Why in 3+1D QFT a scalar field has (mass) dimension $1$ but in 3+1D QM the wavefunction has dimension $3/2$?

Question

In this question we assume as usual that $\hbar = c =1$, so the word "dimension" means the dimension in mass.

From the fact that an action has the same unit as $\hbar$, or dimensionless under the assumption above, we deduce that the dimension of a scalar field $\phi$ in QFT is $1$. The simplest example of this is the $\phi$ in Klein-Gordon equation. $$ (\square+m^2) \phi=0. $$

On the other hand, we may deduce from the fact that probability is dimensionless that the dimension of a wavefunction is $3/2$. See this question for an explanation.

One thing I do not understand about this is that, according to the usual derivation of Klein-Gordon equation (see for example here), it looks as if the scalar field $\phi$ comes from the wavefunction $\psi$ in Schrodinger equation. So, why do they have different dimension?

Of course, the scalar field does not have a probabilistic interpretation, which is the apparent reason why there is a such a difference. But is it possible that we could understand this a little bit deeper than this? After all, trying to get the dimension of $\phi$ back from the action integral does not seem to be a very satisfying approach to me - how can we be sure that the dimension of action must be the same as in classical physics?

Of course in QFT, $\phi$ is actually an operator, and we are working under Hamiltonians rather than Lagrangians and actions - but this should not affect its dimension.

Answer 1

Let spacetime have $d$ dimensions.

1. In non-relativistic QM the wave function $\psi$ is normalized $$\int \! d^{d-1}\vec{x} ~|\psi|^2~=~1,\tag{1}$$ which means that the wave function $\psi$ has mass-dimension $\frac{d-1}{2}$.

2. In relativistic FT a complex scalar field $\phi$ is usually normalized so that the kinetic term is $$S_{\rm kin}~=~ \int \! dt~ d^{d-1}\vec{x} ~|\partial\phi|^2.\tag{2}$$ This means that $\phi$ has mass-dimension $\frac{d-2}{2}$.

3. We can take the non-relativistic limit by identifying $$ \phi(\vec{x},t)~=~\frac{\hbar}{\sqrt{2m}}\exp\left(-\frac{imc^2t}{\hbar}\right)\psi(\vec{x},t),\tag{3}$$ see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post. Note that eq. (3) is dimensionally consistent with eqs. (1) & (2).

Answer 2

Well that's sort of the point with the KG equation. The Schrödinger equation has a first-order time derivative but a second order spatial derivative, which is inconsistent with relativistic invariance. The Schrödinger equation for a free particle can be derived by quantizing $$ E = \frac{p^2}{2m} $$ Whereas the KG equation for a free particle is found by quantizing $$ E^2 = p^2 c^2 + m^2 c^4 $$ Note that $E$ and $p$ have the same exponent.

To sensibly work in natural units, you have to be able to relate various dimensionful quantities using only fundamental constants. The relation $E = mc^2$ directly relates the dimension of energy and mass because $c$ is a fundamental constant. But $c$ is not a fundamental constant in theories which are not relativistically invariant. Of course no one is stopping you from treating $c$ as a fundamental constant, and using natural units, so you can rewrite the Schrödinger equation in natural units. But you should not expect this to agree with the the KG equation.

EDIT: More concretely, the QFT scaling dimension of $1$ for the scalar field theory takes both space and time into account, whereas the argument you linked only relates to the spatial dimension of the wavefunction. But asking what the QFT scaling dimension for a solution to the Schrödinger equation is, is not well defined as the scaling is not homogeneous in space and time.

Answer 3

In the free scalar QFT the field $\phi$ can have any dimension we want because it is always possible to provide the Lagrange density with a scaling factor any desired dimension which would change the dimension of $\phi$. So I found on the wikipedia page you give as a reference in your post that the Lagrange density looks like this (which still shows similarities with the Schroedinger-theory):

$$\cal{S} = \int \cal{L} d^4 x = \int \left(-\frac{\hbar^2}{m}\partial^\nu \bar{\phi} \partial_\nu \phi -mc^2 \bar{\phi}\phi\right) d^4 x $$

In this case $[\phi]=L^{-3/2}$ as it can be easily checked (assuming$[\cal{S}$]=J$\cdot$s). However, browsing the QFT-literature the following Lagrange density ${\cal{L}}$ is much more common:

$${\cal{L}} = -\frac{1}{2}\partial^\nu\bar{\phi} \partial_\nu \phi - \frac{1}{2}\frac{m^2 c^2}{\hbar^2}\bar{\phi}\phi$$

and here the $[\phi] = L^{-1}$. As already said, as long as the theory is free one could choose one or the other, there is no convincing reason of a specific dimension of $\phi$. The perspective, however, changes if the interacting $\phi^4$-theory is considered (now $c=\hbar=1$ is assumed):

$${\cal{L}} = -\frac{1}{2}\partial^\nu\bar{\phi} \partial_\nu \phi - \frac{1}{2}m^2\bar{\phi}\phi - \frac{\lambda}{4!}(\bar{\phi}\phi)^2$$

In this theory it turns out that with the chosen scaling factors (no change with respect to the commonly used Lagrangian in QFT-literature) the coupling constant $\lambda$ is dimensionless, so it should be scale-independent (QFT actually shows that this only approximately true, but this detail is now out of scope). As Srednicki explains in chapter 12 in his book the dependence of a scattering amplitude on the coupling should be a function of a dimensionless parameter, otherwise, the scattering amplitude would blow up at high energy or would go to zero. These two options are undesired. So by using ${\cal{L}}=-\frac{1}{2}\partial^\nu\bar{\phi} \partial_\nu \phi - m^2\bar{\phi} \phi$ without any rescaling we get to the interacting theory by just adding the $\phi^4$-term. I think, this is a very good reason for no longer sticking to the old dimension of $[\phi]=L^{-3/2}$. The $\phi^4$-term fixes actually the dimension of $\phi$ and in a way we want. Of course, going to other dimensions $d\neq 4$ things get another time a bit more involved.