Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have read through the wikipedia page and several lecture notes/arxiv papers from my google search (and several related P.SE questions), but I'm still hopelessly confused.

  1. Consider a 'classical Schrodinger field in a box' problem. Since the field evolves just like a normal wavefunction, I could still extract the x-projection of any operators I want from it, in addition to the field momentum $\frac{\partial\mathcal{L}}{\partial\dot{\phi}}$ or any other field variables.

  2. From my understanding, there are two ways to arrive at QFT:

    • from QM/RQM: we change our basis into particle-number basis (fock space).
    • from CFT: we "2nd quantize" the field and field-momentum.

Looks like the "particle creation" interpretation of the field in QFT can be deduced only from the first approach. Which means, the interpretation of the latter approach can't be completely independent from the former. But they don't give me any hint on how to interpret the classical field. Or am I missing something? Or should I just give it up because the quantum field is not an observable?

Random comment: I could still see $\hbar$ lurking inside CFT (some people might argue that it's there to keep the dimension consistent, but I could apply the same argument to Schrodinger equation, too).

share|improve this question
    
"and several related P.SE questions":physics.stackexchange.com/questions/4156/… –  pcr Apr 7 '11 at 22:05
    
Schrodinger field in QFT: $\left[\psi^{\dagger}(x),\psi(y)\right] = \delta(x-y)$, while in CFT this becomes zero. Which means, CFT is not strictly a classical limit of QFT (no $\hbar \rightarrow 0$ involved) –  pcr Apr 7 '11 at 22:24
1  
Your $\delta(x-y)$ should have a factor $\hbar$, $\hbar\delta(x-y)$. The limit $\hbar\rightarrow 0$ is then a clear possibility, but to construct a non-trivial state over the classical random field theory $[\psi^\dagger(x),\psi(y)]=0$ -- something other than "the expected values of all observable quantities are zero in all states"-- needs some kind of fluctuations to be introduced. A state over a field of operator-valued distributions that satisfies a trivial commutator is still a presentation of a system of probability measures and expected values, not a system of classical fields. –  Peter Morgan Apr 8 '11 at 2:03
    
@Peter Morgan I've checked the dimension, the equation comes from $\left[ \psi(x),\pi(y) \right] = i\hbar\delta(x-y)$. But I've just realized the flaw in my previous argument, sorry about that. I need to think about your final remark (and also, it's weird to find that QFT has two classical limits: newtonian and CFT, so the latter one should be something else). –  pcr Apr 8 '11 at 4:47

2 Answers 2

up vote 4 down vote accepted

The classical field has a straightforward interpretation in the bosonic case--- it is determined by the density and phase of the superfluid condensate of the particles and both are simultaneously measurable in the thermodynamic limit. If there is no superfluid, the field is zero. Where there is a superfluid, it is the square-root of the density, with a phase whose gradient is the local superflow.

You can measure the value of psi-squared at x, that's determining the superfluid condensate density, which can be done by shining light through the fluid to get the index. You can also measure the value of the velocity by the exact way light refrects (or if it is dense enough, by putting little dust specks in the fluid). A gaseous Bose-Einstein condensate, or even liquid Helium, is described precisely by the classical limit of this formalism in the thermodynamic limit.

If you take the classical limit wrong, by insisting that the particle number N is fixed, you do still get hbars lurking around. In order to take a good classical field limit, you need enormous occupation numbers for the field, which requires that you take N to infinity and the mass to zero keeping the density fixed. In this limit, the conjugate variables density/phase become commuting.

As for measuring the "x-projection of operators", I couldn't figure out what that meant. You can measure the field momentum too, of course, by just taking the complex conjugate of the measurement of the field (the imaginary part of the field and the real part are not independent, and you can describe the whole system using only the real part and its time derivative, as described in the Wikipedia article).

share|improve this answer
    
"If you take the classical limit wrong, by insisting that the particle number N is fixed, you do still get hbars lurking around. In order to take a good classical field limit, you need enormous occupation numbers for the field, which requires that you take N to infinity and the mass to zero keeping the density fixed. " This clears the fog, thanks! So we can observe single-particle classical fermion, but not classical single-particle boson. (I see, classical field theory of fermion = particle!) –  pcr Nov 4 '11 at 7:06

I am not sure I am getting the source of the difficulty, but let me try:

  1. The usual object described by a Schrodinger equation is the wavefunction, which is not observable. There are some classical systems which involve continuous media (= fields) which evolve according to classical equations similar in form to the Schrodinger equation. Be that as it may, the physics is all in interpreting the results of the mathematical framework, and classical fields are different objects than quantum wave functions, even if they obey similar equations.

  2. When you quantize a free classical field theory you obtained results that are isomorphic to a multi-particle Fock space. But, quantum field theory is more general than that - the physics of strongly coupled QFTs can be very different from that of weakly interacting particles, for example they may well not have any sort of particles involved. The distinction between the two ways of approaching QFT is historical: the first way you quote is the older and less general way to approach the subject, and the second way is more precise and general (reproducing the results of the first one in a specific limit of free fields).

  3. In general field theory the field operators are observable, but not always easy to interpret as results of concrete measurements, one usually constructs correlation functions which are more directly related to simple experiments.

  4. Finally, as for the random comment: if you phrase you classical theory in terms of quantities which stay finite in the classical limit, the Planck constant $\hbar$ ought to completely disappear. If you keep an inherently quantum mechanical quantity (say the mass or charge of individual particle) fixed, there will be (misleading) factors of $\hbar$ left.

share|improve this answer
    
Thanks for your point#1, I won't mix it with the usual Schrodinger wavefunction anymore. But for "general field theory the field operators are observable", what about the nonzero commutation value between bosonic $\psi$ and $\psi^{\dagger}$? Then how should I interpret the classical Schrodinger field itself? I also have one more related subquestion (see my 2nd comment). Sorry for my barrage of questions. –  pcr Apr 7 '11 at 23:08
    
Quantum field operator is observable in the same sense that momentum and position are observable in quantum mechanics. They represent a measurable quantity, the result of each measurement is random but there is some probability for obtaining each possible value etc. –  user566 Apr 7 '11 at 23:49
    
I used to think that $\psi^{\dagger}\psi$ is an observable, but $\psi$ by itself isn't. I'll try to convince myself by rereading my book. If it is so, then I'll be able to construct a classical field interpretation from it. Thanks anyway. back to reading –  pcr Apr 8 '11 at 0:20
    
This is a matter of terminology: the square of the wavefunction is the probability of getting some result for some measurement - the measurement itself is what is called observable in quantum mechanics, –  user566 Apr 8 '11 at 14:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.