When is operator phi(x) an observable in QFT?

In summary, the field operator φ(x) in QFT can be an observable in the case of a real Klein-Gordon field due to its Hermitian nature. However, in the case of a complex field, such as one for a charged particle, the operator is not Hermitian and therefore does not correspond to an observable. This is because measuring the field would result in a superposition of states with different numbers of particles, which is unphysical. Additionally, the complex field has a physical meaning in symmetry broken or superfluid phases.
  • #1
Sonderval
234
11
In QFT of a real Klein-Gordon-Field, the field operator
[itex]\phi(x)[/itex] is an observable.
Mathematically, this is the case because it is a sum (over all k) of [itex]a[/itex] and [itex]a^\dagger[/itex] and this yields a Hermitian operator.
Physically, I can understand this because this equation would describe, for example, a membrane treated with QFT.

However, if the field is complex (charged particle) [itex]\phi(x)[/itex], the operator is the sum of particle lowering and anti-particle raising operators (or vice versa). Mathematically, I can see that it is not Hermitian and therefore it will not have Eigenvalues and does not correspond to an observable.

But I don't really understand this physically: Why can I (in principle) measure the uncharged pion field but not the charged pion field? I suspect this has to do with the scalar particle being its own anti-particle, but I don't really see how this affects the measurability of the field or the possibility to create a local field excitation (which is what operating with [itex]\phi(x)[/itex] should do)?
 
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  • #2
Sonderval, Even though in the real case φ(x) is Hermitian, that's no guarantee that its eigenvectors are in any way useful. For example since [H, φ(x)] ≠ 0, eigenvectors of φ(x) cannot be eigenvectors of energy. Also, since φ(x) is a linear combination of creation and annihilation operators, applying it to a state either raises or lowers the number of particles - unless the number of particles was infinite! So any eigenvector of φ(x) must be built from states with an infinite number of particles.
 
  • #3
Of course you are right, an eigenvector to phi(x) will be a mixture of different particle numbers. This is completely analoguous to the QM harmonic oscillator: the eigenvectors position operator also is a combination of all energy eigenfunctions. Nevertheless, x is a valid observable.

The same should be the case in QFT - after all, the QM harm. osc. is equivalent to QFT in 0+1 dimensions. Measuring phi(x) will create an eigenstate at point x with a certain field value phi, similar to the way measuring position will create a position eigenstate (if I measure position within a small region, I'll create a wave packet which can be a coherent state in the case of the QM HO, which is also a mixture of infinitely many energy eigenstates.).

I understand that this is the case and am intuitively quite happy with that. I just don't have an intuitive idea why the situation should be so different for a charged particle.
 
  • #4
Sonderval, complex scalar field can be represented by a collection of two real mutually commuting scalar fields, each of which can (at least in principle) be measured separately.

However, an eigenstate of such a hermitian field operator would be a superposition of states with different numbers of charged particles. Superpositions of different charges certainly exist mathematically in the Hilbert space, but there is a superselection rule that prevents the existence of such states in reality. This superselection rule can in fact be explained by the theory of decoherence. Essentially, such superpositions are unphysical for the same reason for which a superposition of a live and a dead cat is unphysical.
 
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  • #5
Sonderval said:
However, if the field is complex (charged particle) [itex]\phi(x)[/itex], the operator is the sum of particle lowering and anti-particle raising operators (or vice versa). Mathematically, I can see that it is not Hermitian and therefore it will not have Eigenvalues and does not correspond to an observable.

But I don't really understand this physically: Why can I (in principle) measure the uncharged pion field but not the charged pion field? I suspect this has to do with the scalar particle being its own anti-particle, but I don't really see how this affects the measurability of the field or the possibility to create a local field excitation (which is what operating with [itex]\phi(x)[/itex] should do)?
You are right, but, then again, the ladder operators a and a+ are not hermitian either.

The physical meaning of a complex [itex]\phi(x)[/itex] is that it destroys a charged scalar particle at space-time point x, and the adjoint creates one. For a real scalar field, the particles are uncharged and equal to their own antiparticles. For them, the field operator creates/destroys.
 
  • #6
Bill_K said:
Sonderval, Even though in the real case φ(x) is Hermitian, that's no guarantee that its eigenvectors are in any way useful. For example since [H, φ(x)] ≠ 0, eigenvectors of φ(x) cannot be eigenvectors of energy. Also, since φ(x) is a linear combination of creation and annihilation operators, applying it to a state either raises or lowers the number of particles - unless the number of particles was infinite! So any eigenvector of φ(x) must be built from states with an infinite number of particles.

That's not an argument. Eigenstates of the electric or magnetic field also don't commute with the Hamiltonian and are coherent states formed from in principle all particle numbers. Yet they are very useful.
 
  • #7
@Demystifier
However, an eigenstate of such a hermitian field operator would be a superposition of states with different numbers of charged particles
D'oh. I didn't think of that, that's really a good physical argument.

I'll need to think a bit further about the implications, but that's some help for sure.

@Dickfore
The physical meaning of a complex ϕ(x) is that it destroys a charged scalar particle at space-time point x
I don't think so - as Demystifer said, actually applying ϕ(x) will select an eigenvector of it, which is a superposition of all particle-number eigenstates.
 
  • #8
One place where the complex scalar field has a nice physical meaning is in symmetry broken or superfluid phases. In this case, for example, its phase (or really a phase difference) can be measured by doing Josephson type experiments.
 
  • #9
@Physics Monkey
Thanks for the hint, I'll look at that.
 

1. What is an operator in quantum field theory?

An operator in quantum field theory (QFT) is a mathematical representation of a physical quantity that can be measured in a system. In QFT, operators act on quantum fields and can create or destroy particles, as well as measure properties such as position, momentum, and energy.

2. What does it mean for an operator to be an observable in QFT?

An observable in QFT is an operator that corresponds to a measurable physical quantity. This means that the operator can be used to calculate the expectation value of the physical quantity in a quantum state, which can then be compared to experimental results.

3. How is the operator phi(x) related to observables in QFT?

The operator phi(x) is a quantum field operator that represents the field value at a specific point in space-time. In QFT, observables are usually expressed in terms of field operators, so phi(x) can be used to calculate the expectation values of physical quantities at a particular location in space and time.

4. When is the operator phi(x) an observable in QFT?

The operator phi(x) is an observable in QFT when it satisfies certain mathematical requirements. These include being Hermitian, meaning that its adjoint is equal to its conjugate transpose, and being self-adjoint, meaning that it is equal to its own adjoint. These conditions ensure that the operator has real eigenvalues and is suitable for calculating measurable physical quantities.

5. Can the operator phi(x) be an observable in all QFT systems?

No, the operator phi(x) is not always an observable in QFT systems. In certain cases, such as when the system has a continuous symmetry, the operator may not correspond to a measurable physical quantity. In these cases, a different operator may need to be used as the observable for that particular system.

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