What is the interpretation of a mass term in field theory?

[parton]

Hello!

I've some problems understanding why in field theory a term (appearing in a Lagrangian) like \propto \phi^{2} is called "mass"-term (whereas \phi denotes a real field). Is there any interpretation? And is this a general expression for a mass term or could it be of any other shape?

Thanks

 

[Avodyne]

It's called the mass term because, in free field theory, its coefficient turns out to be equal to the mass-squared of the particle. This is a general expression for the mass term for scalar fields. For fermions, the mass term is \overline{\psi}\psi, and its coefficient is the mass (not the mass-squared) of the particle.

 

[parton]

Thanks for your answer.

But could you tell me how exactly it turns out that the coefficient can really be identified with the mass-squared of the particle?

Thanks again

 

[samalkhaiat]

But could you tell me how exactly it turns out that the coefficient can really be identified with the mass-squared of the particle?

write

L = \frac{1}{2}\left( \partial_{\mu}\phi \partial^{\mu}\phi - a \phi^{2} \right)

for some constant a. Clearly it has a dimension of mass squared (the action is dimensionless)

Now put

\phi = \exp (ipx)

in the equation derived from L, i.e.,

\left( \partial_{\mu}\partial^{\mu} + a \right) \phi = 0

you get

p^{2} (= m^{2}) = a

regards

sam

 

[Avodyne]

Well, there's a little more to it than that. You've assumed that the wavevector and frequency in a plane-wave solution to the equation of motion for the field are equal to the momentum and energy of a particle (in units with \hbar=c=1). This is of course correct, but there are a few steps missing.

This topic is covered in detail in every QFT book, though.

 

[samalkhaiat]

Given the equation

\left( \partial^{2} + b \right) \phi(x) = 0

it is easy to show that

\phi^{-}(x) = e^{ip.x} a^{\dagger}

is a solution, if and only if

b = m^{2}

The mass of any representation of the Piocare' group is given by the eigenvalue of its Casimir operator

P^{2}|p \rangle = m^{2}|p \rangle

<br /> P^{2}|p \rangle = P^{2} \phi^{-}(x)|0 \rangle = - \partial^{2} \phi^{-}(x) |0 \rangle = b \phi^{-}(x) |0 \rangle = b |p \rangle<br />

 

[ismaili]

write

L = \frac{1}{2}\left( \partial_{\mu}\phi \partial^{\mu}\phi - a \phi^{2} \right)

for some constant a. Clearly it has a dimension of mass squared (the action is dimensionless)

Now put

\phi = \exp (ipx)

in the equation derived from L, i.e.,

\left( \partial_{\mu}\partial^{\mu} + a \right) \phi = 0

you get

p^{2} (= m^{2}) = a

regards

sam

Another point of view, in the language of path integral formalism. The mass is defined as the pole of the propagator, where the propagator is the inverse of the quadratic operator in the Lagrangian. Hence, the quadratic term like m^2\phi^2 is a mass term.