Relations between classical and quantum time-evolution of fields


by kof9595995
Tags: classical, fields, quantum, relations, timeevolution
kof9595995
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#1
Feb21-12, 08:21 AM
P: 679
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.
[tex]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[/tex]
Where [itex]U(t)[/itex] is the time evolution on the classical field, [itex]\hat{\psi}(x)[/itex] is the field operator, and [itex]|\alpha\rangle[/itex] 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.
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strangerep
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#2
Feb21-12, 06:25 PM
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Quote Quote by kof9595995 View Post
This question is gonna be a bit vague [...]
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
[tex]
\dot f ~:=~ \frac{df}{dt} ~=~ \left\{f,H \right\}_{PB}
[/tex]
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
[tex]
\dot f ~=~ \frac{i}{\hbar} \, \left[f,H \right]
[/tex]
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
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 [itex]\omega[/itex]), the general solution is of the form
[tex]
q(t) ~=~ e^{-i\omega t}a_0 + e^{i\omega t}a_0^* ~=:~ a(t)+a^*(t)
[/tex]
where [itex]a_0[/itex] is an arbitrary complex number, and [itex]a^*_0[/itex] 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., [itex]\{q,p\}=1[/itex] remains satisfied for all time.

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

but the field expectation value of Fock states is 0
I don't know what you're talking about here. But I'll carry on anway...

[...] not an eigenvector of the classical time-evolution, since the eigenvectors of classical evolution are plane-waves. So what is happening here mathematically?
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.
kof9595995
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#3
Feb22-12, 03:30 AM
P: 679
Quote Quote by strangerep View Post
I don't know what you're talking about here. But I'll carry on anway...
I meant that if you take the field expectation value of any Fock state it'll give you 0, i.e.
[tex]\langle n_{k_1},n_{k_2}\ldots|\psi(x)|n_{k_1},n_{k_2} \ldots \rangle=0[/tex]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.

Oudeis Eimi
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#4
Feb22-12, 06:18 AM
P: 66

Relations between classical and quantum time-evolution of fields


Quote Quote by kof9595995 View Post
I meant that if you take the field expectation value of any Fock state it'll give you 0, i.e.
[tex]\langle n_{k_1},n_{k_2}\ldots|\psi(x)|n_{k_1},n_{k_2} \ldots \rangle=0[/tex]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.
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).
kof9595995
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#5
Feb22-12, 08:33 AM
P: 679
Quote Quote by Oudeis Eimi View Post
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.
I'm aware of that, but I think coherent states aren't eigenstates of time-evolution.
kof9595995
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#6
Feb28-12, 07:02 AM
P: 679
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?
A. Neumaier
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#7
Feb28-12, 07:23 AM
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Quote Quote by kof9595995 View Post
I'm aware of that, but I think coherent states aren't eigenstates of time-evolution.
Neither are number states. In a scalar neutral field theory, the number operator is not conserved.
A. Neumaier
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#8
Feb28-12, 07:24 AM
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Quote Quote by kof9595995 View Post
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?
In the classical limit of a quantum system (if this limit exists), the spectrum always becomes continuous.
kof9595995
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#9
Feb28-12, 07:54 AM
P: 679
Quote Quote by A. Neumaier View Post
Neither are number states. In a scalar neutral field theory, the number operator is not conserved.
I was taking free field as an example.
kof9595995
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#10
Feb28-12, 07:59 AM
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Quote Quote by A. Neumaier View Post
In the classical limit of a quantum system (if this limit exists), the spectrum always becomes continuous.
Emm, I'm not quite interested in the classical limit. For quantized free fields, one-particle spectrum is just [itex]\sqrt{p^2+m^2}[/itex], 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.
A. Neumaier
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#11
Feb28-12, 10:01 AM
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Quote Quote by kof9595995 View Post
I was taking free field as an example.
The simplest free field _is_ that of a scalar neutral field.
A. Neumaier
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#12
Feb28-12, 10:09 AM
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Quote Quote by kof9595995 View Post
Emm, I'm not quite interested in the classical limit. For quantized free fields, one-particle spectrum is just [itex]\sqrt{p^2+m^2}[/itex], 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.
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.
kof9595995
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#13
Feb28-12, 10:25 AM
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Quote Quote by A. Neumaier View Post
The simplest free field _is_ that of a scalar neutral field.
You mean for example a real KG field? But aren't number states the eigenstates of the Hamiltonian? i.e.[itex] \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[/itex]
kof9595995
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#14
Feb28-12, 10:38 AM
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Quote Quote by A. Neumaier View Post
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.
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?
A. Neumaier
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#15
Feb28-12, 10:51 AM
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Quote Quote by kof9595995 View Post
You mean for example a real KG field? But aren't number states the eigenstates of the Hamiltonian? i.e.[itex] \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[/itex]
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.
A. Neumaier
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#16
Feb28-12, 10:57 AM
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Quote Quote by kof9595995 View Post
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
No. He was solving a reduced 1-particle quantum problem for an electron in the potential of the nucleus plus its quantum field corrections.

Quote Quote by kof9595995 View Post
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.
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.
kof9595995
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#17
Feb28-12, 11:14 AM
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Quote Quote by A. Neumaier View Post
No. He was solving a reduced 1-particle quantum problem for an electron in the potential of the nucleus plus its quantum field corrections.
Wasn't he solving the equation[itex]i \gamma^\mu \partial_\mu \psi - m \psi = e \gamma_\mu A^\mu \psi[/itex] with [itex]A^\mu=(-\frac{e^2}{r},0,0,0)[/itex],where [itex]\psi[/itex] is complex-valued 4-spinor(not an operator)? In a QFT viewpoint isn't it just a classical field?
A. Neumaier
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#18
Feb29-12, 09:04 AM
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Quote Quote by kof9595995 View Post
Wasn't he solving the equation[itex]i \gamma^\mu \partial_\mu \psi - m \psi = e \gamma_\mu A^\mu \psi[/itex] with [itex]A^\mu=(-\frac{e^2}{r},0,0,0)[/itex],where [itex]\psi[/itex] is complex-valued 4-spinor(not an operator)? In a QFT viewpoint isn't it just a classical field?
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.


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