# 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