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A Field strength measurement and continuity

  1. Jun 28, 2017 #1

    hilbert2

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    Something about the theory of quantum measurement/collapse in the case of quantum fields... Suppose I have a field, either a scalar, vector or spinor, that I want to describe as a quantum object. For simplicity let's say that it's a scalar field ##\phi (\mathbf{x})##, where the ##\mathbf{x}## is a 3D position vector. The corresponding field operator is denoted ##\hat{\phi}(\mathbf{x})##.

    If I do a measurement of the field strength ##\phi (\mathbf{x})## at some point ##\mathbf{x}##, does this also affect the field at points neighboring ##\mathbf{x}## ? If the field is required to be continuous, the "collapse" of the field strength to some value at point ##\mathbf{x}## is bound to limit the field strengths in some open 3-sphere containing ##\mathbf{x}## to an arbitrarily small neigborhood of the measured value.

    The differential forms of Maxwell, Klein-Gordon, Schrödinger equations etc. obviously assume that the field is a continuous and differentiable function, but it's possible to circumvent this limitation by converting the eqs to integral form. Has the quantum theory of measurement even been developed to the point of being able to handle field strength measurements?

    Edit: Clearly one can't have an arbitrarily small "test charge" for measuring the field without affecting it significantly at the same time, as electric charge is quantized.
     
    Last edited: Jun 28, 2017
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  3. Jun 29, 2017 #2

    Vanadium 50

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    As asked, your question is "what is the value of φ when I am not measuring it". That is, of course, unanswerable. An answerable question is, "if I measure φ at r1 and t1, what will I measure at r2 and t2?" However, I suspect that, just like in the classical world, by itself this isn't all that constraining.
     
  4. Jun 29, 2017 #3

    hilbert2

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    Sorry, the thing that I had in mind was whether there could be some rule, similar to HUP, that limits how accurately the field strength at a given position ##\mathbf{x}## can be measured without affecting the probability distribution of the field strengths at a nearby point ##\mathbf{x} + \mathbf{\delta x}##, measured a short time ##\delta t## after the first measurement.

    This probably wasn't as good a question as I first though it was, because I wasn't really thinking about the practicalities of how electric or other field strengths are measured.

    Edit: If I have a 1D Klein-Gordon field and act on its vacuum with a creation operator, getting a state ##\hat{a}_p^\dagger \left|0\right>##, is that state an eigenfunction of the field operator at every point x: ##\hat{\phi} (x)\hat{a}_p^\dagger \left|0\right> \propto e^{ipx}\hat{a}_p^\dagger \left|0\right>##, or is there an uncertainty in the phase of the plane wave? If the state is an eigenfunction of ##\hat{\phi} (x)## for every value of ##x##, then I could in principle talk about the actual configuration of the field, not just probabilities of different measured results.
     
    Last edited: Jun 29, 2017
  5. Jun 29, 2017 #4

    Vanadium 50

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    There is an uncertainty in the phase of the plane wave. This is the basis of local gauge invariance. I can insert an unobservable phase at every point in spacetime, and since it's unobservable, all functions of this phase have to disappear from the equations of motion. This severely constrains the allowed behavior of entities in this theory.
     
    Last edited: Jun 29, 2017
  6. Jun 29, 2017 #5

    hilbert2

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    Ok, thanks. So if I do field strength measurements at some point ##\mathbf{x}## for a state ##\hat{a}_p^\dagger \left|0 \right>##, I will get random numbers between some limits determined by the amplitude of the field.

    Does there exist any state vector ##\left|\psi \right>## that's simultaneously an eigenfunction of the field operator at every point?
     
  7. Jun 29, 2017 #6

    Vanadium 50

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    I don't think so. Isn't this how we get into second quantization?
     
  8. Jun 29, 2017 #7

    hilbert2

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    Good. I didn't get very far in QFT when I was an undergraduate, and later I've been trying to learn it occasionally just for fun.

    It's often said that the operator ##\hat{\phi}(x)## creates a localized particle at point ##x## when it acts on the vacuum ##\left| 0\right>##. Is the state that's produced there actually an eigenfunction of the particle number operator with eigenvalue ##1## (as in the case of the state obtained by applying a creation operator ##\hat{a}_p^\dagger## on ##\left| 0\right>##) or is the particle number uncertain with one of the particles being at point ##x## with 100% probability?

    Edit: Apparently it is, as ##\hat{\phi}(x)\left| 0\right> = \int \frac{dp}{2\pi}\frac{e^{-ipx}}{\sqrt{2\omega_p}}\left| p\right>##, where ##\left| p\right>## is a state with one particle of momentum ##p##. Acting on this with the number operator I get

    ##\hat{N}\hat{\phi}(x)\left| 0\right> = \int \int \frac{dpdp'}{(2\pi )^2}\frac{e^{-ip'x}}{\sqrt{2\omega_p'}}\hat{a}_p^\dagger\hat{a}_p \left| p'\right>##

    and the action of the lowering operator ##\hat{a}_p## on the state ##\left| p'\right>## produces a ##\delta (p-p' )## which makes the integral same as ##\hat{\phi}(x)\left| 0\right>##...
     
    Last edited: Jun 29, 2017
  9. Jul 10, 2017 #8

    hilbert2

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  10. Jul 11, 2017 #9

    vanhees71

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    Well, that's a bit subtle. First note that
    $$\langle \vec{p}|\vec{p}' \rangle=\delta^{(3)}(\vec{p}-\vec{p}').$$
    So it's a generalized eigenstate not a normalizable true single-particle state. To get one you need to "smear it" with a square-integrable function ##A(\vec{p})##
    $$|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} A(\vec{p}) |\vec{p} \rangle.$$

    Now you get
    $$\hat{N} |\psi \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}' \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} A(\vec{p}) \hat{a}^{\dagger}(\vec{p}') \hat{a}(\vec{p}') \hat{a}^{\dagger}(\vec{p})|\Omega \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p}' \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} A(\vec{p}) \hat{a}^{\dagger}(\vec{p}') [\hat{a}(\vec{p}'), \hat{a}^{\dagger}(\vec{p})]|\Omega \rangle,$$
    because ##\hat{a}(\vec{p}') |\Omega \rangle=0## but now ##[\hat{a}(\vec{p}') ,\hat{a}^{\dagger}(\vec{p})] = \delta^{(3)}(\vec{p}-\vec{p}')##. Plugging this into the equation gives that for all ##A(\vec{p})## you have
    $$\hat{N} |\psi \rangle=|\psi \rangle,$$
    i.e., it's an eigenstate of ##\hat{N}## with eigenvalue 1, i.e., a state with the total particle number determined to be 1. So all these states are single-particle states.

    Note: I assumed the normalization to be such that
    $$\hat{\phi}(x)=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi)^3 2 E}} [\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x) + \hat{a}^{\dagger}(\vec{p}) \exp(+\mathrm{i} p \cdot x)]_{p^0=E}, \quad E=\sqrt{m^2+\vec{p}^2}.$$
    Then the ##\hat{a}^{\dagger}(\vec{p})## create a single-particle generalized momentum eigenstate, "normalized to a ##\delta## distribution" as written above.
     
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