Units of Scalar Field \phi & Lagrangian Density

[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.

 

[Fredrik]

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

 

[RedX]

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

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.

 

[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.

 

[Fredrik]

The dimensions of a derivative would be 1/L

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).

 

[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.

 

[RedX]

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.

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.

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).

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.

 

[ansgar]

mark Srednicki textbook

http://www.physics.ucsb.edu/~mark/qft.html

chapter 12

 

[Fredrik]

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.

Yes, this is right. My mistake.

 

[RedX]

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.

Your Lagrangian leads to the canonical commutation relation:

[\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.

 

[pellman]

Your Lagrangian leads to the canonical commutation relation:

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

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.

 

[RedX]

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.

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.