Units of Scalar Field \phi & Lagrangian Density

In summary, the dimensions of a scalar field \phi should be \sqrt{\frac{1}{L^3}} in order for it to have the appropriate units for position eigenstates and wavefunctions. However, in most books, the coefficient of the mass term in the Lagrangian is chosen such that the field has dimensions of mass^(1/2). This leads to a slight difference in the canonical commutation relation, but ultimately gives the same results.
  • #1
RedX
970
3
What are the dimensions of a scalar field [tex]\phi [/tex]? The Lagrangian density is:

[tex]\mathcal L= \partial_\mu \phi \partial^\mu \phi - m^2 \phi \phi [/tex]

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

[tex]\mathcal L=\frac{\hbar^2}{c^2} \partial_\mu \phi \partial^\mu \phi - m^2 \phi \phi [/tex]

or

[tex]\mathcal L= \partial_\mu \phi \partial^\mu \phi - \frac{c^2}{\hbar^2}m^2 \phi \phi [/tex]

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

[tex]\mathcal L=\frac{1}{c} \partial_\mu \phi \partial^\mu \phi - \frac{c}{\hbar^2}m^2 \phi \phi [/tex]

Once both terms are of the same units, [tex]\phi [/tex] 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 [tex]\phi(x) [/tex], you can find the units of the source current for the field [tex]J(x) [/tex].

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

I want to follow the [tex]\hbar[/tex]'s really closely, because they are small quantities and 1-loop diagrams are smaller than tree diagrams by a factor of [tex]\hbar[/tex], but they're not really that much smaller (they're 1/137 smaller for QED, not [tex]\hbar [/tex] smaller), so something has to happen.
 
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  • #2
Don't forget that a derivative operator counts as a factor with dimensions of momentum.
 
  • #3
Fredrik said:
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.
 
  • #4
I think a scalar field should have units of [tex]\sqrt{\frac{1}{distance^3}}[/tex]. That way position eigenstates

[tex]|x\rangle = \phi^\dag (x)|0\rangle[/tex]

can be used to give wavefunctions [tex]\Phi(x)=\langle x|\Phi\rangle[/tex] 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

[tex]\mathcal L= \frac{\hbar^2}{m}\partial_\mu \phi \partial^\mu \phi - mc^2 \phi \phi [/tex]

would have the correct units of energy-density.
 
  • #5
RedX said:
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 [itex]\hbar=1[/itex]:

[tex]\partial_\mu e^{ipx}=p_\mu e^{ipx}[/tex]

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 [itex]\hbar=1[/itex] 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 [itex]m^2\phi^2[/itex] has mass dimension 3).

(I didn't see diazona's post below until I was done editing this).
 
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  • #6
Actually it is 1/L, since
[tex]\partial_\mu = \frac{\partial}{\partial x^{\mu}}[/tex]
That [itex]x^{\mu}[/itex] 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 [itex]\hbar[/itex], it works out that way.
 
  • #7
pellman said:
Therefore

[tex]\mathcal L= \frac{\hbar^2}{m}\partial_\mu \phi \partial^\mu \phi - mc^2 \phi \phi [/tex]

would have the correct units of energy-density.

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

[tex]\mathcal L= \frac{1}{m}\partial_\mu \phi \partial^\mu \phi - m \phi \phi [/tex]

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.

Fredrik said:
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 [tex] m^2\phi^2 [/tex] 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 [tex] m^2\phi^2*d^4x [/tex] would require [tex]\phi [/tex] to have dimension 1 to make the action have mass dimension 0.
 
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  • #9
RedX said:
Technically speaking, in 4-dimensional spacetime, the Lagrangian density would have dimension 1/L^4, and [tex] m^2\phi^2*d^4x [/tex] would require [tex]\phi [/tex] to have dimension 1 to make the action have mass dimension 0.
Yes, this is right. My mistake.
 
  • #10
pellman said:
I think a scalar field should have units of [tex]\sqrt{\frac{1}{distance^3}}[/tex]. That way position eigenstates

[tex]|x\rangle = \phi^\dag (x)|0\rangle[/tex]

can be used to give wavefunctions [tex]\Phi(x)=\langle x|\Phi\rangle[/tex] 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

[tex]\mathcal L= \frac{\hbar^2}{m}\partial_\mu \phi \partial^\mu \phi - mc^2 \phi \phi [/tex]

would have the correct units of energy-density.

Your Lagrangian leads to the canonical commutation relation:

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

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
RedX said:
Your Lagrangian leads to the canonical commutation relation:

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

which of course is also

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

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

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

If you don't work in these units, that relation is not so explicit.
 
  • #12
pellman said:
which of course is also

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

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

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

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

Indeed, at first glance, the relation:

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

doesn't seem like it can tell you the units of the field [tex]\phi[/tex], since there is a [tex]\phi [/tex] in the numerator and the denominator. This is not true however, because of the Lagrangian in the numerator (indeed, you don't get a [tex]1/\phi[/tex] 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.
 

What are the units of scalar field φ?

The units of scalar field φ depend on the specific physical quantity it represents. For example, if φ represents electric potential, its units would be volts. If φ represents temperature, its units would be degrees Celsius or Kelvin. In general, the units of φ are determined by the equation in which it appears.

What is a Lagrangian density?

A Lagrangian density is a mathematical expression that describes the dynamics of a physical system in terms of its constituent parts. It is defined as the Lagrangian function per unit volume and is used in the field of theoretical physics to describe the time evolution of a system.

How is the Lagrangian density related to the Lagrangian?

The Lagrangian density is the integral of the Lagrangian function over the entire volume of the system. In other words, it is the total energy of the system per unit volume. The Lagrangian density is used to derive the equations of motion for the system, which ultimately determine its behavior over time.

What are the advantages of using Lagrangian density over other methods?

One of the main advantages of using Lagrangian density is that it allows for a more elegant and intuitive formulation of the equations of motion for a system. It also allows for the use of symmetries and conservation laws, making it a powerful tool for analyzing physical systems. Additionally, the Lagrangian density approach is often more convenient and efficient for certain types of calculations compared to other methods.

Can Lagrangian density be used for both classical and quantum systems?

Yes, Lagrangian density can be used for both classical and quantum systems. In classical mechanics, it is used to describe the motion of particles and objects, while in quantum mechanics, it is used to describe the behavior of subatomic particles. However, the specific form and application of the Lagrangian density may differ between classical and quantum systems.

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