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Let us consider two different scalar fields $\phi$ and $\chi$. The commutation relations for the creation and annihilation operators of the scalar field $\phi$ are given by $$ [a(\textbf{k}), a(\textbf{k}') ]= 0 $$ $$ [a^\dagger(\textbf{k}), a^\dagger(\textbf{k}') ]= 0 $$ $$ [a(\textbf{k}), a^\dagger(\textbf{k}') ]= (2\pi)^3 2\omega \, \delta^3(\textbf{k} - \textbf{k}'). $$

For $\chi$ similarly we have $$ [b(\textbf{k}), b(\textbf{k}') ]= 0 $$ $$ [b^\dagger(\textbf{k}), b^\dagger(\textbf{k}') ]= 0 $$ $$ [b(\textbf{k}), b^\dagger(\textbf{k}') ]= (2\pi)^3 2\omega \, \delta^3(\textbf{k} - \textbf{k}'). $$

Are there any commutation relations among the operators of the two different fields upon any condition?

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  • $\begingroup$ Try writing ladder operators in terms of the field operator and its conjugate momentum! $\endgroup$ – zzz May 16 '14 at 20:15
  • $\begingroup$ Of course there are other very intuitive physical answers to this, such as: if you first create a particle in the first field, and then annihilate a particle in the second field, does it matter what order you should do this in? (once again, non-rigorous, handwaving answer purely for physical intuition) $\endgroup$ – zzz May 16 '14 at 20:17
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The commutator measures the degree to which states can't have definite values of two observables. (Creation operators are not observables but their commutation relations follow from the commutators for the field and fields are observables.) Since the scalar fields can presumably both have definite values, they should commute. From this it follows that their creation operators do, too.

Another way to think about it is the analogy with classical Hamiltonian mechanics. Here the commutator is the Poisson bracket and $$\{ q_i, p_j\} =\delta_{ij} $$ where $p_j$ is the momentum conjugate to $q_j$ and $\delta_{ij}$ is the Kronecker delta. If you take this prescription you see that different fields commute.

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