Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One of methods of introducing of virtual particles is using perturbation theory. We say that scattering matrice amplitude $M_{in \to out}$ contains of $\delta(P_{out} - P_{in})$, which realizes energy-impulse conservation and relativistic connection $E^{2} - p^{2} = m^{2}$. When we leave only a few terms of the expansion this leads to unconservation energy at the interaction stage. But in final we obtain that energy-impulse is conserved.

So my question is following: how exactly the dropping of terms of higher order by the coupling constant leads to energy-impulse non-conservation and how exactly do we obtain that in final energy-impulse is conserved?

Also there is a question about delta-function. Whether it is administered solely by requiring the conservation of energy-momentum? Or it arises naturally from the very formulation of the problem?

share|cite|improve this question
It arises naturally order by order. If you are computing by Feynman diagrams then every diagram already has an energy-momentum conserving delta function in it. Where did you hear that dropping higher order terms leads to non-conservation of energy-momentum? You may be stuck using the "old" (i.e. before the 1950s) perturbation theory, which wasn't fully covariant. Energy-momentum conservation is due to spacetime translational symmetry, which is manifest at every stage of a covariant perturbation theory calculation (in vacuum). – Michael Brown Oct 22 '13 at 1:30
@MichaelBrown . But how arises possibility of the existence of virtual particles when neglecting higher terms? For virtual particles we may assume that they don't conserve energy, because they aren't on mass surface of their real analog, don't it? – Andrew McAddams Oct 22 '13 at 3:41
Or this doesn't say about non-conservation of energy? – Andrew McAddams Oct 22 '13 at 3:47
The thing is(as an elaboration of Michael Brown), energy-momentum is conserved at every intermediate step, because there is always a $\delta(p-k)$ whenever there is an energy-momentum exchange. The only unusual thing is that in intermediate steps, -energy-momentum $p,k$ do not obey the "relativistic connection" as you mentioned. – Jia Yiyang Oct 22 '13 at 4:54
@JiaYiyang is correct. Popularizers who say virtual particles "borrow" energy from the vacuum or anything like that are being misleading, at least from the standpoint of covariant perturbation theory. In the old non-covariant perturbation theory it is true to say that energy conservation is violated in intermediate states, only being restored when all intermediate times are integrated over. But hardly anyone uses that theory to compute the S matrix anymore, mainly because its good properties - like energy conservation - are hidden. – Michael Brown Oct 22 '13 at 5:12
up vote 3 down vote accepted

It is better to begin with amplitudes calculus in position space, because things become more clear (usual "Feynman diagrams" correspond to momentum space).

Suppose, for instance, a massless scalar field theory, with a $\phi^4$ interaction. In this theory, each vertex has 4 "legs", which may be real particles or field perturbations/field correlations (also unfortunately called "virtual particles")

Now, look at the corresponding process : $2$ in-particles "beginning" at $x_1, x_2$, joining at a first vertex in $z$, then two field perturbations beginning at $z$ and ending at a new vertex $w$, plus $2$ out-particles beginning at $w$ and ending at points $x_3$ and $x_4$.

We are interested in the amplitude of the process $A(x^1,x^2,x^3,x^4)$. The amplitude for a field perturbation beginning at $x$ and ending at $x'$ is $D(x-x')$, the propagator. The amplitude for a real particle (the in- and out- particles) is $\delta(x-x')$

So, we have :

$A(x^1,x^2,x^3,x^4) \sim \int dz ~dw ~\delta(x^1 - z)\delta(x^2 - z) [D(z-w)]^2 \delta( w - x^3 )\delta(w - x^4) \tag{1}$

(here all the integrations and $\delta$ functions are $4$-dimensional)

Now, look at the Fourier transform : $\tilde A(p_1,p_2,p_3,p_4)\sim \int dx^1dx^2dx^3dx^4 ~e^{-i(p_1x^1 +p_2x^2+p_3x^3+p_4x^4)}A(x^1,x^2,x^3,x^4)$

We suppose here that the in and out particles are real particles, so that $p_i^2=0$ $1 \leq i\leq 4$ (for simplicity we do not put explicitely the $\delta(p_i^2)$ functions, this is just a global term)

You get :

$\tilde A(p_1,p_2,p_3,p_4) \sim \int dz dw\int e^{-i((p_1 +p_2) z -(p_3+p_4)w)}[D(z-w)]^2 \tag{2}$

Let $u=z+w, v=z-w$, so that $z =\frac{u+v}{2}, w =\frac{u-v}{2}$, we have :

$\tilde A(p_1,p_2,p_3,p_4) \sim \int e^{-i((p_1 +p_2) +(p_3+p_4))\large \frac{v}{2}}[D(v)]^2 \int du e^{-i((p_1 +p_2) -(p_3+p_4))\large \frac{u}{2}} \tag{3}$

The integration on $u$ gives a $\sim \delta((p_1 +p_2) -(p_3+p_4))$, so you have :

$\tilde A(p_1,p_2,p_3,p_4) \sim \delta(p_1 +p_2- p_3-p_4)\int dv ~e^{-i(p_1 +p_2)v}[D(v)]^2 \tag{4}$

Let $\tilde D(p)$ be the Fourier transform of the propagator $D(x)$, we have finally :

$\tilde A(p_1,p_2,p_3,p_4) \sim \delta(p_1 +p_2- p_3-p_4)\int dp \tilde D(p) \tilde D(p_1+p_2-p) \tag{5}$

Now, we are in momentum space, the domain of Feynman diagrams.

We see the difference between real particles and fields perturbations. Real particles, (in and out) have a on-shell momenta $p_i^2=0$, and there is a global conservation of momentum-energy.

There is also a conservation of momentum/energy between real particles and field perturbations, this may be seen, in the last expression in the term $\tilde D(p)\tilde D(p_1+p_2-p)$. At the first (momentum) vertex, the total momentum is $p_1+p_2$, and there is a repartition of the momentum into the two legs (fields perturbations). But you may see that the integration is on all $p$, so there is no mass-shell conditions for the fields perturbations. Field perturbations are not particles.


If you are not satisfyed with formula $(1)$,it could be obtained from the LSZ reduction formula, which states, roughly :

$$A(x^1,x^2,x^3,x^4) \sim \square_{x_1}\square_{x_2}\square_{x_3}\square_{x_4} \int dz ~dw ~D(x^1 - z)D(x^2 - z) [D(z-w)]^2 \\ D( w - x^3 )D(w - x^4) \tag{6}$$

The connection with the formula $(1)$ is done with $\square_{x_1}D(x^1 - z) = \delta(x^1 - z)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.