Due to the Wiki article, "...In quantum mechanics and quantum field theory, the propagator gives the probability amplitude for a particle to travel from one place to another in a given time, or to travel with a certain energy and momentum...".

Let's have the expression for the propagator of some field $\hat {\Psi}_{l}$ in the free theory: $$ \tag 1 D_{lm}(x - y) = \langle |\hat{T}\left( \hat {\varphi}_{l}(x)\hat {\varphi}_{m}^{\dagger}(y)\right) |\rangle = -\frac{i}{(2 \pi )^{4}}\int \frac{F_{lm}(p)e^{-ip(x - y)}d^{4}p}{p^{2} - m^{2} - i\varepsilon}. $$ Let's use two examples: spinor and scalar fields: for them $(1)$ takes the form $$ \tag 2 D_{lm}(x - y) = D(x - y) = -\frac{i}{(2 \pi )^{4}}\int \frac{e^{-ip(x - y)}d^{4}p}{p^{2} - m^{2} - i\varepsilon}, $$ $$ \tag 3 D_{lm}(x - y) = -\frac{i}{(2 \pi )^{4}}\int \frac{(\gamma^{\mu}p_{\mu} + m)_{lm}e^{-ip(x - y)}d^{4}p}{p^{2} - m^{2} - i\varepsilon} $$ respectively.

How to convert $(2), (3)$ into the probability (which lies at interval $[0, 1]$)? Particularly i don't understand what to do with spinor indices in $(3)$ (the probability must be Lorentz scalar, so I need to sum over the indices (?)). And also, why doesn't probability depends on momentum (the propagator doesn't contain info about momentum)?

  • 1
    $\begingroup$ See Feynman diagrams. You must take in account vertex factors, and incoming/outgoing particles. Propagator (in momentum space) does contain information about momentum. $\endgroup$
    – Trimok
    Commented Jul 7, 2014 at 8:24

1 Answer 1


The propagator is a different object than the probability, even though they are essentially the same in time-independent quantum mechanics. In quantum field theory, $D(x-y) = \langle \psi(y)\psi^{*}(x)\rangle$ which is not a conserved quantity; in fact, it is badly divergent on the light-cone even for non-relativistic fields. On the other hand, for fermions and scalars, we take the conserved charge current $j_{\mu} = \frac{\hbar}{2 i m}(\psi^* \partial_{\mu} \psi - \psi \partial_{\mu}\psi^*)$ and take its zeroth component $j_0$ which is conserved separately assuming you are integrating over all of space and your fields are decaying fast enough at infinity since its rate of change is a total divergence. This is proportional to the probability assuming you are interested in equal-time probability (if you are in a boosted frame, you might be interested in a different part of $j_{\mu}$ that is normal to your hyperplane). Probability is not a Lorentz invariant quantity, so you are going to see the energy $E$ without its $\vec p$ counterpart. In fact, the probability in momentum space is related to the propagator as $$P(p) \propto j_0(p) = \frac{E}{m}D(p)$$ Since the only dimensional number that can make this dimensionless is its mass, we have $$P(p) = \frac{E m}{p^2 - m^2 - i\epsilon}$$ up to a constant c-number.

To answer your question about spinor indices, you trace over them, which is why there is no $\not \!p$ and the answer is the same for scalars and fermions.

Aside: For general hypersurfaces $\Sigma$ that have a normal direction at each point $n^{\mu}$ (for example an equal-time surface such that $n^{\mu} = (1,0,0,0)$), $j_{\mu}n^{\mu}$ is your probability (assuming the surface is flat - see this for details). Also, for a fun quantum mechanics-only motivation of this current see here.

Note that this current only works for massive, uncharged fields. For charged fields, see this. For a probability current for the photon, only in the last 4 years has there been one proposed.


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