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

  • Thread starter Sonderval
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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)?
 

Bill_K

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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.
 
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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.
 

Demystifier

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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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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.
 

DrDu

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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.
 
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@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.
 

Physics Monkey

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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.
 
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@Physics Monkey
Thanks for the hint, I'll look at that.
 

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