Statistical phase space distributions are related with Wigner functions defined by:

$f(x,k,t) = \int \frac{d^3x'}{(2 \pi)^3}e^{ikx'}\psi^*(x+x'/2,t)\psi(x-x'/2,t)$.

This definition holds only for nonrelativistic quantum mechanics. In relativistic quantum field theory I can't use this definition, since it is not Lorentz invariant. Now I have seen the phase space distribution function (hats denote operators)

$f(\phi,\Pi,t)=\int \frac{d^3\phi'}{(2 \pi)^3}e^{i\Pi \phi'}<\phi+\phi'/2,t|\phi-\phi'/2,t>=<0|\delta(\phi-\hat{\phi})\delta(\Pi-\hat{\Pi})|0>$

I can use the time evolution operator $U(t,t+T) = e^{-i \int_t^{t+T}dt'H(t')}$ to compute $f(\phi,\Pi,t)$ at a later time. After computing $(f(\phi,\Pi,t+T)-f(\phi,\Pi,t))/T = \partial_tf(\phi,\Pi,t)$ I can get a Boltzmann-like transport equation; there will occur integrals over Haar measure generated by $\phi,\Pi$.

Question: It is convenient to set $\phi = \frac{e^{i kx}}{\sqrt{2k_0 V}}, \Pi = \frac{\partial \phi}{\partial t}$ for a particle state in momentum $k$ and energy $k_0$. Is it possible to transform the phase space integrals over $\phi,\Pi$ to ordinary phase space integrals running over $x,p$ (as it is used for Statistical Physics)? Is it possible to make the Haar measure over quantum fields to a Lebesgue measure over $x,p$?

  • $\begingroup$ 1. What do you mean by "Haar measure"? A Haar measure is an invariant measure on a topological group, what is that topological group here? 2. What do you mean by "ordinary phase space"? The difference between field theory and point particle theories is precisely that the finite-dimensional phase space spanned by $x,p$ must be replaced by an infinite-dimensional space spanned by field configurations $\phi,\pi$. There is no "ordinary phase space" in a field theory, what are your $x,p$ supposed to be? $\endgroup$
    – ACuriousMind
    Oct 28, 2016 at 20:12
  • $\begingroup$ With Haar measure I mean the integral over all possible $\phi,\Pi$. Can I transform $d[\phi]d[\Pi]$ in terms of $d^3xd^3k$? $\endgroup$
    – kryomaxim
    Oct 28, 2016 at 20:13
  • $\begingroup$ That's not a Haar measure (where did you read that it is?), that measure is even in general mathematically ill-defined (not that that stops any physicist from using it). I repeat, in the setting of fields $\phi(x)$ and $\pi(x)$, what is your $k$ (or $p$) or whatever even supposed to be? $\endgroup$
    – ACuriousMind
    Oct 28, 2016 at 20:16
  • $\begingroup$ $p$ is momentum (I denoted also quantum momentum by $k$) and $x$ is space coordinate $\endgroup$
    – kryomaxim
    Oct 28, 2016 at 20:38

2 Answers 2


There are two possible objects that you can study, the Wigner function (which reduces to the ordinary distribution function), and the Wigner functional (which is a functional on the space of fields and their conjugate momenta).

To get to ordinary kinetics and the Boltzmann equation we study $$ W(x,p) = \int d^4y\, \exp(-ip\cdot y) \langle \bar\psi(x+y/2)\psi(x-y/2)\rangle $$ and derive an equation of motion for $W$. For Dirac fermions $W$ is a matrix in spin space. In gauge theory we have to put in gauge links. In scalar field theories the density matrix is of the form $\phi^\dagger\partial_\mu\phi$.

In order to get to ordinary kinetic theory we have to show that in the semiclassical limit $$ W(x,p) = A \delta(p^2-m^2) [\Theta(p_0)f_+(x,p) + \Theta(-p_0)(f_-(x,p)-1)] $$ where $A$ is a spin matrix ($(\gamma\cdot p+m)$ in Dirac theory), and $f_\pm$ satisfy the Boltzmann equation. This is described in standard text books, for example de Groot, van Leeuven, and van Weert, "Relativistic Kinetic Theory". The result is manifestly covariant, but the on-shell projectors ensure that $f_\pm$ is only a function of $\vec{p}$.

The Wigner functional is (for a scalar field theory, where $\pi$ is conjugate to $\phi$) $$ W[\phi,\pi] = \int d\psi \exp\left(-i\int \psi\pi\right) \langle \rho[\phi-\psi/2,\phi+\psi/2]\rangle $$ where $\rho$ is a density matrix in field space (see text books like Calzetta and Hu for definitions of $\rho$). The semi-classical limit of this object can be used to study thermalization of classical fields (which is a thorny subject).


Yes, what you are suggesting has been done, in Wigner trajectory characteristics in phase space and field theory, by T Curtright and myself, JPhysA:Math Gen 32 (5).

QFT need not be manifestly covariant, as it is an infinite array of oscillators after all, so manifest covariance is sacrificed here. I don't believe you have seen the 2nd expression just as you wrote it, as it precisely fails to have the functional measure [dφ'(x)] instead of a simple integral in one variable (remember, there is an infinity of φ(x)s, an infinity of oscillators!), and the simple product in the exponent should be an integral in dx, to reflect that infinity... the exponent is an infinite dimensional dot product. Cf. (35) , a functional $$W [\phi,\pi] =\int\!\left[ \frac{d\eta}{2\pi}\right] ~\Psi^{\ast}\left[ \phi-{\frac{\hbar}{2}\eta}\right] \, e^{-i\int dx\,\eta\left( x\right) \pi\left( x\right) }~ \Psi\left[ \phi+{\frac{\hbar }{2}\eta}\right] . $$

But the spirit of your expression is essentially sound, and the time evolution equation is indeed the infinite-dimensional phase space Moyal equation, (49), and you are on your way. Morphing it to Boltzmann through approximations depends on the nature of the Hamiltonian.

Free scalar field theories are duck soup, (48), $$ \partial_{t}W = -\int dx\,\left( \pi\left( x\right) \frac{\delta}{\delta \phi\left( x\right) }-\phi\left( x\right) \left( m^{2}-\nabla_{x}% ^{2}\right) \frac{\delta}{\delta\pi\left( x\right) }\right) W~ , $$ so pretty much some sort of functional rigid rotation; but interactions are a pain... the reason they don't much teach wavefunctional field theory in most places, anymore... A more intuitive talk version of the problem might be friendlier to start with.

  • $\begingroup$ I'm talking here from complete ignorance. Feynman propagators is a way to "get rid" of operators. Is is not just possible to simply try to construct a distribution function from this formalism? $\endgroup$ Apr 14, 2020 at 22:45
  • 1
    $\begingroup$ Yes, wherever you have an infinity of d.o.f., you end up with distribution functions... $\endgroup$ Apr 14, 2020 at 22:55

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