# Relations between classical and quantum time-evolution of fields

1. Feb 21, 2012

### kof9595995

This question is gonna be a bit vague and might lead to nowhere, but still I'll take the risk and try to ask it here.
I know in general how to quantize a field, and from the quantized field one gets the quantized Hamitonian thus the time-evolution operator. However, I wonder what're the precise relations between the classical and quantum time evolution. I can think of one, i.e.
$$U(t)\langle\alpha|\hat{\psi}(x)|\alpha\rangle= \langle\alpha|e^{i\hat{H}t}\hat{\psi}(x)e^{-i\hat{H}t}|\alpha\rangle$$
Where $U(t)$ is the time evolution on the classical field, $\hat{\psi}(x)$ is the field operator, and $|\alpha\rangle$ is some quantum state.
However if one looks into more details, there is some thing odd(I think) happening: Take a free field for example, the eigenvectors of quantum time-evolution are Fock states, but the field expectation value of Fock states is 0, which is not an eigenvector of the classical time-evolution, since the eigenvetors of classical evolution are plane-waves. So what is happening here mathematically?
Also I'd like to know more about the relations classical and quantum time-evolution, any reference is also appreciated.

2. Feb 21, 2012

### strangerep

Then I guess it'll be ok if I offer a vague answer... :-)

Consider a classical quantity f, i.e., a function on phase space: f(q,p), where typically both q and p are functions of time. In this context, the time evolution of f is given by
$$\dot f ~:=~ \frac{df}{dt} ~=~ \left\{f,H \right\}_{PB}$$
where the "PB" subscript stands for Poisson Bracket. Typically f is some element constructed from the generators of the dynamical algebra for the (class of) system under study.

To quantize that system, one seeks a representation of the same dynamical algebra as operators on a Hilbert space. (If the dynamical algebra involves quadratic and higher products of the generators, then modifications are often needed. E.g., one might have to symmetrize the generators in the product.) But assuming that has been done, f and H now correspond to Hilbert space operators, and the time evolution of f is given by
$$\dot f ~=~ \frac{i}{\hbar} \, \left[f,H \right]$$
That's only because you constructed the Fock space by tensor products of the 1-particle Hilbert space for a free particle. For a classical harmonic oscillator with fixed mass and stiffness (hence fixed angular frequency $\omega$), the general solution is of the form
$$q(t) ~=~ e^{-i\omega t}a_0 + e^{i\omega t}a_0^* ~=:~ a(t)+a^*(t)$$
where $a_0$ is an arbitrary complex number, and $a^*_0$ its conjugate.
The momentum p(t) can be calculated from the time derivative of q(t) and one can easily show that the Poisson bracket relationship between q and p, i.e., $\{q,p\}=1$ remains satisfied for all time.

For a quantum harmonic oscillator, the solution is very similar:
$$Q(t) ~=~ e^{-i\omega t} a_0 + e^{i\omega t} a^*_0$$
except that now $a_0, a^*_0$ are operators satisfying
$$[a_0 , a^*_0] ~=~ \hbar$$
and one can show from the above that this also remains satisfied for all time.

I don't know what you're talking about here. But I'll carry on anway...

In the free 1-particle case, the eigenstates of the quantum Hamiltonian are also not normalizable, and live in a rigged Hilbert space. Ballentine gives a useful introduction to the latter.

Summary: the classical and quantum cases are more similar than different.

3. Feb 22, 2012

### kof9595995

I meant that if you take the field expectation value of any Fock state it'll give you 0, i.e.
$$\langle n_{k_1},n_{k_2}\ldots|\psi(x)|n_{k_1},n_{k_2} \ldots \rangle=0$$Which means the classical correpondence of Fock state is a trivial classical field (0 field value everywhere) ,because the field operator is linear w.r.t creation and annhilation operators.
This seems a bit weird to me, because naively one expects an eigenvector of quantized time-evolution(i.e. eigenvector of Hamiltonian) should correspond to a classical field which is an eigenvetor of the classical time evolution. I'm just wondering what happened during the quantization so that this naive expectation fails.

4. Feb 22, 2012

### Oudeis Eimi

I believe this is only true for the simplest Fock states, i.e., those with a sharp number of quanta (eigenstates of the relevant number operator), but not for other states. For instance, <ψ>≠0 for coherent states.

But even in the above mentioned simplest states, other field observables will have non vanishing expectation values. Try for instance computing the expectation value for the energy or momentum density for a scalar field in a 1-quantum state (it's a pretty simple exercise).

5. Feb 22, 2012

### kof9595995

I'm aware of that, but I think coherent states aren't eigenstates of time-evolution.

6. Feb 28, 2012

### kof9595995

And I have a follow up question: Is there any relation between the spectrum of classical field equation and quantum Hamiltonian? For free field the spectrum of classical field equation is just the one-particle spectrum quantum Hamiltonian, so is there any principle about quantization saying similar things in general?

7. Feb 28, 2012

### A. Neumaier

Neither are number states. In a scalar neutral field theory, the number operator is not conserved.

8. Feb 28, 2012

### A. Neumaier

In the classical limit of a quantum system (if this limit exists), the spectrum always becomes continuous.

9. Feb 28, 2012

### kof9595995

I was taking free field as an example.

10. Feb 28, 2012

### kof9595995

Emm, I'm not quite interested in the classical limit. For quantized free fields, one-particle spectrum is just $\sqrt{p^2+m^2}$, which is exactly the same as the eigenvalues of the classical field equation. I'm just wondering if something similar is ubiquitous for most of the quantum field theories, including interacting ones.

11. Feb 28, 2012

### A. Neumaier

The simplest free field _is_ that of a scalar neutral field.

12. Feb 28, 2012

### A. Neumaier

You were asking about the classical limit, but it seems you are really interested in the dispersion relation.

The dispersion relation of a classical linear field theory becomes the 1-particle energy operator in the corresponding free quantum field theory. The Hamiltonian itself is the second-quantized form of the 1-particle energy operator.

A classical nonlinear field theory doesn't have a fixed dispersion relation, and the corresponding quantum field theory has no well-defined single-particle energy operator,
as it is an interacting theory.

13. Feb 28, 2012

### kof9595995

You mean for example a real KG field? But aren't number states the eigenstates of the Hamiltonian? i.e.$\hat{H}|n_{k_1},n_{k_2}\ldots\rangle=(n_{k_1} \omega_{k_1}+n_{k_2}\omega_{k_2}+\ldots)|n_{k_1},n_{k_2}\ldots \rangle$

14. Feb 28, 2012

### kof9595995

Thanks for the information, I'm not sure if I fully get you, but let me use one more example to explain my question:For example, if we think about the way Dirac solved for the spectrum of hydrogen atom using Dirac equation, he was really solving a eigenvalue problem of the classical field equation, but it turns out the solved spectrum agrees quite well with experiment. And I think, in a QFT perspective, one can roughly interpret Dirac's method as a result of ignoring all multi-particle contributions from QED. So in this rough interpretation, it seems again eigenvalues of classical field equation conicides with one-particle spectrum. Am I making some sense?

Last edited: Feb 28, 2012
15. Feb 28, 2012

### A. Neumaier

Yes, in this very special case the Number operator commutes with the Hamiltonian. Nevertheless, coherent states are in practice far more important than number states, as they are far easier to prepare, while the latter can be prepared only with very special precautions.

Also, they exhibit the closeness to the classical situation much better than number states, which have no classical analogue.

Last edited: Feb 28, 2012
16. Feb 28, 2012

### A. Neumaier

No. He was solving a reduced 1-particle quantum problem for an electron in the potential of the nucleus plus its quantum field corrections.

To solve the ground state problem for N electrons in the central field of an atom, you need to solve the an eigenvalue problem for an N-electron Hamiltonian operator, which can be derived from QED in some approximation (Dirac-Fock is typical). For hydrogen, one gets N=1; so the Hamilton operator happens to have the form H_0(p)+corrections with H_0 being the Dirac operator. There is nothing classical in this procedure.

17. Feb 28, 2012

### kof9595995

Wasn't he solving the equation$i \gamma^\mu \partial_\mu \psi - m \psi = e \gamma_\mu A^\mu \psi$ with $A^\mu=(-\frac{e^2}{r},0,0,0)$,where $\psi$ is complex-valued 4-spinor(not an operator)? In a QFT viewpoint isn't it just a classical field?

18. Feb 29, 2012

### A. Neumaier

You may of course regard every 1-particle Schroedinger equation as a classical field equation. This doesn't change the fact that he was solving a quantum problem.

19. Feb 29, 2012

### kof9595995

I totally agree with you on this. But my point is, thinking of these equations as classical field equations, we can solve for eigenvalues of the classical field equation, and then if we quantize the classical field in to operators, in principle we can solve for one-particle spectrum from the quantum Hamitonian(for Dirac hydrogen I don't know if anybody can do this explicitly, but I presume this is the logical procedure in QFT frame work).
My examples(free fields and Dirac Hydrogen atom) seem to suggest eigenvalues of field equation are exactly(more or less, if one ignores multiparticle process e.g. lamb shift) one-particle spectrum, and this doesn't look like a mere coincidence to me, so I was wondering what's the principle behind this.

20. Mar 1, 2012

### A. Neumaier

This doesn't make sense. A classical field equation has no eigenvalues; you need an operator to speak of eigenvalues. And this operator is the quantum Hamiltonian.
The principle behind this is that every linear homogeneous classical field equation has the same form D(t,x,p_0,\p) psi =0 as every 1-particle quantum problem. (The lamb shift only changes the detailed form of D.)

If you can solve D(t,x,p_0,\p) =0 for p_0=H(t,x,\p) you get an operator H, which is the quantum Hamiltonian. Its eigenvalues lambda(\p) give you the dispersion relation omega=lambda(\p) of the classical field equation, or the other way around.
Solving one or the other is exactly the same thing.

However, there is not much classical about the field equations coming from a quantum system, except for this formal correspondence.

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