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  How do we compute correlation function in the Schrodinger picture?

+ 4 like - 0 dislike
418 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 (13,989 points) [ revision history ]
edited Jun 23, 2016 by Arnold Neumaier

The Schroedinger equation can be safely solved with help of Green's function containing two times, no problem.

If the Heisenberg picture is fundamental, then any other derived picture is as fundamental as the original one. No variable change can make the resulting picture less fundamental.

The Heisenberg picture is also "derived" when the second quantization is considered (introduced).

@VladimirKalitvianski: The Greens function is known only for exactly solvable problems. Most problems don't belong to this class.

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.
 

@wzdlc1996: Are you serious?

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.

Indeed, we need the initial state, just like in any other QM representation which all are thus equivalent.

@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.

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