ such that $e^{-iytB}|x;t\rangle=|x+y;t\rangle$. Just define $|x;t\rangle=\int dp e^{-iptx}|pt\rangle$ where $tB|pt\rangle=tB|pt\rangle$. Then $\langle x;t|x^\prime;t\rangle=\int dp \exp[i pt(x-x^\prime)]=t^{-1}\int d(pt) \exp[i (pt)(x-x^\prime)]$ where the latter integral takes the domain of $pt$ which is infinite even if $B$ has compact support | |
| Sep 24, 2023 at 13:41 | comment | added | Quantum Mechanic | @Hldngpk all I need is an infinite set of eigenstates of $\hat{\mathbf{B}}$ such that $\langle p|p^\prime\rangle=\delta(p-p^\prime)$ and $\int dp |p\rangle\langle p|=\mathbb{I}$, then I just define $|x\rangle=\int dp e^{-ixp}|p\rangle$. Now you might say $\hat{\mathbf{B}}$ has compact support so this is not true, and then my suggestion would be to instead look at the eigenstates of $t\hat{\mathbf{B}}$ and those will have eigenvalues spanning $\mathbb{R}$ and so the Fourier transform should be nice | |
| Sep 24, 2023 at 8:08 | comment | added | Hldngpk | Mechanics. The problem is that by assuming that $e^{-ity\mathbf{\hat{B}}}\big|x\big\rangle$ you have already constrained your model to the one I described, i.e. the one satisfying the Heisenberg commutation relations. What you provided is a special case of what I am after and not the full general setting. | |
| Sep 24, 2023 at 3:02 | comment | added | Quantum Mechanic | @Hldngpk does it help that I did not actually define $|x\rangle$ as eigenstates of anything? I simply formed them from a continuous superposition of eigenstates of $\hat{\mathbf{B}}$, and that state will obey the shift property and orthonormality property required so long as the eigenstates of $\hat{\mathbf{B}}$ are delta-orthogonal | |
| Sep 23, 2023 at 22:00 | comment | added | Quantum Mechanic | @Hldngpk I have nothing to add! Your comments are all spot-on: 1) is well taken, I have only proven for a certain class of operators; 2) is helpful, I appreciate you tightening that up; 3) yes since this method proves your relation for more than just pure states I feel that it should be general enough. If not, and you need another method to prove the general case, this is at least stronger evidence that your relation is valid | |
| Sep 23, 2023 at 3:30 | comment | added | Hldngpk | Sorry, in my second to last comment, I meant to write "It had not occurred to me to use the spectral decomposition of the conjugate operator to $\mathbf{\hat{B}$... | |
| Sep 23, 2023 at 3:16 | comment | added | Hldngpk | continued... we start with the spectral decomposition $e^{-itx_{i}\mathbf{\hat{B}}} = \int_{\sigma(\mathbf{\hat{B}})} e^{-itx_{i}\lambda}d\mathbf{\hat{E}}_{\lambda}$ . For the case you have worked out the spectral projector family $\mathbf{\hat{E}}_{x} = \int_{-\infty}^{x}\big|x\big\rangle\big\langle x\big| dx$ . I will think about this more. Thank you again for your ideas. | |
| Sep 23, 2023 at 3:15 | comment | added | Hldngpk | Comment 3) This variational bound to the Fidelity (Nielsen and Chuang 9.2.2) is indeed an interesting approach and it had not occurred to me to use the spectral decomposition of $\mathbf{\hat{B}}$ as the POVM used for the bound. Unless I have overlooked something your approach seems to solve my problem for the special case of $\mathbf{\hat{B}}$ being a position or momentum operator but only for this case. I am wondering if this approach may now be adapted to a general $\mathbf{\hat{B}}$ with purely absolutely continuous spectrum. i.e. we start with the spectral decomposition ... | |
| Sep 23, 2023 at 3:08 | comment | added | Hldngpk | Comment 2) The integral you present following the line "Putting these together, our inequality becomes" decays to zero via first noting that the convolution of two L1 functions is yet again an L1 function and then by the usage of Tonneli's theorem and a Dominated Convergence theorem argument. | |
| Sep 23, 2023 at 2:53 | comment | added | Hldngpk | Continuation of comment 1) What you seem to imply is that there always exists $\mathbf{\hat{X}}$ for any $\mathbf{\hat{B}}$ with purely absolutely continuous spectrum such that $\big[\mathbf{\hat{B}},\;\mathbf{\hat{X}}\big]$. This, however, is not so! Consider the case where $\mathbf{\hat{B}} = f(\mathbf{\hat{P}})$, where $\mathbf{\hat{P}}$ is the momentum operator and $f(x)$ is a compactly support continuous fucntion. Can you find an $\mathbf{\hat{X}}$ satisfying $\big[\mathbf{\hat{X}},\; f(\mathbf{\hat{P}})\big] = i \mathbb{I}$ ? I think not. | |
| Sep 23, 2023 at 2:49 | comment | added | Hldngpk | Comment 1) It is not the case that one, as you say, may always find a basis $|x\rangle$ such that $e^{itx_{i}\mathbf{\hat{B}}}\big|x\big\rangle = \big|x+x_{i}t\big\rangle$. When $\mathbf{\hat{B}}$ is the laplacian or, any quadrature momentum, this is true. In general, this is not true however since the commutation relations $\big[ \mathbf{\hat{B}}, \mathbf{\hat{X}}\big] = i\mathbb{I}$ would need to be satisfied, where $\mathbf{\hat{X}}$ is the operator with generalized eigenvectors $\big|x\big\rangle$. | |
| Sep 23, 2023 at 2:44 | comment | added | Hldngpk | thank you very much for your response. I have a few comments. I'll present them separately. | |
| Sep 22, 2023 at 16:26 | history | answered | Quantum Mechanic | CC BY-SA 4.0 |