Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,103 questions , 2,249 unanswered
5,355 answers , 22,797 comments
1,470 users with positive rep
820 active unimported users
More ...

  How do we compute correlation function in the Schrodinger picture?

+ 4 like - 0 dislike
3333 views

From concreteness' sake consider $\phi^4$ theory with a real scalar (even though the choice of the theory has nothing to do in principle with what I am going to ask).

Consider thefollowing correlation function
$$
<\Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega>
$$
My question is, do we have to understand the formula above as given in the Heisenberg picture, the Schrodinger picture, or it doesn't matter? What motivates this question is that if we were in the Schrodinger picture i wouldn't know how to take the time ordering of the fields. On the other hand, if it really is meant to be understood in the Heisenberg picture, it is dissatisfying for me not to have a means to compute correlation functions in the Schrodinger picture. So which one is it?

asked Jun 22, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

In order to wonder about such an expression, you have to obtain (derive, encounter) it naturally in your calculations. Then you is certain what picture is used.

1 Answer

+ 3 like - 0 dislike

The fact that $\phi(x)$ has the time in the operator identifies the formula as one in the Heisenberg picture. In the Schroedinger picture, operators are time independent, unless time occurs explicitly as parameter in the interaction.

Even outside of field theory, time correlations are awkward to represent in the Schroedinger picture, where only a single time is natural. One can put one of the times into the Schroedinger state, and the time difference remains in the operator expression, as a conjugation by exponentiated Hamiltonians.

Moreover, the meaning as a time correlation is apparent only in the Heisenberg picture in the Schroedinger picture it is just an obscure formal expression without ready interpretation. Thus anything using multiple times is natural only in the Heisenberg picture (although formally one can rewrite it also in the Schroedinger picture).

This shows that the Heisenberg picture is the fundamental one, and the Schroedinger picture is derived.

answered Jun 22, 2016 by Arnold Neumaier (15,787 points) [ revision history ]
edited Jun 23, 2016 by Arnold Neumaier
Most voted comments show all comments

@wzdlc1996: In the Heisenberg picture, one has a time-independent state which provides expectation values to all space-time dependent operators and operator products. The dynamics is in the operators. 

In the Schrödinger picture, the dynamics is in the states, which breaks manifest Poincare invariance, and does not allow one to describe time correlations - unless one introduces these in the Heisenberg picture and then transforms the resulting expressions to the Schrödinger picture. Thus one needs the Heisenberg picture within the Schrödinger picture, which makes the Heisenberg picture fundamental. 

@ArnoldNeumaier @VladimirKalitvianski

No, may be you misunderstood my mean. I think there are many cases in which we need a time-dependent wavefunction. And in fact these cases are directly related to the fundamental of quantum mechanics. Like measurement, quantum statistical mechanics(especially thermalization). Time-independent wavefunction and time-dependent operator, or correlation function are not complete either.

@wzdlc1996: For problems not involving time correlations, the Schrödinger and the Heisenberg picture are equivalent. In this case, the former is often easier to use, but not necessary! Thus time-dependent wave functions are never essential.

Quantum statistical mechanics is never done with wave functions but always with density operators, and the technically most versatile approach to time-dependent statistical mechanics uses only the Heisenberg picture. See the book by Calzetta and Hu.

@ArnoldNeumaier Sometimes one need to discuss the thermalization of isolated system, especially in canonical typicality, in which pure state but not density operator is involved. I think time-dependent wavefunction is essential in some situation but correlation function is not, like the usage of Many-world interpretation. (but this is still an open issue)

But anyway I finally got your point. In the sense of the application and theory of correlation function Heisenberg Picture is fundamental one but Schrodinger Picture is not. And now I think these two play different roles in quantum mechanics but it is hard to say which one is universally fundamental or which one is more fundamental than the other in any situation. What do you think about my opinion?

@wzdlc1996: For a closed system the true thermalization may never happen. For example, you excite one "energy level" in a many-level internally interacting system, and with time the other energy levels get "occupied", but with the occupation number amplitudes still time-dependent and far from a thermal (Gibbs) distribution.

Most recent comments show all comments

Exact (but unknown) Green's function $G(x_1,x_2)$ is expressed via exact (but unknown) solutions $\phi_n(x)$ where you can see the same structure as a "correlation function". There is a perturbation theory for GF similar to that for $\phi_n(x)$, namely, it contains nothing else but the perturbative expansion of each exact $E_n$ and $\psi_n$ in the exact $\phi_n(x)=e^{-iE_n t}\psi_n({\bf{x}}) $.

As far as I know, Schrodinger Picture is the way to describe the Quantum dynamics in which all dynamics regulation is "encoded" by the time-dependent wave function. And it is directly related to the measurement theory in Quantum mechanics. Moreover, many theories shows that the wavefunction-formalism is deeply related to the classical-quantum corresponding (e.g. KvN classicle mechanics) and statistical mechanics (e.g. entropy and thermalization for isolated system) . I think it is reasonable to say that Schrodinger Picture is a more direct way to understand the quantum mechanics, and it is likely to be the fundamental one. 

Then, if the Heisenberg picture is the fundamental one, how does it perform a complete dynamics theory? We know that operator are not complete to describe a system. We still need the (initial) state in Heisenberg picture or we can extract nothing!. In this view, Heisenberg picture looks like not a fundamental one.
 

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$y$\varnothing$icsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...