- #1
orentago
- 27
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Hi,
I haven't posted this in the homework section, as I don't really see it as homework as such.
I'm trying to derive the Heisenberg equations of motion for the Klein Gordon field (exercise 2.2 of Mandl and Shaw).
I'm trying to derive the commutator of the Hamiltonian and canonical momentum, i.e. [tex][H,\pi(x)][/tex]
For Klein Gordon, we have:[tex]H={1 \over 2}\int\mathrm{d}^3\mathbf{x}\left[c^2\pi^2(x) + (\mathbf{\nabla}\phi)^2+\mu^2\phi^2\right][/tex]
Inserting this into the commutator give several sub-commutators:
[tex][c^2\pi^2(x),\pi(x)][/tex]
[tex][(\mathbf{\nabla}\phi)^2,\pi(x)][/tex]
[tex][\mu^2\phi^2,\pi(x)][/tex]
I have evaluated the first and last of these commutators, but I'm really struggling with the second. The first is of course zero, and the last can be evaluated practically directly using the canonical commutation relations to give:
[tex][\mu^2\phi^2,\pi(x)]=2\mathrm{i}\hbar\mu^2\delta(x-x')\phi(x)[/tex]
My attempt at evaluating the second gives: [tex][(\mathbf{\nabla}\phi)^2,\pi(x)]=2\mathrm{i}\hbar\delta(x-x')\nabla^2\phi[/tex]
However, the result given indicates that [tex][(\mathbf{\nabla}\phi)^2,\pi(x)]=-2\mathrm{i}\hbar\delta(x-x')\nabla^2\phi[/tex]. I'm not sure where I've gone wrong. I've used the commutation relation:
[tex][AB,C]=A[B,C]+[A,C]B[/tex]
to expand the commutator as much as possible, but I'm not sure this is the best method.
Does anyone know how to derive this commutator?
Any help is most appreciated.
EDIT: I'm using the commutation relation: [tex][\phi(x),\pi(x)]=\mathrm{i}\hbar\delta(x-x')[/tex]
I haven't posted this in the homework section, as I don't really see it as homework as such.
I'm trying to derive the Heisenberg equations of motion for the Klein Gordon field (exercise 2.2 of Mandl and Shaw).
I'm trying to derive the commutator of the Hamiltonian and canonical momentum, i.e. [tex][H,\pi(x)][/tex]
For Klein Gordon, we have:[tex]H={1 \over 2}\int\mathrm{d}^3\mathbf{x}\left[c^2\pi^2(x) + (\mathbf{\nabla}\phi)^2+\mu^2\phi^2\right][/tex]
Inserting this into the commutator give several sub-commutators:
[tex][c^2\pi^2(x),\pi(x)][/tex]
[tex][(\mathbf{\nabla}\phi)^2,\pi(x)][/tex]
[tex][\mu^2\phi^2,\pi(x)][/tex]
I have evaluated the first and last of these commutators, but I'm really struggling with the second. The first is of course zero, and the last can be evaluated practically directly using the canonical commutation relations to give:
[tex][\mu^2\phi^2,\pi(x)]=2\mathrm{i}\hbar\mu^2\delta(x-x')\phi(x)[/tex]
My attempt at evaluating the second gives: [tex][(\mathbf{\nabla}\phi)^2,\pi(x)]=2\mathrm{i}\hbar\delta(x-x')\nabla^2\phi[/tex]
However, the result given indicates that [tex][(\mathbf{\nabla}\phi)^2,\pi(x)]=-2\mathrm{i}\hbar\delta(x-x')\nabla^2\phi[/tex]. I'm not sure where I've gone wrong. I've used the commutation relation:
[tex][AB,C]=A[B,C]+[A,C]B[/tex]
to expand the commutator as much as possible, but I'm not sure this is the best method.
Does anyone know how to derive this commutator?
Any help is most appreciated.
EDIT: I'm using the commutation relation: [tex][\phi(x),\pi(x)]=\mathrm{i}\hbar\delta(x-x')[/tex]
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