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meowzz
meowzz
12:18 AM
@glS would u mind explaining why u removed the qubyte tag from my qubyte post?
 
 
6 hours later…
glS
glS
6:45 AM
@meowzz because "qubyte" is not a thing.. Also it was used only for that question, making it pretty useless imo
 
 
4 hours later…
Quantum Computing Meta
Quantum Computing Meta
11:07 AM
0
Q: Why was John Duffield suspended?

GravitonI came across John Duffield Quantum Computing SE via this hot question. I was curious to see an account with 1 reputation and a question with hundreds of upvotes. It turned out that the reason why he has so little reputation despite a massively popular question is that he was suspended. May I ...

discussion
 
Nelimee
Nelimee
11:28 AM
@Blue Could you merge the tags "phase-estimation" and "quantum-phase-estimation"?
 
Anonymous
11:56 AM
@Nelimee Do we need to merge? Currently, there's just one question with "phase-estimation" and another question with "quantum-phase-estimation". Might we as well use just one tag? (say just "phase-estimation")
 
Nelimee
Nelimee
I think they both convey the same meaning
 
Anonymous
@Nelimee I know. I'm saying that I could delete the "quantum-phase-estimation" tag instead
 
Anonymous
And replace it with "phase-estimation"
 
Nelimee
Nelimee
As you want :)
 
Anonymous
Okay, cool! Actually there are just too many tags beginning with "quantum"
 
Anonymous
11:58 AM
It becomes a mess when you're searching for a particular tag
 
Anonymous
So I wanted to minimize the number of tags beginning with "quantum"
 
Anonymous
The quantum part is obvious since the site is about quantum computing ;)
 
Nelimee
Nelimee
Yup you are right :) I also added 3 tags excerpt, waiting for review =)
 
Anonymous
 
Mithrandir24601
Mithrandir24601
@Blue 'merging', if I'm getting the terms right, is a specific single action that does exactly that and is generally preferable to editing tags on questions. Having said that, if it's just one question, it doesn't really matter although performing a proper merge is still probably preferable
 
Anonymous
12:01 PM
@Mithrandir24601 Well, merging does create permanent tag synonyms I guess? Do we really need more synonyms beginning with "quantum"? :P
 
Anonymous
And yeah, it was simply a matter of 1 question, and that too, my question
 
Anonymous
@Nelimee In the review queue I approved all of them. Just approval from one more user is needed I suppose
 
Anonymous
Or @Mithrandir24601 could do it
 
Nelimee
Nelimee
Ha ok! :) Thanks =)
 
Mithrandir24601
Mithrandir24601
@Blue ah no, that's a different thing :P
You can do that as part of the merge process but it's optional
 
Anonymous
12:08 PM
@Mithrandir24601 Do what?
 
Mithrandir24601
Mithrandir24601
@Blue create synonyms
 
Anonymous
@Mithrandir24601 Oh, I didn't know that
 
Anonymous
@Mithrandir24601 Are synonymous tags on a question automatically replaced by the master tag? If so, you could add "quantum-phase-estimation" as a synonym for "phase-estimation"
 
Mithrandir24601
Mithrandir24601
Merging is taking all the questions with a specific tag and replacing that tag with a different one, on all those questions, on a tag level, without permanently changing anything about the underlying tags
@Blue yeah, you could do that. It generally requires votes, so it's probably not worth bothering when only one question has that tag
 
Nelimee
Nelimee
12:35 PM
@Blue You have a typo in quantumcomputing.stackexchange.com/questions/2564/… : the last Rzz is $R_zz$ and should be $R_{zz}$ :p
 
Anonymous
@Nelimee Corrected!
 
Anonymous
@glS That paper does have a few mistakes. We were discussing that. But on making some corrections, it does give the correct solution for the system of linear equations
 
Anonymous
The basic idea in the paper seems to be correct though
 
glS
glS
@Blue I'm puzzled about the removing of details in the latest version though. Also, they published the work in a weird journal.
 
Anonymous
I was puzzled too :P
 
Nelimee
Nelimee
12:42 PM
Yep, they removed nearly all the interesting material x)
 
Anonymous
But well, if we can make it work, we're done ;)
 
Nelimee
Nelimee
And their justification is not really clear
 
glS
glS
@Nelimee is it included in the published version (if you checked that)?
 
Nelimee
Nelimee
But after a few corrections, it worked so it's fine x)
What is included?
 
glS
glS
@Nelimee the details removed in the third arxiv version
 
Nelimee
Nelimee
12:44 PM
@glS "Every hermitian matrix satisfy this property: more specifically, all and only Hermitian matrices have this property" ha? I though it was only a subset of the set of valid matrices ^^ Thanks for the precision :)
 
Anonymous
 
Anonymous
Based on version 2 it seems
 
Nelimee
Nelimee
I'm not sure I understood your question... Are the removed parts published with the new paper? <- this is your question?
@Blue "Server is busy"
 
Anonymous
@Nelimee Try scihub
 
Anonymous
It's behind a paywall
 
Anonymous
12:47 PM
@glS The journal published version of the paper seems to be exactly same as version 2 on arXiv
 
Anonymous
They didn't publish version 3 in a journal it seems
 
glS
glS
@Nelimee if you think about it it's quite easy to see. Unitary matrices are the ones with phases as eigenvalues, while Hermitians have real eigenvalues. Therefore, if a matrix is not Hermitian (does not have real eigenvalues), then its exponential will not have eigenvalues of the form $e^{i\phi}$ with $\phi\in\mathbb R$. Although I'm not sure whether there could be exceptions for non diagonalizable matrices (if $A$ is not diagonalizable, then the above argument doesn't work)
 
Nelimee
Nelimee
A matrix is not Hermitian does not mean that it does not have real eigenvalues. Or I completly missed something
Hermitian => real eigenvalues
real eigenvalues =/> Hermitian
 
Anonymous
Right, the converse isn't true
 
glS
glS
@Nelimee if a matrix $A$ is unitarily diagonalizable with real eigenvalues, then $A=U\Lambda U^\dagger$ with $\Lambda$ diagonal and real. Then, $A^\dagger = U\Lambda^\dagger U^\dagger = A$
 
Nelimee
Nelimee
12:49 PM
And... Every matrix is diagonalisable in $\mathbb{C}$ doesn't it?
 
Anonymous
Not every
 
Anonymous
There's some condition
 
Anonymous
But "most" are
 
Anonymous
Yes
 
Anonymous
8
Q: Proving "almost all matrices over C are diagonalizable".

AnweshiThis is an elementary question, but a little subtle so I hope it is suitable for MO. Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$. The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form: $$ J = \begin...

linear-algebra measure-theory
 
Nelimee
Nelimee
12:51 PM
Linear algebra courses were 3 years ago, I may need a refresh x)
 
glS
glS
@Nelimee no! unitarily diagonalizable matrices are all and only the normal ones (satisfying $AA^\dagger =A^\dagger A$). For general diagonalizability if I'm not mistaken onecharacterization is that the sum of the dimensions of the eigenspaces has to match the total dimension
 
Anonymous
Eh, defining eigenspaces will be tricky if you're allowing for complex eigenvalues
 
Nelimee
Nelimee
$$A = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 2 & -4 & 2\end{pmatrix}$$ has real eigenvalues $\{1, 2, 3\}$ from en.wikipedia.org/wiki/… but is not Hermitian
I don't know if $e^{iAt}$ is unitary though
 
glS
glS
@Blue how so? given $\lambda\in\mathbb C$ then the associated eigenspace is $\{v: \,\, (A-\lambda)v=0\}$
@Nelimee indeed, my proof used unitarily diagonalizable matrices. You are right that the argument fails for generally diagonalizable ones
 
Nelimee
Nelimee
The matrix $e^{iAt}$ is not unitary
Do you have any proof of your "only hermitian matrix works"?
 
glS
glS
12:57 PM
ok, let's rephrase. If $A$ and $e^{iA}$ are normal, then $e^{iA}$ is unitary iff $A$ is Hermitian
 
Nelimee
Nelimee
I still don't see why "iff", but this is what you wrote in the comment yes!
Wait
Nope, still don't see it
 
Anonymous
@glS Ah, you're right indeed
 
Anonymous
@Nelimee If $U=e^{iAt}$, then what would $U^{\dagger}$ be if $A$ is not Hermitian?
 
Nelimee
Nelimee
$U^\dagger = e^{(iA)^\dagger t}$
 
Anonymous
We could expand $e^{iAt}=I+it A + (it)^2A^2/2! + ...$ if I'm not wrong
 
Anonymous
1:05 PM
So, all the $A$'s get conjugate transposed
 
Anonymous
@Nelimee Looks right
 
Anonymous
Then $UU^{\dagger}= ?$
 
Anonymous
$U^{\dagger} = e^{-iA^{\dagger}t}$
 
glS
glS
@Blue I actually agree with Nelimee here that it's not that easy. You get $UU^\dagger = e^{iA} e^{-iA^\dagger}$, but if $A$ and $A^\dagger$ do not commute it's not straightforward that this doesn't give you an identity
 
Semiclassical
Semiclassical
A and A^dagger don’t commute in general tho
Sniped
 
Nelimee
Nelimee
1:07 PM
And applying an operator on an infinite sum needs some checks
Infinites sums are bad boys, you don't want to manipulates them too much :p
 
glS
glS
@Nelimee the exponential map converges with all the nice properties so usually that's not a problem
 
Anonymous
@glS I was seeing how far we can go with that :P
 
Nelimee
Nelimee
Yup, right for the exponential, I forgot it was this one ^^
So it's not obvious that only hermitian matrices works
 
Semiclassical
Semiclassical
You also get a fairly nice product if for instance $[A,A^\dagger]=id$
 
glS
glS
I'm getting confused. I remember there being some theorem about one-to-one mappings between unitaries and hermitians provided by the exponential, but it was some time ago and may be confusing things in my head
 
Semiclassical
Semiclassical
1:10 PM
with the raising/lowering operators of a QHO being the easiest example
 
Nelimee
Nelimee
@Semiclassical what does the notation with the square brackets means?
Matrix product?
 
Semiclassical
Semiclassical
Commutator
[A,B]=AB-BA
 
Nelimee
Nelimee
Nice! Thanks :)
Hummm, it should be $0$ instead of $Id$ no?
 
Semiclassical
Semiclassical
No
 
Nelimee
Nelimee
I checked, it's OK, I need to re-do some calculus x)
 
Anonymous
1:15 PM
There's a theorem that every unitary matrix $U$ can be written in the form $U=e^{iA}$ where $A$ is Hermitian though. But $A$ isn't unique
 
Anonymous
Got to invoke the Spectral theorem it seems
 
glS
glS
@Nelimee if there is a $0$ there then it becomes the normality condition. Otherwise it means that the matrix is not normal, therefore not unitarily diagonalizable, but still the product of exponentials is relatively easy to write
 
Anonymous
But our question is precisely: Can exponential of non-Hermitian matrices be unitary?
 
glS
glS
@Blue you are right indeed. If $U$ is unitary then for sure you can write it as exponential of an Hermitian (time $i$). This is easily proven because $U$ is ensured to be unitarily diagonalizable, so you can simply compute it's logarithm through the eigenvalues. However, logarithms are tricky and multivalued, and there may be logarithms which are not diagonalizable at all.
I've actually recently asked some questions on math.SE on related topics
 
Nelimee
Nelimee
I will need to go (I need to work a little before France-Uruguay :p), but I follow the discussion here, even if I do not speak :p
 
glS
glS
1:20 PM
for example, Are all square roots of diagonalizable matrices diagonalizable?. I've yet to fully understand the last answer though, which I believe is very interesting
given that the square root in the complex plane is essentially defined through the logarithm, there probably are similarities between the two questions
I would argue that if the logarithm of a unitary $U$ is diagonalizable, then it is ensured to be (anti-)Hermitian
 
Mithrandir24601
Mithrandir24601
@glS not quite - non-Hermitian matrices can still have real eigenvalues
 
Semiclassical
Semiclassical
@Blue presumably you mean that in the sense of $e^{i A}$ with $A$ non-hermitian. ($A$ skew-hermitian $\implies e^A$ is unitary)
 
Mithrandir24601
Mithrandir24601
Ah, that's already been mentioned
 
glS
glS
@Mithrandir24601 indeed, that was also what @Nelimee showed with an example above. I believe my argument holds for unitarily diagonalizable matrices. If a matrix is only generally diagonalizable (so it's not normal) then it's not true
 
Semiclassical
Semiclassical
fair enough
 
Mithrandir24601
Mithrandir24601
1:33 PM
@Blue nope
 
Semiclassical
Semiclassical
hmm, suppose $U(t)=e^{i A t}$ and $U^\dagger(t)U(t)=id$. Then $\frac{d}{dt}U^\dagger(t)U(t) = e^{-i A^\dagger t}(A-A^\dagger)e^{i A t}=\frac{d}{dt}id=0$
and in particular for $t=0$ one has $A=A^\dagger$
so if that differentiation is legal then I'm not sure one can avoid it
 
glS
glS
@Semiclassical nice argument, that seems to work. Isn't this the same as proving that $e^{iA}e^{iB}=1$ implies $A=-B$?
 
Semiclassical
Semiclassical
quite probably
i mean, if you modify that to $e^{i A t}e^{i B t}=1$ then the same argument should apply
 
glS
glS
also probably even more generally without $i$ factors
so, in conclusion, it does indeed seem that $e^{iA}$ unitary implies $A$ Hermitian. It therefore also seems that $e^{iA}$ unitary implies $A$ normal, so that also my argument passing through the spectra works (though one has to show that $A$ is ensured to be normal)
 
Semiclassical
Semiclassical
1:49 PM
though, my argument only works if $e^{i A t}e^{i B t}$ for all $t$ (or at least all $t$ within a nbhd including $t=0$)
 
glS
glS
@Semiclassical isn't that the same? If $e^{iA}$ is unitary then $e^{iAt}$ is unitary for all $t$
or isn't it?
 
Semiclassical
Semiclassical
hmm
that should be right, yes
...hmm
something feels suspicious there
 
glS
glS
does $(U^t)^\dagger = (U^\dagger)^t$ hold for all $t\in\mathbb R$?
 
Semiclassical
Semiclassical
For integer $a$, definitely. For non-integers...hmmmm
 
Semiclassical
Semiclassical
2:23 PM
@glS yeah, no. see Emilio's example in h bar
 
Anonymous
@Mithrandir24601 For real matrices $A$, it is true that if $e^{iA}$ is unitary, it implies $A$ is Hermitian. But I don't think we can say the same for complex matrices
 
Anonymous
@Semiclassical Yup, I meant that
 
Semiclassical
Semiclassical
the fact that $e^{2\pi i}=1$ doesn't imply $2\pi i=0$ made me wonder if there was an escape clause
and EP's example bears that out
 
Anonymous
in The h Bar, 44 mins ago, by Emilio Pisanty
$$\exp\left(i\begin{pmatrix} 2\pi & 1 \\ 0 & 0 \end{pmatrix}\right) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ according to Mathematica
 
Anonymous
Nice example, hmm
 
Semiclassical
Semiclassical
2:33 PM
yeah
it's a good one
note, though, that if that's $e^{i A}$ then $e^{i A t}\neq 1$ in general
 
Quantum Computing
Quantum Computing
0
Q: What are the possible ways to visualise large, entangled states?

SLesslyTallWhat are the prominent visualisations used to depict large, entangled states and in what context are they most commonly applied?

quantum-entanglement quantum-state
 
Anonymous
8 mins ago, by Blue
@Mithrandir24601 For real matrices $A$, it is true that if $e^{iA}$ is unitary, it implies $A$ is Hermitian. But I don't think we can say the same for complex matrices
 
Anonymous
Ah, so this is wrong
 
Semiclassical
Semiclassical
@QuantumComputing my snarky response to this would probably be "Does sufficiently-complicated algebra count as visualization?"
 
Anonymous
What does hold is: If $e^A$ is unitary, then $A$ is Hermitian (assuming $A$ is real)
 
Anonymous
2:39 PM
But that $i$ messes it up
 
glS
glS
@Blue right. This also reminds me a lot of the other example I was given in the math.SE post
 
Anonymous
Interesting
 
Semiclassical
Semiclassical
well, if $A$ is real then $A$ being hermitian just means it's symmetric
and $e^A$ is orthogonal I guess
 
Anonymous
@Semiclassical Umm, but the requirement is skew-symmetry? I'm messing it up again
 
Semiclassical
Semiclassical
derp
yeah
@Blue but then in this case the statement should be that A is skew-hermitian
 
Anonymous
2:42 PM
Right, makes sense
 
Anonymous
Now what we need to look for is 1) The exact set of conditions for which the matrix exponential $e^A$ of a complex matrix $A$, is unitary 2) The exact set of conditions for which the matrix exponential $e^{iA}$ of a real matrix $A$ is unitary
 
Anonymous
Of course (1) will be a stronger version
 
Anonymous
11 mins ago, by Semiclassical
note, though, that if that's $e^{i A}$ then $e^{i A t}\neq 1$ in general
 
Anonymous
This is again an interesting point
 
Anonymous
What should be the constraint on $t$ here? ^@Semiclassical
 
Semiclassical
Semiclassical
2:48 PM
dunno. my story was in the opposite direction: If it's true that $e^{i At}$ is unitary for all $t$, then $A$ is hermitian.
 
Anonymous
@Semiclassical Proof? (sorry I'm a bit hazy here)
 
Anonymous
1 hour ago, by Semiclassical
hmm, suppose $U(t)=e^{i A t}$ and $U^\dagger(t)U(t)=id$. Then $\frac{d}{dt}U^\dagger(t)U(t) = e^{-i A^\dagger t}(A-A^\dagger)e^{i A t}=\frac{d}{dt}id=0$
 
Semiclassical
Semiclassical
yeah, that
 
Anonymous
3:07 PM
@Semiclassical Shouldn't that be $e^{-i A^\dagger t}(iA-iA^\dagger)e^{i A t}$? I think there's an $i$ missing there
 
Semiclassical
Semiclassical
yeah
sloppy
 
Anonymous
Got it, no prob
 
Mithrandir24601
Mithrandir24601
3:25 PM
@Blue fair enough - as with @Semiclassical I was thinking about it with the t parameter, as that's what we care about in physics :P I can possibly come up with a number of non-Hermitian matrices that gives unitary evolution for a specific t
Or rather, the exponential of which is unitary for $t+n\tau$, although I'd need to check
 
Semiclassical
Semiclassical
Unitary time evolution ftw
 
Mithrandir24601
Mithrandir24601
@Semiclassical exactly
 
Semiclassical
Semiclassical
3:40 PM
Unless your Hamiltonian is time-dependent. Then I just get a headache
 
Mithrandir24601
Mithrandir24601
Although there's something to be said for quasi-unitary evolution as well :P
 
Semiclassical
Semiclassical
(H at time 1 not commuting with H at time 2 is really a hassle)
 
Mithrandir24601
Mithrandir24601
@Semiclassical my PhD is dealing with non-Hermitian stuff, so you'll have to try harder if you want any sympathy :P
 
Semiclassical
Semiclassical
Lol
My first project was semiclassical quantization / level-splitting of periodic pt-symmetric potentials
Which somehow worked very well with numerics
 
Mithrandir24601
Mithrandir24601
3:55 PM
@Semiclassical part of mine is simulating them, so fun times...
(PT-symmetric Hamiltonians, that is)
 
Semiclassical
Semiclassical
Oof
The nice thing about the periodic case is that numerics are tractable (Fourier basis ftw)
 
Mithrandir24601
Mithrandir24601
The moment you first discover that Green's functions break down is somewhat alarming, yeah
 
Semiclassical
Semiclassical
4:21 PM
Oof
Yeah, sounds right
Non-hermitian spectral problems are generically pretty goofy
 
glS
glS
4:51 PM
$X$ is skew-Hermitian iff all $\exp(sX)$ are unitary
 
Nelimee
Nelimee
Nice!
 
glS
glS
I believe this is the result I was remembering
 
Nelimee
Nelimee
Is there any proof?
 
glS
glS
@Blue @Semiclassical basically it holds exactly in the case as per @Semiclassical proof
@Nelimee the excerpt is from pag. 7 of pi.math.cornell.edu/~hatcher/Other/Samelson-LieAlg.pdf
@Nelimee they say it's because $\exp\operatorname{tr}(X)=\det \exp X$, but they don't prove this. This fact is easily shown for diagonalizable matrices but I'm not sure how to prove it in general
actually on second thought, maybe the proof is exactly the one @Semiclassical gave above?
 
Anonymous
9
A: Determinant of matrix exponential?

JulienIf you're afraid of the density of diagonalizable matrices, simply triangularize $A$. You get $$A=P^{-1}UP,$$ with $U$ upper triangular and the eigenvalues $\{\lambda_j\}$ of $A$ on the diagonal. Then $$ \mbox{det}\;e^A=\mbox{det}(P^{-1}e^UP)=\mbox{det}\;e^U. $$ Now observe that $e^U$ is upper ...

 
Anonymous
4:58 PM
Seems to be covered here ^
 
glS
glS
@Blue ah, great find!
 
Anonymous
@glS I was sort of expecting that after seeing SemiC's proof. I'll see it in more detail later, but thanks :)
 
glS
glS
@Blue yes it's basically what was found above, but it's good to see it mentioned in an authoritative reference I guess
 
 
7 hours later…
Quantum Computing Meta
Quantum Computing Meta
11:51 PM
0
Q: Can a suspended user be allowed to award a bounty?

user1271772There's 15 hours left on a bountied question, but the person who offered the bounty is suspended and his suspension doesn't expire until about 2 days, meaning he may not be able to award the bounty himself? That's not fair: It's a 300 point bounty. The largest bounty ever offered on QCSE. Let h...

discussion bounty account-suspension
 

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