Quantum field theory is usually expressed in natural units in which $\hbar=c=1$. This simplifies equations and one can always get back to other units by inserting $\hbar$ and $c$ in appropriate places. However, to me this is not always straightforward.

In the second edition of the book of Quantum Field Theory in a Nutshell by Zee we find on page 22 the equation for the Klein-Gordon (KG) equation \begin{equation} -(\partial^2+m^2)D(x-y) = \delta^{(4)}(x-y) \end{equation} One possible way to go to other units is to set \begin{equation} -(\partial^2+\left(\frac{mc}{\hbar}\right)^2)D(x-y) = \delta^{(4)}(x-y) \end{equation} in which case $D$ has dimensions of $L^{-2}$, where $L$ is length, but is this logical?

On page 24 we learn that the propagator describes the amplitude for a disturbance in the field to propagate from $y$ to $x$. With this interpretation in mind, what should be the logical units of the KG or any other propagator?

In quantum mechanics the wave function $\psi(\mathbf{r})$ has dimension $L^{-3/2}$ and is interpreted as a probability amplitude. This interpretation leads us to require that $\int\psi^{\ast}(\mathbf{r})\psi(\mathbf{r}) d^3\mathbf{r}=1$. Are there similar "sum rules" for propagators, reflecting an interpretation in terms of probabilities?


1 Answer 1


Yes, it's logical. With natural units $c=\hbar=1$ we usually work in mass dimension, so lengths and times are $-1$ (in other words, change signs in what follows if you prefer to think in terms of length), $\partial$ is $+1$, $\delta^{(4)}$ is $+4$, and $D$ is $+2$. This is the only way to determine the propagator dimension; in particular, there isn't an alternative rooted in unitarity, because the role of propagators is to invert differential operators.

I appreciate the comparison to e.g. $(i\partial_t+\nabla^2/2m-V-E)\psi=0$, which can't determine $\psi$'s dimension, whereas $\int|\psi|^2d^3\vec{r}=1$ can. But therein lies the difference: $D$ satisfies an inhomogeneous equation with $\delta^{(4)}$ on the RHS, which is what sets its scale (i.e. prevents us from just doubling it or multiplying it by a length or whatever), not an integral equal to $1$. Of course, we can interpret it in integral terms viz.$$-\int(\partial^2+m^2)Df(x)dx=\int\delta^{(4)}(x-y)f(x)dx=f(y).$$

  • $\begingroup$ Just to be sure: In my second equation the operator on the left is the one of the Klein-Gordon equation on standard form. However, suppose I did not know the standard form, when putting back $\hbar$ and $c$ on the left-hand side of the KG propagator equation, how do I know that there should not be for instance an overall factor of $c^2$ on the left-hand side ? This would indeed change the dimension of D. $\endgroup$
    – Trond Saue
    Mar 2 at 16:01
  • 1
    $\begingroup$ @TroudSaue This is why mass dimension is more helpful than length/time dimension, which is agnostic as to $c$ powers. Were it not for $\hbar$ one couldn't reduce to a single dimension type anyway. The rule of thumb is get the $\hbar$ power from the mass dimension, then work out the $c$ power with either the length or time dimension afterwards. $\endgroup$
    – J.G.
    Mar 2 at 16:34

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