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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:
<br /> [\phi(x,t),\Pi(y,t)]=i\delta^3(x-y)<br />
then unless your canonical momentum \Pi(x,t) is equal to \dot{\phi}(x,t), then the commutation relations of the Fourier components of \phi(x,t) no longer obey equations like:
<br /> [a(k,t),a^\dagger(q,t)]=\delta^3(k-q)<br />
or using a different normalization scheme:
<br /> [a(k,t),a^\dagger(q,t)]=\delta^3(k-q)(2\pi)^32E_k<br />
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:
<br /> [\phi(x,t),\Pi(y,t)]=i\delta^3(x-y)<br />
then unless your canonical momentum \Pi(x,t) is equal to \dot{\phi}(x,t), then the commutation relations of the Fourier components of \phi(x,t) no longer obey equations like:
<br /> [a(k,t),a^\dagger(q,t)]=\delta^3(k-q)<br />
or using a different normalization scheme:
<br /> [a(k,t),a^\dagger(q,t)]=\delta^3(k-q)(2\pi)^32E_k<br />