Talk:Heisenberg picture

I've added the technical tag, as this article seems to be too technical. There's no introduction to the subject, and it launches into the mathematical formulism without too much background. I'm tempted to add a "clean up" tag as well, as the article seems somewhat confusing and hard to follow. Thoughts? Privong 17:25, 24 August 2005 (UTC)Reply

I agree that the article is too advanced for the casual reader. But this is organized in as orderly a way as mathematics will allow. I will add an introduction and a brief historical section and remove the tag. Beyond the history of Matrix mechanics nothing can be said about it without "advanced" mathematics. Such is the nature of quantum physics. --Hfarmer 18:09, 25 August 2005 (UTC)Reply

Yea. I agree that, by the nature of the topic it's very technical. I was more looking for an introduction like the one you provided, so that the casual reader could at least get a sense of what the Heisenberg picture is, even if they don't understand the mathematics behind it. Thanks for adding the introduction! Privong 15:13, 1 September 2005 (UTC)Reply

To the original Author of the page I feel it would be best to give a more basic but mathematical presentation of the matterial. By more basic what I mean is a presentation of how to solve for the state vector given the Hamiltonian operator of the interaction. I will work on this and add such a section latter. --Hfarmer 18:42, 25 August 2005 (UTC)Reply

I am moving this material over to Matrix mechanics. There is no need for two articles and the topic is more commonly known by that name. DV8 2XL 20:50, 2 November 2005 (UTC)Reply

I probably don't agree with that. Matrix mechanics is really just a historical theory these days, whereas the Heisenberg picture is still used and discussed. -Lethe | Talk 06:44, 3 November 2005 (UTC)

I also disagree. The Heisenberg picture and Schrodinger picture (and the interaction picture) are different from any distinction between matrix mechanics and wave mechanics. --MarSch 13:00, 3 November 2005 (UTC)Reply

Heisenberg's quote "famous 'commutation relation' for the quantization condition that is at the basis of quantum mechanics" as shown on the web page:

[1]

I can't add it because I don't know how to make the math symbols, but I think it is important because it shows the difference between the Fourier series of amplitudes of position and momentum not to commute by a value of h/2pi. (h-bar) of intensity.--Voyajer 02:36, 27 December 2005 (UTC)Reply

Hi. I've noticed that this article's author have used Schrödinger equation to derive Heisenberg's. But as long as heisenberg's picture of quantum mechanics is complete by it's own, this is not really necessary...

I agree, deriving the Heisenberg picture from the Schrodinger picture doesn't make any more sense than deriving the Schrodinger picture from the Heisenberg picture. The article assumes there's something fundimental about attaching time dependence to psi, rather than the operator, and their isn't. Maybe somebody can replace this with something better. —Preceding unsigned comment added by 128.211.179.162 (talk) 22:58, 30 June 2010 (UTC)Reply

The derivation, is somewhat circular. Also, interestingly, there is no mention of where the (del A / del t)classical comes from, and, as it appears in the derivation, I think its wrong! Not to say its wrong but it comes up differently I think. —Preceding unsigned comment added by 131.94.41.41 (talk) 13:21, 4 January 2008 (UTC)Reply

Is this correct? « noting that ${\displaystyle {\frac {\partial A}{\partial t}}}$  is the time derivative of A(t), the transformed operator, not the one we started with. » I think ${\displaystyle {\frac {\partial A(t)}{\partial t}}}$  is not the same as ${\displaystyle e^{iHt/\hbar }{\frac {\partial A}{\partial t}}e^{-iHt/\hbar }}$  189.179.243.82 (talk) 06:55, 14 January 2011 (UTC)Reply

You are right, I fixed the mistake. Tank00 (talk) 09:56, 5 April 2012 (UTC)Reply

I hope you are satisfied hy the way I fixed it. The initial A is basically Schroedinger's A, a function of x, p, and explicit t. The partial derivative always indicates derivation with respect to this explicit t--In the overwhelming bulk of applications it vanishes; it is only there for time-dependent external fields, etc. Now after ∂A(x,p,t)/∂t is taken, the similarity transformation by exp(iHt) and its inverse on it converts all xs and ps to x(t)s and p(t)s, and thus the last term to ∂A(t)/∂t as asserted. I would recommend switching off all explicit time dependence first, and thus all partial time derivatives, to appreciate the convective canonical structure involved, and only then introduce it to take care of the special cases mentioned. Of course it is a technical issue, and unfortunately I cannot see how it could be made less technical, but the two QM texts referenced should settle any questions or misgivings. Cuzkatzimhut (talk) 21:02, 3 December 2012 (UTC)Reply

"Lorentz invariance is manifest in the Heisenberg picture."

Really? Like the Schrödinger equation, the Heisenberg equation has first order derivatives in t and second order derivatives in space (${\displaystyle H=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}}$ ), which doesn't look too good. I think this could use some backing up. 195.37.186.62 (talk) 03:05, 15 September 2010 (UTC)Reply

I put in a small clarification to the self-evident point: the state vectors |ψ> in this picture are Lorentz scalars, as they have no explicit time, or space!, dependence. The expressions one ends up getting are as Lorentz covariant or not as the respective transformation properties of the operators one chooses to hit them with. You are not going to get a Lorentz invariant equation by hitting them with a nonrelativistic Hamiltonian. If, on the other hand, you acted on them through a relativistically invariant QFT,..... Cuzkatzimhut (talk) 15:06, 18 October 2013 (UTC)Reply

The content is very poor and too informal. Can someone take the time to give a hand?88.243.167.81 (talk) 17:39, 10 September 2011 (UTC)Reply

In agreement, I tried something. Properly, the article is a footnote in a standard text; however, it is mostly of interest to quantum field theory students and those of the classical correspondence. If they are here, it is probably because their texts failed them, and skipped it by overexposing them to the Schroedinger picture. This is the (valid) reason for introducing it through the Schroedinger equation/picture, contrasting to complaints above.Cuzkatzimhut (talk) 21:52, 2 December 2012 (UTC)Reply

The article lacks results from recent researches, which could be useful for the readers. It's too retrospective. — Preceding unsigned comment added by Waterwizardm (talk • contribs) 11:26, 9 June 2020 (UTC)Reply

(edit conflict) Like what?--ReyHahn (talk) 11:34, 9 June 2020 (UTC)Reply

I'm discussing this mathematical box:

"In the Heisenberg picture of quantum mechanics the state vectors |ψ〉 do not change with time, while observables A satisfy

where H is the Hamiltonian and [•,•] denotes the commutator of two operators (in this case H and A). Taking expectation values automatically yields the Ehrenfest theorem, featured in the correspondence principle."

After a google search, I've decided that the H and S subscripts probably stand for "Heisenberg" and "Schrödinger", and that this equation is relating definitions of variables between the two pictures, but that is not at all clear from the article, and I don't want to add this in, as I have no experience with these concepts and I'm just guessing based on a google search and common sense. If someone can confirm that I've guessed correctly, I would strongly suggest that this be clarified in the text. Equations are often more informative than text, but not defining all of the variables is the easiest way to make equations absolutely useless to the reader.98.17.108.86 (talk) 06:43, 14 January 2021 (UTC)Reply

You are right, the labels refer to the name of the picture. Thanks for noticing.--ReyHahn (talk) 19:15, 14 January 2021 (UTC)Reply