- #1
RedX
- 970
- 3
I heard that in classical field theory, terms in the Lagrangian cannot have more than two derivatives acting on them. Why is this?
In quantum field theory, I read somewhere that having more than two derivatives on a term in the Lagrangian leads to a violation of Poincare invariance. Is this true?
One thing I derived is that, for a scalar field, if you accept the canonical commutation relations as true:
[tex]
[\phi(x,t),\Pi(y,t)]=i\delta^3(x-y)
[/tex]
then unless your canonical momentum [tex]\Pi(x,t) [/tex] is equal to [tex]\dot{\phi}(x,t) [/tex], then the commutation relations of the Fourier components of [tex]\phi(x,t) [/tex] no longer obey equations like:
[tex]
[a(k,t),a^\dagger(q,t)]=\delta^3(k-q)
[/tex]
or using a different normalization scheme:
[tex]
[a(k,t),a^\dagger(q,t)]=\delta^3(k-q)(2\pi)^32E_k
[/tex]
In quantum field theory, I read somewhere that having more than two derivatives on a term in the Lagrangian leads to a violation of Poincare invariance. Is this true?
One thing I derived is that, for a scalar field, if you accept the canonical commutation relations as true:
[tex]
[\phi(x,t),\Pi(y,t)]=i\delta^3(x-y)
[/tex]
then unless your canonical momentum [tex]\Pi(x,t) [/tex] is equal to [tex]\dot{\phi}(x,t) [/tex], then the commutation relations of the Fourier components of [tex]\phi(x,t) [/tex] no longer obey equations like:
[tex]
[a(k,t),a^\dagger(q,t)]=\delta^3(k-q)
[/tex]
or using a different normalization scheme:
[tex]
[a(k,t),a^\dagger(q,t)]=\delta^3(k-q)(2\pi)^32E_k
[/tex]