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What is the interpretation of the vacuum correlation function in QFT

  1. May 9, 2012 #1
    I'm currently trying to make some intuitive sense out of Quantum field theory, but I'm not really understanding the vacuum.

    Consider a real (or complex, with + in the right places) scalar
    particle (a Klein-Gordon field).

    Now consider the propagator (or correlation function)
    G(x-y)= <0|phi(y) phi(x) |0>
    (where I assume that the two operators are correctly time-ordered already)

    My questions are related to how exactly to interpret this.

    A standard interpretation seems to be to say that since phi(x) is (or
    contains) a creation operator for a virtual particle, this describes
    the creation of a virtual particle at x and its destruction at y.

    My first problem is that intuitively this does not make much sense,
    because the vacuum is completely Lorentz invariant, so for each particle
    created at x (and later destroyed at y), there would be another
    virtual particle that was created at an earlier time and is destroyed
    at x. After all, the vacuum is absolutely stationary.

    So here is my Question 1: In what sense is this interpretation to
    be understood?

    I could try to understand it in a sense from a path integral
    interpretation. I look at the vacuum-to-vacuum amplitude <0|0>. There
    I have to integrate over all field configurations that have different
    contributions according to their action. For each configuration, I
    could calculate phi(y)phi(x), and summing over all configurations with
    the appropriate weights, I could see the correlation. This would give
    meaning to the term "vacuum fluctuation" if applied to each
    configuration separately - if the field value is very large at x and
    if y is a near-by point, the field will with higher probability be
    large there as well because otherwise the action would be very
    large. I think this is how correlation functions are used in lattice
    gauge theory.

    So Question 2 is simply: Is this interpretation within the path
    integral formalism correct? If so, is there a way to transfer it to
    the canonical formalism?

    Furthermore, on the other hand, standard lore is that for any operator Q
    is the vacuum expectation value of Q, i.e., it is the value I would
    get if I were to perform a very large number of measurements of the
    variable Q.

    So if I now set Q=phi(y)phi(x), this would tell me that G(x-y) is the
    expectation value I would get if I measure first phi(x), then phi(y)
    and then calculate the product of both.

    So let's create infinitely many copies of the vacuum (so I can perform
    enough measurements to get an expectation value). In each of my
    copies, I now perform a measurement of the field at x. This will
    create a new state of my universe. In quantum mechanics, I would say
    that this new state would be an eigenstate to phi(x) (which would then
    evolve), but - here is my Question 3 - does phi(x) actually *have*
    eigenstates? (If so, what are they, especially if phi is a complex
    field?) And is there actually any way I could perform this
    measurement of phi? How do you measure a field value of a quantum
    field anyway? (Again, especially in the case of a complex field?)

    Since I'm now in a new state phi(x)|0>, I could interpret
    <0|phi(y) phi(x) |0>
    as the matrix element of getting from my new state back to the vacuum
    state by applying another field measurement phi(y). This would then
    mean that <0|phi(y) phi(x) |0> is something like the probability of
    getting from vacuum back to a vacuum if I apply two subsequent
    measurements of phi (similar to Fermi's golden rule, where a matrix
    element also gives a transition probability). So finally, my question
    4: Would this be a correct interpretation of <0|phi(y) phi(x) |0>?

    This is my first post here - I've read a lot hereabouts, but did not really find an answer to these question. Thanks for any help.
  2. jcsd
  3. May 9, 2012 #2
    The vacuum is Lorentz invariant, but the state phi(x)|0> is not Lorentz invariant; the Lorentz-invariant vacuum has been hit with an operator that is explicitly associated with a specific spacetime point. To be clear, the operator phi(x) hits a state with no particles, |0>, and turns it into a state with one particle (if we're talking about a free theory here). The state |0> contains no particles for phi(x) to destroy.

    Lorentz invariance tells you thinks like

    [itex]\left<0|\phi(y)\phi(x)|0\right> = \left<0|\phi(y-x)\phi(0)|0\right>[/itex]

    i.e., the amplitude to propagate from x to y is the same as the amplitude to propagate from the spacetime origin to y-x, and perhaps this is what your intuition corresponds to?

    Yes, your description sounds good. Are you asking for the steps used to derive the path integral from the operator formalism, and vice versa? Or are you looking for something else?

    There are some issues here. For instance, phi(y)phi(x) is not in general Hermitian even if we are talking about a real scalar field (which is why its vacuum expectation value, G(x-y), is in general complex and not real). So phi(y)phi(x) apparently not a good observable.


    I personally don't have particularly good answers for your questions about measuring the field operator, but I think again you should not expect it to be meaningful to talk about measuring a operator that is not Hermitian. Mathematically, at least, if phi is real then phi(x), being Hermitian, should indeed have a complete set of eigenstates, but I'm not aware that these eigenstates have any particularly straightforward physical interpretation.

    However, if we decide that there is indeed some physical operation that corresponds to "measuring" phi(x), note that doing this measurement puts the system in an eigenstate of phi(x), not the state phi(x)|0>, which is not an eigenstate of phi(x). I don't think that <0|phi(x)phi(y)|0> should be interpreted as a series of two measurements of phi(x) at different spacetime points.
  4. May 10, 2012 #3
    Hi TheDuck,

    thanks for the answer.

    Sigh of relief. I still have trouble to transfer this intuitive picture to the canonical picture - there, I should have states that are superpositions of field configurations resulting in the same things as in the path integral picture (similar to psi(x) being a superposition of the particle position states in normal QM), but I have no clue how to write this down mathematically.

    I'm also still confused by the meaning of phi(x).

    If phi(x) does not correspond to any physical observable, then what is meant by calling <0|phi(y)phi(x)|0> a "correlation function"? What is the thing that is correlated?

    I'm especially puzzled when I think of a quantum field theory of, for example, an elastic membrane or a mattress (an example used in the book by Stone, "Physics of Quantum fields").
    Here, the classical field phi(x) is directly the displacement of the medium which is surely a good and valid observable. How can quantising this kind of theory lead to an operator phi that does not correspond directly to this observable?

    So at least in this kind of theory (which is a real scalar theory), phi(x) should have a clear physical interpretation, shouldn't it?
  5. May 10, 2012 #4
    Ah, yes, of course in this case phi corresponds to exactly the physical observable it represents classically. I guess I was thinking about the case of the fields used to describe fundamental particles, where I don't know what it would mean exactly to measure the field operator.
  6. May 10, 2012 #5
    O.k., thanks for that clarification.
    Neither do I know, that's probably part of my problem with the VEV.
    If the particle were a real scalar Klein-Gordon-particle, like a neutral pion, would phi classically be a kind of particle density?
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