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Probability of states in QFT

  1. Mar 6, 2014 #1
    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. jcsd
  3. Mar 6, 2014 #2

    Avodyne

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    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
    [tex]\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][/tex]
    where ##K(\vec x,\vec y)## is formally defined by something like
    [tex]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][/tex]
    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.
     
  4. Mar 6, 2014 #3

    samalkhaiat

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    You find what you need in

    www.physicsforums.com/showthread.php?t=388556&page=2

    Read post#5, then jump to posts# 26 to 31

    Sam
     
  5. Mar 7, 2014 #4

    naima

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    You wrote in your link that [itex]\psi(f)[/itex] 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 [itex]\psi(f)[/itex] 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?
     
  6. Mar 7, 2014 #5

    samalkhaiat

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    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
     
  7. Mar 7, 2014 #6

    Avodyne

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    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.
     
  8. Mar 7, 2014 #7

    naima

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    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
  9. Mar 7, 2014 #8

    samalkhaiat

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    I believe because it is old.

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

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

    Sam
     
  10. Mar 8, 2014 #9

    naima

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    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
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