# Setting c=h=1

1. Dec 26, 2009

### RedX

What are the dimensions of a scalar field $$\phi$$? The Lagrangian density is:

$$\mathcal L= \partial_\mu \phi \partial^\mu \phi - m^2 \phi \phi$$

So in order to make all the terms have the same units, you can try either:

$$\mathcal L=\frac{\hbar^2}{c^2} \partial_\mu \phi \partial^\mu \phi - m^2 \phi \phi$$

or

$$\mathcal L= \partial_\mu \phi \partial^\mu \phi - \frac{c^2}{\hbar^2}m^2 \phi \phi$$

But once both terms are the same units, you can multiply it by any other unit, for example, 1/c:

$$\mathcal L=\frac{1}{c} \partial_\mu \phi \partial^\mu \phi - \frac{c}{\hbar^2}m^2 \phi \phi$$

Once both terms are of the same units, $$\phi$$ takes on whatever units required to make the Lagrangian density have units of Planck's constant (units of the action) divided by the units of a volume of space (i.e., depends on how many dimensions of spacetime you specify).

Once you specify units of the field $$\phi(x)$$, you can find the units of the source current for the field $$J(x)$$.

Also, what are the units of the propagator $$1/(p^2-m^2)$$?

I want to follow the $$\hbar$$'s really closely, because they are small quantities and 1-loop diagrams are smaller than tree diagrams by a factor of $$\hbar$$, but they're not really that much smaller (they're 1/137 smaller for QED, not $$\hbar$$ smaller), so something has to happen.

2. Dec 26, 2009

### Fredrik

Staff Emeritus
Don't forget that a derivative operator counts as a factor with dimensions of momentum.

3. Dec 26, 2009

### RedX

The dimensions of a derivative would be 1/L , where L is length. But it can really be a momentum, or an energy, or a mass...or it can really be just reciprocal length (these things are all related by h's and c's). There are various quantities in field theory, such as the field, the propagator, the green's functions, the source functions, and I don't even know what their units are! If they had kept all the h's and c's in the expressions for the propagator or any such quantity, then I could figure it out. Now it's a mess. I'm even unsure on how to write the Lagrangian in proper units.

4. Dec 26, 2009

### pellman

I think a scalar field should have units of $$\sqrt{\frac{1}{distance^3}}$$. That way position eigenstates

$$|x\rangle = \phi^\dag (x)|0\rangle$$

can be used to give wavefunctions $$\Phi(x)=\langle x|\Phi\rangle$$ like in the non-relativistic Schrodinger case and have the appropriate units. While this doesn't make sense in the relativistic case, it must still approach this in the non-relativistic limit and thus have these units.

Therefore

$$\mathcal L= \frac{\hbar^2}{m}\partial_\mu \phi \partial^\mu \phi - mc^2 \phi \phi$$

would have the correct units of energy-density.

5. Dec 26, 2009

### Fredrik

Staff Emeritus
You're right, but I was right too. I said momentum because I'm used to units in which $\hbar=1$:

$$\partial_\mu e^{ipx}=p_\mu e^{ipx}$$

If you also set c=1, there's only one "dimension" left, the mass dimension. The choice c=1 implies L=T, and then the choice $\hbar=1$ implies M=1/L. Momentum has mass dimension 1. The action is dimensionless (by choice) so the Lagrangian density has dimension 1/L3=M3, i.e. mass dimension 3, and the field has mass dimension 1/2 (because $m^2\phi^2$ has mass dimension 3).

(I didn't see diazona's post below until I was done editing this).

Last edited: Dec 26, 2009
6. Dec 26, 2009

### diazona

Actually it is 1/L, since
$$\partial_\mu = \frac{\partial}{\partial x^{\mu}}$$
That $x^{\mu}$ in the denominator still has to have the dimensions of length, even though it's a differential variable. Plus, when you put in the proper factor of $\hbar$, it works out that way.

7. Dec 27, 2009

### RedX

I agree that the Lagrangian you specified would make the most sense. Unfortunately, setting $$\hbar=c=1$$ in that Lagrangian gives you

$$\mathcal L= \frac{1}{m}\partial_\mu \phi \partial^\mu \phi - m \phi \phi$$

which differs from the Lagrangians given in books by a factor of 1/m. I'm not sure what's going on, as your Lagrangian makes sense, but the books aren't using it.

It's sort of confusing when you say the action is dimensionless. It should have dimensions of Planck's constant (kg*m^2/s), so you'd think that it would have mass dimension 1 since there is a kg there. But Planck's constant is dimensionless because (kg*m^2/s)=(M*L^2/L)=(M*L)=(M*1/M)=1.

Technically speaking, in 4-dimensional spacetime, the Lagrangian density would have dimension 1/L^4, and $$m^2\phi^2*d^4x$$ would require $$\phi$$ to have dimension 1 to make the action have mass dimension 0.

Last edited: Dec 27, 2009
8. Dec 27, 2009

### ansgar

9. Dec 27, 2009

### Fredrik

Staff Emeritus
Yes, this is right. My mistake.

10. Dec 27, 2009

### RedX

$$[\phi(x), \frac{\hbar^2}{c^2m} \dot{\phi}(x')]=i\hbar \delta^3(x-x')$$

which is another way to see that the field would then have dimensions of Sqrt[1/L^3] (the delta function has units of 1/L^3 since it removes three units of length upon integration).

However, most books like the canonical commutation relation to not have a mass of 'm' in it. Therefore the coefficient of the kinetic term in the Lagrangian would not have an '1/m', and therefore the coefficient of the mass term would have a coefficient of 'm^2'. This means that the field would have dimensions with a factor of 1/kg^(1/2), and indeed, if you look at the Fourier expansion of a scalar field, usually there is a funny 1/E^(1/2) term out in front (sometimes you'll see a 1/E term instead, but then the destruction and creation operators have units of E^(1/2)).

Anyways, I got it now.

11. Dec 28, 2009

### pellman

which of course is also

$$[\phi(x), \pi(x')]=i\hbar \delta^3(x-x')$$ where $$\pi$$ is the canonically conjugate momentum density.

Notice that this expression is the continuum analog to the QM relation

$$[x_i, p_j] = i\hbar\delta_{ij}$$

If you don't work in these units, that relation is not so explicit.

12. Dec 28, 2009

### RedX

Indeed, at first glance, the relation:

$$[\phi(x), \frac{\partial \mathcal L}{\partial \dot{\phi}(x')}]=i \hbar \delta^3(x-x')$$

doesn't seem like it can tell you the units of the field $$\phi$$, since there is a $$\phi$$ in the numerator and the denominator. This is not true however, because of the Lagrangian in the numerator (indeed, you don't get a $$1/\phi$$ from differentiating the Lagrangian). Once you choose a Lagrangian, then of course the units of the field are fixed.

I looked at Wikipedia's Lagrangian of the Klein-Gordan field, and it's the same as yours. I think that Lagrangian is used in relativistic QM, but in QFT, the one where the mass is squared is used.