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In what sense do static field configurations enter into perturbative QFT?

+ 6 like - 0 dislike

Taking the Klein-Gordon equation with a Yukawa coupling we have $$ \Box \phi - m^2 \phi \sim \bar{\psi}\psi$$ (up to a sign/phase convention). Imposing $\partial_t \phi = 0$ and $\bar{\psi}\psi \sim \delta$ we obtain the classical solution $$\phi \sim \frac{e ^{-m r}}{r}. $$ I.e., we can understand the Yukawa potential as the static field of a static particle in a certain classical limit. To leading order the scattering amplitude of $\psi$ is $\sim 1/(\vec{k}^2 + m^2)$ which is exactly the scattering amplitude obtained from the Yukawa potential in the Born approximation.

The question is: How is it that the QFT computation and the diagrammatic representation of the amplitude suggest that the process is fully mediated by a single and fully dynamical (time-like) particle event though the non-QFT picture describes the fact in terms of a fully static (space-like) interaction?

I have some thoughts on this, but not a concise picture (I would not be asking the question otherwise). My hand-waivy idea is that the static-field part is actually hidden in the off-shell part of the propagator. Since the propagator is the Green's function of the respective differential equation, it includes all kinds of ways the field can be correlated, even in the "static" (space-like correlated) configuration. Hence, by not cutting out the off-shell, we simply include other configurations established otherwise than by travelling waves.

On the other hand, it is hard to reconcile this notion with a strictly particle-like or "Weinbergian" picture of QFT. The Yukawa potential would imply that the fermion $\psi$ "carries around" a superposition of $\phi$ particles with a momentum distribution $\sim 1/(\vec{k}^2 + m^2)$, however with $k^0 \approx 0$! Furthermore, the off-shell part of the propagator is usually assigned to vacuum fluctuations which does not seem to have any connection to the previous argumentation.

asked Apr 22, 2015 in Theoretical Physics by Void (1,505 points) [ no revision ]

3 Answers

+ 8 like - 0 dislike

The standard answer to this is found by replacing the dynamical sources in the interacting field theory with classical sources, by adding a term $J(x) \phi(x)$ to the otherwise free scalar Lagrangian. The source $J(x)$ models the location of excitations of other quantum fields, in the limit that they are heavy, or moving in a way prescribed externally, and the path integral gives the vacuum-persistence amplitude, which in Minkowski space is $\exp(iEt)$, where E is the vacuum energy, in this case, the energy of the source-vacuum.

The source problem is a simple quadratic path integral, which I write in the Euclidean version:

$$ Z[J] = \int e^{- \int_k (k^2 + m^2)|\phi_k|^2 + J(k)\phi(k) {d^4 k \over (2\pi)^4}} D\phi $$

$$ {Z[J] \over Z[0]} = e^{\int {|J(k)|^2 \over k^2 + m^2} dk} $$

Going back and forth from k-space to x space converts multiplication to convolution, and the free energy log(Z[J]/Z[0]) is:

$$ \log(Z[J]) = \int \int J(x) G(x-y) J(y) d^4x d^4y $$

This path integral is also easily done by summing independent diagrams, where the only diagram describes a particle created by the source, propagating, and annihilating by the source. This instructive exercise is done in an early lecture of Sidney Coleman's Quantum Field Theory course (1975-76 version here: https://www.physics.harvard.edu/events/videos/Phys253 ).

If you want to find the static potential energy between two point sources, substitute $J(x,t) = j(x) = g\delta(x-a) + g\delta(x)$, two static points. To regulate the long-time limit you can consider J(x,t) to turn on and off gradually over the interval [0,T], to regulate the point-source, you can imagine the delta function is fattened up a little. The double time-integral just tells you that the vacuum phase keeps rotating as T gets longer. The energy of the source configuration is the remaining integral:

$$ E[J] = \int J(x) G(x-y,t) J(y) d^3x d^3y dt $$

When you evaluate the delta-functions, you find three contributions. One is equal to the classical self-energy of one delta-function, and is divergent value of G(0,t). The other is the self-energy of the other delta function. The third is the interesting contribution, the part of the energy that depends on the distance $a$ between the two point sources.

$$ E[a] = g^2 \int G(a,t) dt $$

The energy is the integal of the propagator over all time, which is Fourier transforming time at $k_0= 0$. If you Fourier transform space, you just get the usual propagator, except at the special value $k_0=0$, so you reproduce your observation that the $k_0=0$ propagator is: $$ G(k) = {1\over k^2 + m^2}$$ and is the Fourier transform of the Yukawa potential.

You can consider the classical field as forming a cloud of independent scalar particles around the source, and the absorption/emission of the cloud is the potential energy. The scalars wander back and forth in time, but for static situations, you are looking at all contributions, forward, backwards, and spacelike, since you are substituting $k_0 =0$.

Another interesting classical source to consider is the uniformly accelerated delta function. This is treated by Unruh in 1976, and reviewed in "Quantum Fields in Curved Space" by Birrell and Davies. This motion of the source takes a different integral through the Green's function than the constant-motion one for static sources. The evaluation of the behavior here is made easier by considering new particle mode operators for one quadrant. You can also consider oscillating sources as an exercise, etc, to get intuition for the emission statistics in free field theory.

The variations on this toy problem reveals how classical linear field potentials are recovered from quantum field theory. They can be thought of as describing independent particle propagation from sources to sinks, or equivalently (because it is a quadratic path integral) as the (in this case exact) Gaussian approximation of the response of the field to the source. Full quantum field theory is then obtained when you make the physical sources other dynamical fields.

In full quantum field theory, considered in terms of diagrams, you recovering the usual nonrelativistic theory from the ladder approximation, where you consider some particle legs are slowly moving, exchanging particles along horizontal rungs, and the propagation along the rungs pretty much reproduces the static source situation. This is reviewed lots of places in the 1960s literature, but my favorite is the brief telegraphic explanation in Gribov's "The Theory of Complex Angular Momenta".

The limit of potential scattering in full quantum field theory is appropriate when the particles scatter nonrelativistically slowly, as Vladimir Kalitvianski said already. The instantaneous exchange of particles is approximated by the potential energy as a function of distance to a close enough approximation.

The $\psi$ particle, in the final analysis, can be viewed as carrying a cloud of $\phi$ particles around it, all independently propagating. The $k_0=0$ projetion comes from considering slowly moving particles, so that you integrate over all times of emssion and absorption, projecting out to pure spatial momentum. The scattering in terms of Feynman diagrams is a representation of the response of one particle to other particle's clouds. So your intuition is fine, except that the $\phi$ clouds include all virtual momenta, and the lowest order contributions to scattering (rather than self-energy) are those where the $\phi$ links one $\psi$ to another $\psi$.

answered Apr 22, 2015 by Ron Maimon (7,535 points) [ revision history ]
edited Apr 23, 2015 by Ron Maimon

Nice, classical sources and path integrals make this look neat. So say we only have a single static $\delta$ as a classical source, then $\langle \phi \rangle$ would be the classical static field configuration, right?

I am still banging my head against the virtual $k^0$ in the $\phi$ cloud, say we were to try to find $\phi$ particles near the classical source -- what would be the distributions of four-momenta (and most importantly $k^0$) of such particles? I have a feeling that no particles could be found because no momentum can be transferred to/from the static $\delta$, but what what is then the meaning of $\langle \phi \rangle$?

@Void--- the source is static, so if you probe with a second static source, you only see the contributions of $k_0=0$, all other contributions wash out by phases. That's how the static potential arises, and why it's at $k_0=0$. If you probe with a dynamic external source, you see other frequencies to the extent that there are frequencies in the motion of the external source. But then again, that oscillating source would also see particles in a vacuum.

The meaning of $\langle\phi\rangle$ in this problem can be defined by the response to an external source coupled to $\phi$. The change in $\log Z$ with respect to the source at x is the expected value of $\phi(x)$. So you can understand $\langle\phi(x)\rangle$ for the source problem as the amplitude for a new infinitesimal source to absorb the particles emitted by the first source. This source-centeric definition was Schwinger's preferred way, so he called field theory "source theory". You have local field theory as long as you have a reasonable set of local probe sources. You stop having field theory once the sources are restricted, like in string theory, where the sources are not arbitrary fiddlings you can do at a point, but they have to be actual excitations of the theory you can see by scattering to infinity, like a D-brane.

I really recommend getting intuition for this by analyzing the source problem, it's why Coleman puts it first in the lectures, it really un-confuses this issue.

+ 6 like - 0 dislike

In QFT one uses, strictly speaking, the retarded (i.e., time-dependent) potentials. But for a slow motion the main part can be approximated as a "static" or, better, time-dependent "instant" Yukawa (or Coulomb) potential $V(r(t))$ with small velocity-dependent corrections.

The time-dependent field consists of a "near" field and a "propagating" field. The propagating part is relatively small in magnitude for a slow motion. "Vacuum fluctuations" exist in both fields.

Propagators (Green's functions) are expressed via the  solutions of free equation (a spectral representation). But propagators are not gauge invariant, so splitting the propagator may not be too informative.

answered Apr 22, 2015 by Vladimir Kalitvianski (22 points) [ revision history ]
edited Apr 22, 2015 by Arnold Neumaier

+1. Given OP's

The question is: How is it that the QFT computation and the diagrammatic representation of the amplitude suggest that the process is fully mediated by a single and fully dynamical (time-like) particle event though the non-QFT picture describes the fact in terms of a fully static (space-like) interaction?

you might want to expand and elaborate your

But for a slow motion the main part can be approximated as a "static" Yukawa (or Coulomb) potential V(r(t))with small velocity-dependent corrections.

for example, by comparing the QFT tree graph result with Born approximation result in QM. 

@JiaYiyang: There is a good explanation for the classical electromagnetic field in many textbooks (retarded potentials and their expansions). It gives a good physical picture how retarded (time-like) solution can "contain" "instant" parts. The main "instant" part is just not really exhaustive (it is approximate numerically) and does not contain the propagating (radiated) part, but higher orders "remove inexactness".

This is interesting, is there some targeted reference where I could study this into a bit more detail? For example P&S do not discuss such themes as far as I remember.

@JohnK, P&S do talk about such stuff in chap 4.7, where they show how you can reproduce Yukawa or Coloumb potential from tree level QFT.

+ 2 like - 0 dislike

The problem with $k_0$ in QFT is quite similar to the classical field problem. Look, for example, the spectral (Fourier) decomposition of a static and moving charge field in Landau-Lifshitz textbook. (This field gets into the equations of motion of another - probe - charge.) The source field, when static, depends only on the space wave vectors $\mathbf{k}$ with $k_0=0$ (not time-dependence). In a moving reference frame or in case of a uniformly moving source, this field is modified: each space harmonics gets a time-dependent phase $\exp(-i\mathbf{k}\mathbf{V}t)$, where $\mathbf{V}$ is the sourcing charge velocity. It looks as if the field charge coordinate $\mathbf{r}$ were replaced with $\mathbf{r}-\mathbf{V}t$. The field harmonics move with the charge velocity. So $k_0(\mathbf{k})\propto \mathbf{k}\mathbf{V}\ne 0$.

In case of non-uniformly moving charge, there is a radiated addendum with $k_0\propto kc$. This is what the probe charge "feels". Its own field does not influence the source in such a problem setup.

In case of interaction of charges (source and probe), each charge "feels" the retarded field of the other charge (as well as its own self-field, unfortunately, but let us forget it for simplicity). Both charges are sources, according to the Maxwell equations.

Returning to QFT where we have interaction (mutual influence), in QED textbook by Bersetetski-Lifshitz-Pitaevski, there is a short analysis of the latter situation (the Chapter about photon propagator). By the end of the Chapter, there is an expression of propagator in a "mixed representation": $D(\omega,\mathbf{r})=-e^{i|\omega|r}/r$. This function corresponds to the classical retarded field $\propto e^{i\omega r}$, but only for positive frequency. It means that in QED the "source" is the charge that gives away its energy, or more exactly, which "radiates the virtual photon" (i.e. creates it for the other charge). As the frequency integration covers positive and negative values, both charges (electrons on a diagram, to be exact) are "sources" - one for the other.

answered Apr 24, 2015 by Vladimir Kalitvianski (22 points) [ revision history ]
edited Apr 24, 2015 by Vladimir Kalitvianski

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