A probability of field amplitude in QFT

[MichPod]

TL;DR Summary

get a probability density for a quantum field

Per quantized scalar field (quantized Klein-Gordon equation), suppose we act on a vacuum state |0> with some set of creation operators to have some particles.
How then can we calculate a probability density for the field to have a particular value $\psi_0$ (upon measurement) at a specific point (t, x, y, z)?

I am interested in understanding a general formula and approach, not in making a specific calculation in all the details.

 

[PeterDonis]

How then can we calculate a probability density for the field to have a particular value

The question doesn't make sense. To get a meaningful probability density, you need to pick an observable. The field is not an observable.

 

[MichPod]

The question doesn't make sense. To get a meaningful probability density, you need to pick an observable. The field is not an observable.

Why the field is not observable? (I expected the value of the field is observable because it is observable in a classical case).

Just to be on the same page, we can observe a coordinate of a quantum harmonic oscillator at a specific moment of time, right? By analogy, I though, we can observe an amplitude of a field at a particular time and coordinate.

 

[PeterDonis]

Why the field is not observable?

Because it's not a Hermitian operator.

I expected the value of the field is observable because it is observable in a classical case

You expected wrong. A quantum field is not at all the same kind of thing as a classical field.

we can observe a coordinate of a quantum harmonic oscillator at a specific moment of time, right?

What Hermitian operator corresponds to this observable?

By analogy, I though, we can observe an amplitude of a field at a particular time and coordinate.

There is no such analogy for a quantum field.

 

[MichPod]

What Hermitian operator corresponds to this observable?

For a harmonic oscillator? Per Schrodinger picture, it is just 'x'.

How can I see that the field operator is not Hermitian? I though it is for the case of K-G equation.

 

[PeterDonis]

For a harmonic oscillator? Per Schrodinger picture, it is just 'x'.

In the position representation, yes. But this observable is position--the "position of the oscillator" (which is viewed in this representation as a point particle oscillating around in a particular potential). It is not "wave amplitude" or "field amplitude".

 

[MichPod]

In the position representation, yes. But this observable is position--the "position of the oscillator" (which is viewed in this representation as a point particle oscillating around in a particular potential). It is not "wave amplitude" or "field amplitude".

I expected there is an obvious analogy between the x coordinate of an oscillator and the value of a quantum field at particular point. Why was I wrong?

BTW, how can I see that the field operator is not Hermitian, as you said? I though it is for the case of K-G equation.

 

[MichPod]

To continue, as a discrete approximation we can model a K-G field with some mattress of discrete particles connected with springs. Acting with creation operators on the vacuum (ground) state of that mattress would correspond to adding phonons to the "lattice" of mattress. So I think that asking for a probability density of a particular particle of the mattress to have particular displacement at a particular time should be very well posed and make sense. This should also make sense for a continuous limit when we pass from the mattress to a continuous quantized K-G field and want to measure a value of the field at a particular point.

 

[PeterDonis]

I expected there is an obvious analogy between the x coordinate of an oscillator and the value of a quantum field at particular point. Why was I wrong?

Why would you expect not to be wrong? You're just making a hand-waving analogy. You haven't given any references, so I don't know where your hand-waving analogy is coming from.

A quantum field is an operator; it doesn't even have a "value at a particular point", much less one that is analogous to the x coordinate of an oscillator.

how can I see that the field operator is not Hermitian, as you said? I though it is for the case of K-G equation.

What is the field operator for the K-G equation?

 

[PeterDonis]

as a discrete approximation we can model a K-G field with some mattress of discrete particles connected with springs.

Ok.

Acting with creation operators on the vacuum (ground) state of that mattress would correspond to adding phonons to the "lattice" of mattress.

Ok.

So I think that asking for a probability density of a particular particle of the mattress to have particular displacement at a particular time should be very well posed and make sense.

Whether it makes sense or not, it is irrelevant as far as using this toy model as a heuristic for quantum field theory is concerned. The analogue of the quantum field for this toy model is the creation and annihilation operators, not the particles or their positions. The whole toy model itself is not a toy model of a distribution of quantum fields in a region of spacetime; it is a toy model of a quantum field at a single point of spacetime. A toy model of a quantum field theory in a region of spacetime will have one of these mattresses at every point.

 

[MichPod]

What is the field operator for the K-G equation?

 

[PeterDonis]

View attachment 338391

Is this operator Hermitian? Are all of its eigenvalues real?

 

[PeterDonis]

I though it is for the case of K-G equation.

Even if the K-G operator turns out to be Hermitian, that does not mean quantum field operators in general are.

 

[MichPod]

Is this operator Hermitian? Are all of its eigenvalues real?

Providing the Hermitian conjugation of $\hat{a}$ is $\hat{a}^\dagger$, this operator is Hermitian (self-adjoint)

 

[PeterDonis]

Providing the Hermitian conjugation of $\hat{a}$ is $\hat{a}^\dagger$, this operator is Hermitian (self-adjoint)

So if that is the case, what is its physical meaning? What are the eigenstates and eigenvalues of this operator?

 

[Hill]

Acting with creation operators on the vacuum (ground) state of that mattress would correspond to adding phonons to the "lattice" of mattress. So I think that asking for a probability density of a particular particle of the mattress to have particular displacement at a particular time should be very well posed and make sense.

AFAIK, phonons do not describe vibrations of the individual particles of the "mattress", but rather are modes of vibration of the entire collective of the particles, i.e., the entire "mattress."

 

[PeterDonis]

AFAIK, phonons do not describe vibrations of the individual particles of the "mattress", but rather are modes of vibration of the entire collective of the particles, i.e., the entire "mattress."

Yes, that's correct.

 

[MichPod]

AFAIK, phonons do not describe vibrations of the individual particles of the "mattress", but rather are modes of vibration of the entire collective of the particles, i.e., the entire "mattress."

Agree. And I did not mean anything opposite. The mattress has some physical particles connected with springs, then phonons correspond to the modes of the vibration of the mattress as a whole.

 

[vanhees71]

Of course in principle a neutral-Klein-Gordon field operator may represent an observable, i.e., it's a self-adjoint operator, fulfilling the microcausality condition. So they can represent, at least formally, local observables. I only do not have a concrete simple physical example for it.

 

[MichPod]

Of course in principle a neutral-Klein-Gordon field operator may represent an observable, i.e., it's a self-adjoint operator, fulfilling the microcausality condition. So they can represent, at least formally, local observables. I only do not have a concrete simple physical example for it.

Thank you. Thank you for confirming my question has some sense.

 

[PeterDonis]

in principle a neutral-Klein-Gordon field operator may represent an observable

Note, though, that the "neutral" (i.e., uncharged) is key. A charged Klein-Gordon field operator is no longer self-adjoint. Nor is a fermionic field operator (only fermionic bilinears like $\bar{\psi} \psi$ are). I'm not sure about an uncharged spin-1 (Maxwell) field operator.

I only do not have a concrete simple physical example for it.

That is why I asked the OP about the physical meaning of the operator in post #15.

 

[PeterDonis]

in principle a neutral-Klein-Gordon field operator may represent an observable

All along we've been talking about a free Klein-Gordon field operator. However, I'm not sure this still holds for an interacting Klein-Gordon field operator.

 

[Demystifier]

Fields corresponding to uncharged particles, such as electromagnetic field and Higgs field, are hermitian, so these fields are observables. This fact does not depend on whether the fields are free or interacting. Even if one considers a non-hermitian field, such as electron field $\psi$, one can always construct a hermitian field $\phi= \psi+\psi^{\dagger}$.

 

[vanhees71]

In gauge theories there's the additional caveat that only gauge-invariant (composite) local field self-adjoint operators can represent local observables.