2
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

0
$\begingroup$

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.$$-\int(\partial^2+m^2)Df(x)dx=\int\delta^{(4)}(x-y)f(x)dx=f(y).$$

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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