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Do quantum fields vanish at infinity?

  1. Jul 17, 2013 #1
    In Srednicki's textbook (chpt 5) he has an expression:

    [tex]\int d^3k f(k) \int d^4x (\partial^2 e^{ikx}) \phi(x) [/tex]

    and he wants to integrate by parts in order put the Laplacian on the field [tex]\int d^3k f(k) \int d^4x (e^{ikx})\partial^2 \phi(x) [/tex] instead of the the exponential. He says that surface terms vanish because f(k) is chosen so that [tex]\int d^3k f(k)e^{ikx} [/tex] is a wavepacket that will vanish at x=infinity.

    However, doesn't the field [itex]\phi(x) [/itex] and its derivatives vanish at infinity anyway, eliminating the need for f(k) and allowing you to use plane-waves? You don't need a wavepacket to vanish at infinity because [itex]\phi(x) [/itex] will.

    How else can you justify this manipulation in a path integral:

    [tex]e^{i\int d^4x \partial_\mu \phi \partial^\mu \phi}= e^{-i\int d^4x \phi \partial^2 \phi}[/tex]

    other than the field vanishes at infinity? In classical field theory, the reason you can write that is because surface terms don't matter as variations don't occur on the surface. But in quantum field theory, what allows you to integrate by parts like that? Don't you have to stipulate that the field and its derivatives vanish at ∞?
     
  2. jcsd
  3. Jul 18, 2013 #2

    tom.stoer

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    Quantum fields do not vanish at infinity.

    First of all in the canonical formalism they are operators, and you cannot make operators vanish at infinity.
    Then you may want to use perturbation theory in terms of plane wave states, and these do not vanish at infinity, either.

    This is just hand-waving. It seems to work physically, but is not rigorously defined mathematically. The basic argument is that the boundary conditions at infinity should not affect local observations, so setting fields to zero at infinity, applying periodic boundary conditions in finite volume (which is like using a 3-torus for the universe) is OK.
     
  4. Jul 18, 2013 #3
    Another trouble is the Ward Identity:

    [tex] T^{\mu \nu \eta}(p,q,r)=\int dxdy dz \mbox{ } e^{ipx+iqy-irz}<0|J^\mu(x) J^\nu(y) J^{ \eta}(z)|0> [/tex]
    [tex]r_\eta T^{\mu \nu \eta}(p,q,r)=i\int dxdy dz \mbox{ } \partial^z_\eta e^{ipx+iqy-irz} <0|J^\mu(x) J^\nu(y) J^{\eta}(z)|0> [/tex]

    Integrating by parts puts the z-derivative onto the J(z) instead of the plane-wave, and the 4-divergence inside a correlation vanishes, proving the Ward-identity. However, the surface term that is neglected when this is done gives rise to an anomaly that can be very important, and the reasoning given by the book is that the correlation function is infinite so the surface term becomes important.

    However, naively you expect the surface term to vanish, because [tex]J^\eta (z) [/tex] on the z=const hypersurfaces vanishes, assuming fields vanish at the hypersurface.

    But fields don't vanish on hypersurfaces necessarily, although many times it seems like you can say that they do.
     
  5. Jul 18, 2013 #4

    tom.stoer

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    I agree. Sometimes it seems that can do the calculation only if you already know the result.
     
  6. Jul 18, 2013 #5

    Demystifier

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    The question whether the QUANTUM field OPERATOR vanishes at infinity is analogous to the question whether the GENERAL solution of the CLASSICAL field equation of motion vanishes at infinity. Neither of them does.

    What in classical field theory vanishes at infinity is a PARTICULAR solution which describes the ACTUAL state of field, i.e., the actual field configuration. Analogously, in quantum field theory one needs to study the ACTUAL state in the Hilbert space and the corresponding expectation value of the field operator. For a realistic state, the corresponding expectation value of field does vanish at infinity.
     
  7. Jul 18, 2013 #6

    tom.stoer

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    Agreed.

    All I wanted to indicate is that something like

    ##\lim_{x\to\infty}\phi(x) = 0##

    is incompatible with

    ##[\phi(x),\pi(y)] =i\delta(x-y)##
     
  8. Jul 19, 2013 #7

    hilbert2

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    When a field equation is derived variationally from a lagrangian density function, it is usually assumed that the "surface term" vanishes... Isn't this based on an assumption that the field vanishes at infinity?
     
  9. Jul 19, 2013 #8

    tom.stoer

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    Yes, this is a standard assumption in field theory. But in quantum field theory the field equations are of minor importance. Anyway, the Hamiltionian is a formal object for which a rigorous definition cannot be given in many cases. Boundary terms are only one subtlety
     
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