Probability of states in QFT

1. Mar 6, 2014

GarageDweller

Is it actually possible to calculate the probability of field states in QFT? For example the probability of some scalar field being found as some function f(x,t), i find this problem ignored in most texts.

2. Mar 6, 2014

Avodyne

Yes. There is a HW problem about this somewhere in Srednicki's text. For a free scalar field, the ground-state wave functional takes the form
$$\psi(\varphi)\propto \exp\left[-{1\over2}\int d^3\!x\,d^3\!y\,\varphi(\vec x)K(\vec x,\vec y)\varphi(\vec y)\right]$$
where $K(\vec x,\vec y)$ is formally defined by something like
$$K(x,y)=\int {d^3\!k\over(2\pi)^3}\,\sqrt{\vec k{}^2+m^2}\exp\left[i\vec k\cdot(\vec x-\vec y)\right]$$
Then $|\psi(f)|^2$ gives the probability density (in the space of functions) that $\varphi(x)=f(x)$. This is time independent because the ground state (like all energy eigenstates) is time independent.

3. Mar 6, 2014

samalkhaiat

You find what you need in

Sam

4. Mar 7, 2014

naima

You wrote in your link that $\psi(f)$ where f is a function of position and time is a probability amplitude (a number). In your notation, has f to be a Schrodinger solution of the system?.

I ask this question because i try to link this to Wightman axioms:
Look at W1 axiom
here f is just a test function and $\psi(f)$ is (with a spin 0 field) an operator defined on a dense subset D of the Hilbert space containing the vacuum.
How can we find your probability amplitude from these operator valued distributions?
What is the relation to your notation?

5. Mar 7, 2014

samalkhaiat

Where did I say that?

This is the usual setup in ordinary (operator, Heisenberg) formulation of QFT. The formulation I was describing is the coordinate representation of the Schrodinger picture of QFT. The two formulations are equivalent in QFT as they are in QM.

Sam

6. Mar 7, 2014

Avodyne

Actually just position.

No it's any function.

Wightman's $f$ and $\psi$ are not the the same as my $f$ and $\psi$. Let's change the name of the Wightman $f$ to $w$ and the Wightman $\psi$ to $\cal O$. Then ${\cal O}(w)=\int d^4\!x\,w(\vec x,t)\varphi(\vec x,t)$ is an operator. You can now ask for the probability distribution for the value of this operator in a given quantum state (though this is not something that usually gets asked by people interested in the Wightman axioms), and this could certainly be done in free field theory. I think I know how this would work out, but don't have the time to check right now.

7. Mar 7, 2014

naima

Thank you Avodyne
@Samalkhaiat
As the thread was closed (why?) i cannot use the QUOTE tool.
thread 26: "1) the field ϕ in Ψ[ϕ] is an ordinary function"
thread 26: "The wave functional Ψ[ϕ,t]=⟨ϕ|Ψ(t)⟩, represents the probability amplitude for the field to be in the configuration ϕ(x⃗ ) at time t. "

Last edited: Mar 7, 2014
8. Mar 7, 2014

samalkhaiat

I believe because it is old.

You need to understand that in this formalism, $\phi$ is just a coordinate point in the field space, same as x in the wavefunction $\psi ( x )$ of QM. And since we are considering the coordinate representation, the action of the operator $\hat{ \phi }$ on the wavefunctional results in multiplying the wavefunctional with the eigenvalue $\phi ( \vec{ x } )$. This corresponds to the usual coordinate representation of QM where we have $\hat{ x } \psi ( x ) = x \psi ( x )$. In the Schroginger picture the operators are independent of time, so like the time-independent $\hat{ x }$ in ordinary QM, the operator $\hat{ \phi } ( \vec{ x } )$ and its eigenvalue $\phi ( \vec{ x } )$ do not depend on time.

This is exactly like what you do in QM: $\psi ( x , t ) = \langle x | \psi ( t ) \rangle$

Sam

9. Mar 8, 2014

naima

It is interesting to see that ${\cal O}(w)$ does not depend on time.
If ${| \cal O}(w)| \Omega \rangle$ is the result of the measurement, it comes only from $w$ which is null outside of the apparatus and after the apparatus is switch off. $w$ may appear as a hidden variable.

Last edited: Mar 8, 2014