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First, note that a unitary transformation can not modify the commutation relations.. $$AB-BA=C$$ Use the fact that $U^\dagger U=U U^\dagger=1$ to get, $$AU U^\dagger B-BU U^\dagger A=C$$ and then multiply by the conjugate transpose from the left and $U$ from the right, $$U^\dagger AU U^\dagger B U^\dagger- U^\dagger BU U^\dagger AU^\dagger= U^\dagger C ... 3 If A^\prime and B^\prime commute then there exists a set of mutual eigenvectors of A^\prime and B^\prime. For any eigenbasis of A^\prime there exists a unitary transformation W which takes that basis to the mutual eigenbasis of A^\prime and B^\prime. Consequently if there is a unitary operation such that  |\langle \psi | b \rangle |^2 = ... 0 What you want to ask is whether the position operators of a system with two harmonic oscillators , labeled as A and B, commute. If the system is not entangled, so operator \hat{q_a} act only on A, \hat{q_b} act on B individually. Assume that the state of the system is \psi , \psi=\psi_a\psi_b. So ... 1 Besides the truncated BCH formula (which is also proven in this Phys.SE post and mentioned in Steven Mathey's answer), in practice one often wants to normal order the differential operators, i.e. putting all the x's to the left of all the \partial_x. To this end the formula$$\tag{1} e^{a\partial_x}f(x) ~=~ f(x+a)e^{a\partial_x}. $$is useful. Hence ... 3 Suresh is right, the Baker-Campbell-Hausdorff formula will save you. If \left[A,B\right] commutes with A and B you have,$$ \exp\left(A\right)\exp\left(B\right) = \exp\left(A+B+\frac{1}{2}\left[A,B\right]\right) \, .$$Then the answer to your question is$$ \exp\left(\partial_x\right)\exp\left(x\right) = 2 \exp\left(x+\partial_x\right) \sinh(1/2) \, ...
Here is the formal answer on your question based on particular result of Pauli theorem. Calculations are rather cumbersome, but they are general. Arbitrary fermionic field (with invariance under discrete transformations of the Lorentz group) It can be shown that each Poincare-covariant fermion field with spin $s = n + \frac{1}{2}$ and mass $m$ can be ...