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I'm trying to derive the commutation relations of the raising and lowering operators for a complex scalar field and I had a question. Let's start with the commutation relations:
$$[\varphi(\mathbf{x},t),\varphi(\mathbf{x}',t)]=0$$
$$[\Pi(\mathbf{x},t),\Pi(\mathbf{x}',t)]=0$$
$$[\varphi(\mathbf{x},t),\Pi(\mathbf{x}',t)]=i\delta(\mathbf{x}-\mathbf{x}')$$
as well as the expression for the lowering operator:
$$a(\mathbf{k})=\int {d^3 x e^{-ikx}(i\Pi^{\dagger}(\mathbf{x},t)+\omega\varphi(\mathbf{x}',t))}$$
Now we calculate the commutator:
$$[a(\mathbf{k}),a(\mathbf{k}')]=
\int{ d^3 xd^3x' e^{-i(kx+k'x')}\left(-[\Pi^{\dagger}(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)]+\omega^2[\varphi(\mathbf{x},t),\varphi(\mathbf{x}',t)]+i\omega\left([\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)]
-[\varphi(\mathbf{x}',t),\Pi^{\dagger}(\mathbf{x},t)] \right) \right) }$$
The first two commutators clearly disappear. My question is about the other portion:
$$[\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)]
-[\varphi(\mathbf{x}',t),\Pi^{\dagger}(\mathbf{x},t)]$$
So I'm aware that this term also disappears, but I'm a little confused as to how. I understand that ##\varphi## and ##\varphi^{\dagger}## are supposed to be considered "independent" fields, but does that mean that ##[\varphi,\Pi^{\dagger}]=0##? Is this derivable? What makes fields independent of one another?
The other option is that ##[\varphi,\Pi^{\dagger}]\propto\delta(\mathbf{x}-\mathbf{x}')## and the offending term disappears because the delta functions cancel.
$$[\varphi(\mathbf{x},t),\varphi(\mathbf{x}',t)]=0$$
$$[\Pi(\mathbf{x},t),\Pi(\mathbf{x}',t)]=0$$
$$[\varphi(\mathbf{x},t),\Pi(\mathbf{x}',t)]=i\delta(\mathbf{x}-\mathbf{x}')$$
as well as the expression for the lowering operator:
$$a(\mathbf{k})=\int {d^3 x e^{-ikx}(i\Pi^{\dagger}(\mathbf{x},t)+\omega\varphi(\mathbf{x}',t))}$$
Now we calculate the commutator:
$$[a(\mathbf{k}),a(\mathbf{k}')]=
\int{ d^3 xd^3x' e^{-i(kx+k'x')}\left(-[\Pi^{\dagger}(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)]+\omega^2[\varphi(\mathbf{x},t),\varphi(\mathbf{x}',t)]+i\omega\left([\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)]
-[\varphi(\mathbf{x}',t),\Pi^{\dagger}(\mathbf{x},t)] \right) \right) }$$
The first two commutators clearly disappear. My question is about the other portion:
$$[\varphi(\mathbf{x},t),\Pi^{\dagger}(\mathbf{x}',t)]
-[\varphi(\mathbf{x}',t),\Pi^{\dagger}(\mathbf{x},t)]$$
So I'm aware that this term also disappears, but I'm a little confused as to how. I understand that ##\varphi## and ##\varphi^{\dagger}## are supposed to be considered "independent" fields, but does that mean that ##[\varphi,\Pi^{\dagger}]=0##? Is this derivable? What makes fields independent of one another?
The other option is that ##[\varphi,\Pi^{\dagger}]\propto\delta(\mathbf{x}-\mathbf{x}')## and the offending term disappears because the delta functions cancel.