# Commutation Relations for Creation & Annihilation Opertors of Two Different Scalar Fields

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?

• Try writing ladder operators in terms of the field operator and its conjugate momentum! – zzz May 16 '14 at 20:15
• 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) – zzz May 16 '14 at 20:17

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.