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Martin Sleziak
Martin Sleziak
4:24 AM
It might be interesting to check whether this problem appears also on some other sites.
For example, on Physics we might check macros such as \bra, \ket, \braket.
The first query returns also this false positive: Coulumb Integral in Helium Variational Aproximation. The question uses \braket - but it is not defined there.
These answers contain macro defined in the question.
And also the answer "Why can dipole-allowed transitions occur between sets of electron states which differ by only one electronic state?" - it was already mentioned as it contains \braket. (It contains also several other macros.)
The rule you declared as axiom is not a special instance of the rule that states given a state $\ket{\chi}$, the probability to find it in a state $\ket{\psi}$ is given by (for normalized states) $\lvert \bra{\chi} \psi \rangle \rvert^2$. The two are separate rules, because the former gives probability as integral of a square, while the latter gives probability as square of an integral. Unfortunately, they are both referred to as the Born rule. — Ján Lalinský May 26 '15 at 16:38
@user36790 What this shows is that, if you're in state $\ket{1}$ at $t=0$, the amplitude for state $\ket{2}$ at some later time $t=\delta t$ is approximately $-i\bra{2}H\ket{1}$ times the time elapsed. This relationship becomes exact in the limit of very small time elapsed. — Jahan Claes Sep 25 '15 at 19:10
The last query returns this false positive: Adding 3 electron spins
data-explorer I have edited the query, so that it checks for both \newcommand{\ket} and \renewcommand{\ket}: data.stackexchange.com/physics/revision/1174524/1443817/… data.stackexchange.com/physics/revision/1174524/1451495/…
The above queries also found several correct results. I won't repeat the ones which were already mentioned with \bra.
@user36790 What this shows is that, if you're in state $\ket{1}$ at $t=0$, the amplitude for state $\ket{2}$ at some later time $t=\delta t$ is approximately $-i\bra{2}H\ket{1}$ times the time elapsed. This relationship becomes exact in the limit of very small time elapsed. — Jahan Claes Sep 25 '15 at 19:10
The rule you declared as axiom is not a special instance of the rule that states given a state $\ket{\chi}$, the probability to find it in a state $\ket{\psi}$ is given by (for normalized states) $\lvert \bra{\chi} \psi \rangle \rvert^2$. The two are separate rules, because the former gives probability as integral of a square, while the latter gives probability as square of an integral. Unfortunately, they are both referred to as the Born rule. — Ján Lalinský May 26 '15 at 16:38
The qubit notation $\ket0$ $\ket1$ is a "quantum-information standard", so the best book reference is probably the standard textbook : the Nielsen & Chuang ( squint.org/qci ) — Frédéric Grosshans Apr 3 '12 at 14:57
There are several comments under this answer: Quantum Circuit, example of the Bernstein-Vazirani problem
Thank you for this, it is great, just working out exactly what you mean. What do you mean by $N$ & $H^{\otimes N}$, we are applying $H\otimes H\otimes H\otimes 1$ to the system correct? So can we just write $N=3$? Also some of the elements of $x$ are $0$ which will be taken to $\ket{+}$? — Freeman Feb 22 '13 at 14:45
Ok, great, so I guess your description was a general one. So wrt to your second statement, wouldn't $|(-1)^{x_j}\rangle=\ket{0}$ or $\ket{-1}$ not $\ket{+}$ or $\ket{-}$?, so our three string vector $\ket{x}=\ket{x_1 x_2 x_3}$ would go to a product of some combination of $\ket{+}$ and $\ket{-}$ under the Hadamard gate? — Freeman Feb 22 '13 at 14:53
$\ket{-}=\frac{1}{\sqrt2}(\ket{0}-\ket{1}) \Rightarrow (\ket{-})^0=(\frac{1}{\sqrt2}(\ket{0}-\ket{1}))^0$, I don't see how this equals $\ket{+}=\frac{1}{\sqrt2}(\ket{0}+\ket{1})$? Wouldn't it be $\Big(\begin{matrix} 1 \\ 1 \end{matrix} \Big)=\sqrt2 \ket{+}$? — Freeman Feb 22 '13 at 15:48
And a few comments elsewhere:
@NorbertSchuch: You don't have to buy this argument, since we only need a lower bound. However, one can make this more rigorous by looking into the fidelity with a general maximally entangled state of the form $\newcommand\ket[1]{\left|#1\right\rangle} I\otimes U(\ket{\uparrow\uparrow}+\ket{\downarrow\downarrow})/\sqrt2$. The latter form is not too complicated and allows to also have an upper bound on the possible entanglement. — Frédéric Grosshans Apr 26 '13 at 9:56
Here's my logic (pun): $$\begin{bmatrix} 0 & 1 \\ 1 & 0\\ \end{bmatrix} \ket{\downarrow} = \ket{\uparrow}$$ This suggested to me that this was acting as a NOT gate because I can simply repeat the action and get $\ket{\downarrow}$. But, then I reasoned, maybe that's not the same kind of operation as the classical $\mathrm{NOT}$. Because @ACuriousMind asked me to define the NOT gate. And to do that I used set theory, like in my initial example. — Stan Shunpike Mar 25 '15 at 5:57
But consider $$\ket{\downarrow} = \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$$ this is just one point of a set of all possible linear combinations of $a\ket{\uparrow} +b\ket{\downarrow}$. It's a member of a set. So it's complement should be the rest of the points. So then $$\mathrm{NOT}(\ket{\downarrow}) = \text{set of a bunch of points}$$ which might include $\ket{\uparrow}$ but wouldn't be limited to it. — Stan Shunpike Mar 25 '15 at 5:59
I also asked ACuriousMind, given $$\ket{\psi} = \alpha \ket{\uparrow} + \beta \ket{\downarrow}$$, does there exist a $\ket{\phi}$ such that $$\ket{\phi} = \mathrm{QNOT}\left(\ket{\psi}\right)$$? By $\mathrm{QNOT}$, I just mean the 2x2 matrix above. Is it bijective? But he (rightly) wanted me to define $\mathrm{NOT}$ and I couldn't do that in such a way that I could explain how $\mathrm{QNOT}$ works. — Stan Shunpike Mar 25 '15 at 6:11
@ConstantineBlack: Yes, that is the assumption I am referring to. $A\psi$, or $A\ket{\psi}$ refers to the vector that applying the operator $A$ to the vector $\psi$ produces. (In the finite-dimensional case, simply multiply the matrix $A$ with the vector $\psi$ to get this result) The results of measurements are eigenvectors of the observable measured, and not $A$ applied to the state that was measured, since $A\psi$ is not an eigenvector of $A$ unless $\psi$ already were an eigenvector. — ACuriousMind ♦ Apr 18 '15 at 18:05
Thanks for the reply! Two questions: 1) $$\ket {+-} \equiv \ket + \otimes \o + \o \otimes \ket -$$ In this here - is the "+" a regular plus? $ $ 2) Going from the second last line to the last line - how come you can factorise $\frac{\hbar}{2}$ from the first term but not the second? Thanks again! — AXidenT May 14 '15 at 17:02
@SebiSebi: I don't understand your question. They are a basis for the space of states, but e.g. $0.5\ket{\psi_0}+0.5\ket{\psi_1}$ is certainly not an eigenstate. — ACuriousMind ♦ Jun 15 '15 at 16:43
Is $A$ the energy to go from $\ket 1$ to $\ket 2$? Is it the amplitude to go from $\ket 1$ to $\ket 2$? I'm not understanding this as Feynman is referring it as the amplitude but since it is an element of the Hamiltonian matrix, it should be the energy to go from $\ket 1$ to $\ket 2$ like $H_{11}$ being the energy of $\ket 1$. So, is $H_{ij}$ the energy or the amplitude to go from $\ket 1$ ro $\ket 2$? — user36790 Sep 26 '15 at 6:02
Thanks a lot for answering! One doubt that if $\ket{d_i}$ lies in $\mathcal{H}_4$, then in computational basis $\ket{d_1} = [1 0 0 0]^T$, $\ket{d_2} = [0 1 0 0]^T$, $\ket{d_3} = [0 0 1 0]^T$ and $\ket{d_4} = [0 0 0 1]^T$. To write $\ket{d_i}$ as a 9-component vector should I append rest of five values as zer0. I mean should I write $\ket{d_1} = [1 0 0 0 0 0 0 0 0]^T$ and similarly rest of the vectors. — Jitendra Jul 1 '18 at 13:26
Ok, my question has arisen in the context where $\ket{\psi_n}$ are all manymode Gaussian states, and there seem to be formulas much more readily available for Tr[\rho_n \rho_m] =|<\psi_n|\psi_m>|^2 than <\psi_n|\psi_m> itself; although density matrices should in principle contain all relavant information (also corresponding to eg. a W-function). But I suspect the issue is that the vn-entropy depends also on correlators with more than two density matrices... — Wouter Aug 27 '19 at 7:06
As for the question itself: did you change the $\ket{\psi}$ accordingly when you changed the $\phi$'s (especially important for the first plot)? — Ruslan Jan 19 '15 at 11:57
I think it's kinda a relationship of proportionality that is $\biik{1}{U(\delta t)}{2} \propto \biik{2}{H}{1}= k\cdot \biik{2}{H}{1}$ where $k= -\dfrac{i}{\hbar}$ for $\delta t \to 0$. Isn't it? So, more the $H$, more would be the amplitude to go from $\ket 1$ to $\ket 2$ after $\delta t$. Am I concluding right? But what about the '$-$' sign of $k$? Wouldn't it decrease the amplitude $\biik{1}{U}{2}$ if $\biik{2}{H}{1} \gt 0 \implies k\cdot \biik{2}{H}{1} \lt 0$? — user36790 Sep 26 '15 at 16:11
@user36790 If there was already some amplitude on $\ket{2}$, you should keep track of the RELATIVE amplitude. But if not, yeah, signs don't matter so much. — Jahan Claes Sep 26 '15 at 20:04
Also if someone could help me with the bra-ket syntax I would appriciate it. \ket{} did not work. — Tsangares Apr 17 '18 at 2:41
I found two instances where the macro is defined in the question and used in the answer. (And the answer does not contain the macro definition.)
Or, even better, define $U$ so that $U\k{dead}=+\frac1{\sqrt2}\left(\k{alive}-\k{dead}\right)$, then you'd get $U^2\k{alive}=\k{alive}$, i.e. 100% revival :) . — Ruslan Jul 2 '15 at 9:44
@garyp: What I meant to say is that $\k{b}{a}$ is the amplitude of an electron in state $a$ to get into state $b$. — user36790 Oct 11 '15 at 15:46
I think when $$H = \begin{pmatrix} E_{11} & 0 \\ 0 & E_{22} \end{pmatrix}$$, then only $E_{11}$ & $E_{22}$ are the eigenvalues that is the energies of $\k 1$ & $\k 2.$ — user36790 Sep 27 '15 at 11:20
So, $H_{12}$ & $H_{21}$ aren't energy of transitions, is it? Then why do $H_{11}$ & $H_{22}$ are energies for $\k 1$ & $\k 2$? — user36790 Sep 27 '15 at 11:18
search posts physics Posts with \op and without \operatorname, \oplus: data.stackexchange.com/physics/query/1071496/…
search comments physics No comments with \op and without \operatorname, \oplus: data.stackexchange.com/physics/query/1071502/…
 
 
6 hours later…
Martin Sleziak
Martin Sleziak
10:59 AM
I have previously mentioned this problem briefly in one of Physics chatrooms: chat.stackexchange.com/transcript/13775/2019/2/1
in Backup Room – The h Bar, Feb 1 '19 at 14:49, by Martin Sleziak
Actually I originally came here just to mention this: Problem with posts and comments relying on macros defined elsewhere (MathOverflow Meta). I suppose that the same problem might appear in some posts and comments on Physisc, too.
in Backup Room – The h Bar, Feb 1 '19 at 14:51, by Martin Sleziak
Some possible examples: Posts and comments with \ket, ...
 
 
1 hour later…
Martin Sleziak
Martin Sleziak
12:07 PM
This answer uses the macro \fsl defined in the question: Squares difference with gamma matrix - physics.stackexchange.com/posts/407738/revisions
One comment on the question "Squares difference with gamma matrix" where \fsl is not rendered.
Hint: what is $\fsl{p}\;\fsl{p}$? — probably_someone May 23 '18 at 8:37
BTW cannot something similar be achieved using $\require{cancel} \cancel p$ instead of the more complicated `\newcommand{\fsl}[1]{#1\kern-0.4em\raise0.22ex\hbox{/}}?
 
 
8 hours later…
Martin Sleziak
Martin Sleziak
8:02 PM
search posts physics Posts with \d, but without \dd (\ddots), \de (\delta), etc.: data.stackexchange.com/physics/query/1174849/…
search comments physics Comments with \d and without \dd, \de, etc.: data.stackexchange.com/physics/query/1174850/…
I did not find comments with \d - but I stumbled upon some comments with \dv, \dS:
I know $$v_y = \dv{r_y}{t} \hat{j}$$ is 0 in this case. You missed out unit vectors. i wanted to ask what will be walue of $r_x$ so that we differentiate to get the answer. Thanks! — Max Payne Jun 18 '15 at 11:13
@VINAY. In electricity one thinks about one dimensional wires, not three dimensional. The section $\dS$, as a vector, is oriented: there is a normal vector, that I called $\mathbf e_x$, that defines the direction in which the charge is counted. If you change $\mathbf e_x$ into $-\mathbf e_x$, you change the sign of the current densities and therefore the signs of all the currents I have computed. — Tom-Tom Jan 16 '14 at 10:58
 

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