The Classical Channel

meowzz

12:18 AM

@glS would u mind explaining why u removed the qubyte tag from my qubyte post?

glS

6:45 AM

@meowzz because "qubyte" is not a thing.. Also it was used only for that question, making it pretty useless imo

Quantum Computing Meta

11:07 AM

0

Q: Why was John Duffield suspended?

I 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

11:28 AM

@Blue Could you merge the tags "phase-estimation" and "quantum-phase-estimation"?

Blue

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

I think they both convey the same meaning

Blue

@Nelimee I know. I'm saying that I could delete the "quantum-phase-estimation" tag instead

And replace it with "phase-estimation"

Nelimee

As you want :)

Blue

Okay, cool! Actually there are just too many tags beginning with "quantum"

It becomes a mess when you're searching for a particular tag

So I wanted to minimize the number of tags beginning with "quantum"

The quantum part is obvious since the site is about quantum computing ;)

Nelimee

Yup you are right :) I also added 3 tags excerpt, waiting for review =)

Blue

12:00 PM

Done! phase-estimation

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

Blue

@Mithrandir24601 Well, merging does create permanent tag synonyms I guess? Do we really need more synonyms beginning with "quantum"? :P

And yeah, it was simply a matter of 1 question, and that too, my question

@Nelimee In the review queue I approved all of them. Just approval from one more user is needed I suppose

Or @Mithrandir24601 could do it

Nelimee

Ha ok! :) Thanks =)

Mithrandir24601

@Blue ah no, that's a different thing :P

You can do that as part of the merge process but it's optional

Blue

@Mithrandir24601 Do what?

Mithrandir24601

12:09 PM

@Blue create synonyms

Blue

@Mithrandir24601 Oh, I didn't know that

@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

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

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

Blue

@Nelimee Corrected!

@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

The basic idea in the paper seems to be correct though

glS

@Blue I'm puzzled about the removing of details in the latest version though. Also, they published the work in a weird journal.

Blue

I was puzzled too :P

Nelimee

Yep, they removed nearly all the interesting material x)

Blue

But well, if we can make it work, we're done ;)

Nelimee

12:42 PM

And their justification is not really clear

glS

@Nelimee is it included in the published version (if you checked that)?

Nelimee

But after a few corrections, it worked so it's fine x)

What is included?

glS

@Nelimee the details removed in the third arxiv version

Nelimee

@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 :)

Blue

This is the published paper: tandfonline.com/doi/abs/10.1080/00268976.2012.668289

Based on version 2 it seems

Nelimee

12:45 PM

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"

Blue

@Nelimee Try scihub

It's behind a paywall

@glS The journal published version of the paper seems to be exactly same as version 2 on arXiv

They didn't publish version 3 in a journal it seems

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

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

Blue

Right, the converse isn't true

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

12:49 PM

And... Every matrix is diagonalisable in $\mathbb{C}$ doesn't it?

Blue

Not every

There's some condition

But "most" are

Yes

8

Q: Proving "almost all matrices over C are diagonalizable".

This 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

Linear algebra courses were 3 years ago, I may need a refresh x)

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

Blue

Eh, defining eigenspaces will be tricky if you're allowing for complex eigenvalues

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

12:54 PM

@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

The matrix $e^{iAt}$ is not unitary

Do you have any proof of your "only hermitian matrix works"?

glS

ok, let's rephrase. If $A$ and $e^{iA}$ are normal, then $e^{iA}$ is unitary iff $A$ is Hermitian

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

Blue

@glS Ah, you're right indeed

@Nelimee If $U=e^{iAt}$, then what would $U^{\dagger}$ be if $A$ is not Hermitian?

Nelimee

$U^\dagger = e^{(iA)^\dagger t}$

Blue

1:04 PM

We could expand $e^{iAt}=I+it A + (it)^2A^2/2! + ...$ if I'm not wrong

So, all the $A$'s get conjugate transposed

@Nelimee Looks right

Then $UU^{\dagger}= ?$

$U^{\dagger} = e^{-iA^{\dagger}t}$

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

A and A^dagger don’t commute in general tho

Sniped

Nelimee

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

@Nelimee the exponential map converges with all the nice properties so usually that's not a problem

Blue

@glS I was seeing how far we can go with that :P

Nelimee

1:08 PM

Yup, right for the exponential, I forgot it was this one ^^

So it's not obvious that only hermitian matrices works

Semiclassical

You also get a fairly nice product if for instance $[A,A^\dagger]=id$

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

with the raising/lowering operators of a QHO being the easiest example

Nelimee

@Semiclassical what does the notation with the square brackets means?

Matrix product?

Semiclassical

Commutator

[A,B]=AB-BA

Nelimee

1:11 PM

Nice! Thanks :)

Hummm, it should be $0$ instead of $Id$ no?

Semiclassical

No

Nelimee

I checked, it's OK, I need to re-do some calculus x)

Blue

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

Got to invoke the Spectral theorem it seems

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

Blue

But our question is precisely: Can exponential of non-Hermitian matrices be unitary?

glS

1:19 PM

@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

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

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

@glS not quite - non-Hermitian matrices can still have real eigenvalues

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

Ah, that's already been mentioned

glS

1:31 PM

@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

fair enough

Mithrandir24601

@Blue nope

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

@Semiclassical nice argument, that seems to work. Isn't this the same as proving that $e^{iA}e^{iB}=1$ implies $A=-B$?

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

1:43 PM

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

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

@Semiclassical isn't that the same? If $e^{iA}$ is unitary then $e^{iAt}$ is unitary for all $t$

or isn't it?

Semiclassical

hmm

that should be right, yes

...hmm

something feels suspicious there

glS

does $(U^t)^\dagger = (U^\dagger)^t$ hold for all $t\in\mathbb R$?

Semiclassical

For integer $a$, definitely. For non-integers...hmmmm

Semiclassical

2:23 PM

@glS yeah, no. see Emilio's example in h bar

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

@Semiclassical Yup, I meant that

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

Blue

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

Nice example, hmm

Semiclassical

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

0

Q: What are the possible ways to visualise large, entangled states?

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

quantum-entanglement quantum-state

Blue

2:38 PM

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

Ah, so this is wrong

Semiclassical

@QuantumComputing my snarky response to this would probably be "Does sufficiently-complicated algebra count as visualization?"

Blue

What does hold is: If $e^A$ is unitary, then $A$ is Hermitian (assuming $A$ is real)

But that $i$ messes it up

glS

@Blue right. This also reminds me a lot of the other example I was given in the math.SE post

Blue

Interesting

Semiclassical

well, if $A$ is real then $A$ being hermitian just means it's symmetric

and $e^A$ is orthogonal I guess

Blue

2:41 PM

@Semiclassical Umm, but the requirement is skew-symmetry? I'm messing it up again

Semiclassical

derp

yeah

@Blue but then in this case the statement should be that A is skew-hermitian

Blue

Right, makes sense

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

Of course (1) will be a stronger version

11 mins ago, by Semiclassical

note, though, that if that's $e^{i A}$ then $e^{i A t}\neq 1$ in general

This is again an interesting point

What should be the constraint on $t$ here? ^@Semiclassical

Semiclassical

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.

Blue

@Semiclassical Proof? (sorry I'm a bit hazy here)

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

yeah, that

Blue

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

yeah

sloppy

Blue

Got it, no prob

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

Unitary time evolution ftw

Mithrandir24601

@Semiclassical exactly

Semiclassical

3:40 PM

Unless your Hamiltonian is time-dependent. Then I just get a headache

Mithrandir24601

Although there's something to be said for quasi-unitary evolution as well :P

Semiclassical

(H at time 1 not commuting with H at time 2 is really a hassle)

Mithrandir24601

@Semiclassical my PhD is dealing with non-Hermitian stuff, so you'll have to try harder if you want any sympathy :P

Semiclassical

Lol

My first project was semiclassical quantization / level-splitting of periodic pt-symmetric potentials

Which somehow worked very well with numerics

Mithrandir24601

3:55 PM

@Semiclassical part of mine is simulating them, so fun times...

(PT-symmetric Hamiltonians, that is)

Semiclassical

Oof

The nice thing about the periodic case is that numerics are tractable (Fourier basis ftw)

Mithrandir24601

The moment you first discover that Green's functions break down is somewhat alarming, yeah

Semiclassical

4:21 PM

Oof

Yeah, sounds right

Non-hermitian spectral problems are generically pretty goofy

glS

4:51 PM

$X$ is skew-Hermitian iff all $\exp(sX)$ are unitary

Nelimee

Nice!

glS

I believe this is the result I was remembering

Nelimee

Is there any proof?

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?

Blue

9

A: Determinant of matrix exponential?

If 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 ...

Seems to be covered here ^

glS

4:58 PM

@Blue ah, great find!

Blue

@glS I was sort of expecting that after seeing SemiC's proof. I'll see it in more detail later, but thanks :)

glS

@Blue yes it's basically what was found above, but it's good to see it mentioned in an authoritative reference I guess

Quantum Computing Meta

11:51 PM

0

Q: Can a suspended user be allowed to award a bounty?

There'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

Transcript for

The Classical Channel