If two operators commute (eigenvector question)

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The discussion centers on the properties of commuting operators, specifically \Omega_1 and \Omega_2, where [\Omega_1,\Omega_2]=0. It is established that if \Psi_1 and \Psi_2 are eigenvectors of \Omega_1 and \Omega_2, respectively, then their tensor product \Psi = \Psi_1 \otimes \Psi_2 is indeed an eigenvector of \Omega = \Omega_1 + \Omega_2. Furthermore, if the eigenvectors \Psi_i are normalized, the resulting tensor product \Psi is also normalized. The conclusion drawn is that \Omega \Psi = (a_1 + a_2) \Psi holds true, regardless of the commutation relation, assuming the operators are Hermitian.

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Suppose [itex]\Omega_1[/itex] and [itex]\Omega_2[/itex] satisfy [itex][\Omega_1,\Omega_2]=0[/itex] and [itex]\Omega = \Omega_1 + \Omega_2[/itex]. If [itex]\Psi_1[/itex] and [itex]\Psi_2[/itex] are eigenvectors of [itex]\Omega_1[/itex] and [itex]\Omega_2[/itex], respectively, don't we know that the (tensor?) product [itex]\Psi = \Psi_1 \Psi_2[/itex] is an eigenvector of [itex]\Omega[/itex]? Also, if the [itex]\Psi_i[/itex] are normalized, isn't [itex]\Psi[/itex] automatically normalized?
 
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Well, I've just worked through this, and I think I've determined the following: If [itex]\Omega_i \Psi_i = a_i \Psi_i[/itex], we have [itex]\Omega \Psi = (a_1 + a_2) \Psi[/itex], regardless of whether [itex][\Omega_1,\Omega_2] = 0[/itex]. Is this true? (I'm assuming the [itex]\Omega_i[/itex] are Hermitean, but even that might not make any difference.)
 
It seems you are mixing things. Either you are discussing tensor product or not. If you are discussing tensor product and if [itex] \Omega_1[/itex] and [itex] \Omega_2[/itex] refer to two different Hilbert spaces, then they automatically commute.
 

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