Relations between classical and quantum time-evolution of fields

In summary: For more complicated situations, the eigenstates of the quantized Hamiltonian are not normalizable, and live in a rigged Hilbert space.
  • #1
kof9595995
679
2
This question is going to 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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  • #2
kof9595995 said:
This question is going to 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.
 
  • #3
strangerep said:
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.
 
  • #4
kof9595995 said:
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).
 
  • #5
Oudeis Eimi said:
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.
 
  • #6
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
kof9595995 said:
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.
 
  • #8
kof9595995 said:
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.
 
  • #9
A. Neumaier said:
Neither are number states. In a scalar neutral field theory, the number operator is not conserved.

I was taking free field as an example.
 
  • #10
A. Neumaier said:
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.
 
  • #11
kof9595995 said:
I was taking free field as an example.

The simplest free field _is_ that of a scalar neutral field.
 
  • #12
kof9595995 said:
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.
 
  • #13
A. Neumaier said:
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]
 
  • #14
A. Neumaier said:
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?
 
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  • #15
kof9595995 said:
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.
 
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  • #16
kof9595995 said:
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.

kof9595995 said:
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.
 
  • #17
A. Neumaier said:
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?
 
  • #18
kof9595995 said:
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.
 
  • #19
A. Neumaier said:
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.

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 into 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
kof9595995 said:
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,
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.
kof9595995 said:
I
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.

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.
 
  • #21
A. Neumaier said:
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.
H(t,x,\p) is a differential operator, and Dirac indeed viewed it as a quantum Hamitonian. However, I'm told by almost every textbook I've read that after QFT was born, Dirac's interpretation should be abandoned. From the prescription I've learned about QFT, the proper way of solving one-particle spectrum is that, first quantized the fields properly, then you can express quantized Hamiltonian density in terms of field operators, and then integrate the Hamitonian density to get the quantized Hamitonian. Now that you get the Hamiltonian, you look for eigenvalues and eigenvetors in the one-particle sector of Hilbert space.
So my puzzle is, originally you just solve for eigenvalues of the differential operator H(t,x,\p), but later in QFT you must take a big detour before you can write down another eigenvalue problem and solve it, however how do you make sure these two eigenvalue problems give you the same result? I think one must get the same result, since on the one hand QFT is accepted to be the correct relativistic quantum theory, on the other hand, simply solving H(t,x,\p) give good results compared with experiments.
 
  • #22
kof9595995 said:
H(t,x,\p) is a differential operator, and Dirac indeed viewed it as a quantum Hamitonian. However, I'm told by almost every textbook I've read that after QFT was born, Dirac's interpretation should be abandoned.
But not _everything_ from his interpretation. Only the interpretation of the negative energy solutions has become obsolete. The positive energy solutions still have exactly the same meaning as when Dirac introduced them.
kof9595995 said:
From the prescription I've learned about QFT, the proper way of solving one-particle spectrum is that, first quantized the fields properly, then you can express quantized Hamiltonian density in terms of field operators, and then integrate the Hamitonian density to get the quantized Hamitonian. Now that you get the Hamiltonian, you look for eigenvalues and eigenvectors in the one-particle sector of Hilbert space.
This is just one way of presenting things didactically.

But one always builds the Fock space from the 1-particle Hilbert space and one can as well build the Fock Hamiltonian from the 1-particle Hamiltonian. This is the way Weinberg proceeds in his QFT book. First he constructs irreducible representations of the Poincare group (whose P_0 is the 1-particle Hamiltonian) and then he constructs from it the associated Fock space, noting that to get the CCR or CAR he must also introduce antiparticles. The Lagrangian approach comes in his book much later, and he shows that it gives in the free case identical results.
kof9595995 said:
So my puzzle is, originally you just solve for eigenvalues of the differential operator H(t,x,\p), but later in QFT you must take a big detour before you can write down another eigenvalue problem and solve it, however how do you make sure these two eigenvalue problems give you the same result? I think one must get the same result, since on the one hand QFT is accepted to be the correct relativistic quantum theory, on the other hand, simply solving H(t,x,\p) give good results compared with experiments.

In quantum field theory, the Hamiltonian operator acts on a far bigger space, created by second quantization from the 1-particle space. The eigenvalues of this Hamiltonian gives you all bound states of the QFT, not just those of a single particle.

To get the bound states for relativistic hydrogen including radiation corrections, you need to start with this big space, remove the center of mass motion, and project the resulting problem down to a single-particle space. This gives you a reduced 1-particle quantum Hamiltonian, which looks like a Dirac operator but with correction terms. Then one solves this eigenvalue problem. There is nothing classical at all in the whole procedure.
And note that without the QFT approach one doesn't get the same results - the radiative corrections (e.g., lamb shift and anomalous magnetic moment) are missing.

On the other hand, one can take a classical limit of QED. But then one doesn't end up with a classical Dirac-like equation but with a transport equation for the Wigner transform.

The Dirac equation has no classical interpretation, as there is no classical equivalent of the
anticommutation relation. (One could regard a formulation with Grassmann numbers as kind of classical limit, but these don't commute, hence are not really classical.)
 
  • #23
A. Neumaier said:
To get the bound states for relativistic hydrogen including radiation corrections, you need to start with this big space, remove the center of mass motion, and project the resulting problem down to a single-particle space. This gives you a reduced 1-particle quantum Hamiltonian, which looks like a Dirac operator but with correction terms. Then one solves this eigenvalue problem. There is nothing classical at all in the whole procedure.

Thanks, this seems to be quite helpful, and where can I find the details about this approach?
 
  • #24
kof9595995 said:
Thanks, this seems to be quite helpful, and where can I find the details about this approach?

The reduction procedure and the associated approximation schemes are quite involved; the thing to look up in scholar.google.com or the archive is NRQED (=NonRelativistic QED).
I was only describing the qualitative aspects that are easy to understand. I don't know of a simple reference for more details without these being messy.
 
  • #25
Recently I just read Thaller.B "The Dirac equation" Chapter 10. It seems that if taking second quantization approach to quantize Dirac equation, one just takes the solution space of Dirac operator as the one-particle Hilbert space, and forms Fock states and creation/annihilation operators as everyone knows. And if a particle is in an eigenstate of the Dirac Hamiltonian before second quantization, it'll still be in an eigenstate of the second quantized Dirac Hamiltonian. And this actually justifies the way we solve for Dirac hydrogen from QFT view point(still taking Coulomb field classical for simplicity, omitting lamb shift), because one particle sector of QFT is just identical with the solution space of Dirac operator.
I think this implies a single particle Fock state with definite momentum should indeed be identified with a plane-wave solution(in free field case), but this means what I wrote in the original post was just wrong, I thought taking field expectation values would give the classical field, i.e. [itex]\psi_{\alpha}(x)=\langle \alpha|\hat{\psi}(x)|\alpha\rangle[/itex].
However in the mean time, it's stated in Sakurai's Advanced QM(page 33) that for a quantized EM field, field expectation values should be interpreted as the classical EM field. So what causes the difference between Dirac field and EM field?
 
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  • #26
kof9595995 said:
Recently I just read Thaller.B "The Dirac equation" Chapter 10. It seems that if taking second quantization approach to quantize Dirac equation, one just takes the solution space of Dirac operator as the one-particle Hilbert space, and forms Fock states and creation/annihilation operators as everyone knows. And if a particle is in an eigenstate of the Dirac Hamiltonian before second quantization, it'll still be in an eigenstate of the second quantized Dirac Hamiltonian. And this actually justifies the way we solve for Dirac hydrogen from QFT view point(still taking Coulomb field classical for simplicity, omitting lamb shift), because one particle sector of QFT is just identical with the solution space of Dirac operator.
One rather discards negative energy solutions, and takes two copies of the positive energy solution space as the one particle Hilbert space of electron and positron. This gives the free Fermion part of QED.
kof9595995 said:
However in the mean time, it's stated in Sakurai's Advanced QM(page 33) that for a quantized EM field, field expectation values should be interpreted as the classical EM field. So what causes the difference between Dirac field and EM field?
Being Fermions, Dirac fields themselves are unobservable, and the relevant classical (i.e., macroscopic) stuff is the expectation of products of two field operators (or rather their Wigner transform).
 
  • #27
A. Neumaier said:
One rather discards negative energy solutions, and takes two copies of the positive energy solution space as the one particle Hilbert space of electron and positron. This gives the free Fermion part of QED.
Thaller defines and takes the charge conjugation of negative energy solutions and direc sum them with the postive energy solutions, I'm not sure if charge conjugation gives the same copy with the original positive energy solutions.

A. Neumaier said:
Being Fermions, Dirac fields themselves are unobservable, and the relevant classical (i.e., macroscopic) stuff is the expectation of products of two field operators (or rather their Wigner transform).
I agree, but originally I thought c-number solutions of Dirac equation aren't real anyway, so no problem of being some expectation values of the field operator, which is not hermitian.
 
  • #28
kof9595995 said:
Thaller defines and takes the charge conjugation of negative energy solutions and direc sum them with the postive energy solutions, I'm not sure if charge conjugation gives the same copy with the original positive energy solutions.
With this amendment, Thaller's construction is equivalent to the standard construction (e.g., that in Weinberg's book).
 

1. What is the difference between classical and quantum time-evolution of fields?

The classical time-evolution of fields is described by classical mechanics and follows deterministic laws, where the state of a system can be precisely predicted given its initial conditions. On the other hand, the quantum time-evolution of fields is described by quantum mechanics and follows probabilistic laws, where the state of a system can only be predicted in terms of probabilities.

2. How are classical and quantum time-evolution of fields related?

Classical and quantum time-evolution of fields are related through the correspondence principle, which states that the predictions of quantum mechanics should approach those of classical mechanics in the limit of large quantum numbers and energies. This means that classical mechanics can be seen as an approximation of quantum mechanics for macroscopic systems.

3. What is the role of time in classical and quantum time-evolution of fields?

In classical mechanics, time is treated as a continuous variable that flows uniformly and independently of the system. In quantum mechanics, time is treated as an operator that describes the evolution of a system and is often represented as a parameter in the Schrödinger equation.

4. How does the Heisenberg uncertainty principle affect the time-evolution of fields?

The Heisenberg uncertainty principle states that the more precisely we know the position of a particle, the less precisely we can know its momentum and vice versa. This uncertainty in measurements affects the time-evolution of fields in quantum mechanics, as it introduces a fundamental limit to the precision with which we can predict the future behavior of a system.

5. What are some practical applications of studying the relations between classical and quantum time-evolution of fields?

Studying the relations between classical and quantum time-evolution of fields has practical applications in various fields such as quantum computing, quantum information theory, and quantum simulations. It also helps us understand the behavior of systems at the microscopic level and has implications for our understanding of the fundamental laws of nature.

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