Eigenvectors of operators on a tensor product Hilbert Space
Your conjecture is not true. Take e.g. $A=I$. Then every tensor $\sum_i v_i\otimes w_i\in V\otimes W$ is an eigenvector of $A\otimes I=I\otimes I=I$.
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Lammey
Updated on August 22, 2020Comments

Lammey almost 2 years
Suppose I have finite dimensional Hilbert spaces $V$, $W$, and an operator $A$ acting on vectors in $V$ such that it has eigenvectors/values $Ax_a=\lambda_ax_a$. In the tensor product space I want to find the most general eigenvectors of operator $A\otimes I$. I think that they are all of the form $x_a\otimes w$ where $w\in W$ is any vector. It's easy to show that $x_a\otimes w$ is an eigenvector: $$(A\otimes I)(x_a \otimes w)=(Ax_a)\otimes w=\lambda_a (x_a\otimes w)$$
Moreover, if $v\otimes w$ is a general pure tensor, and we suppose it is an eigenvector, then we have $$(A\otimes I)(v\otimes w)=(Av)\otimes w=\lambda v\otimes w=(\lambda v)\otimes w$$ which is iff $v$ is an eigenvector of $A$. But how can I show that the only eigenvectors of $A\otimes I$ are of the form $x_a\otimes w$?

Qmechanic about 7 yearsCrossposted from physics.stackexchange.com/q/170758/2451


Lammey about 7 yearsThis is worrying for me, because in Quantum Mechanics the an entangled system is represented by the tensor product $V\otimes W$ of its constituents. The operator $A$ represents an observable A and its eigenvectors represent the state of the system after measurement. In the entangled system the operator $A\otimes I$ represents the same observable $A$, and only pure states represent nonentangled states of the joint system. So a measurement should always make the system collapse to a pure tensor, and those pure tensors should be eigenvectors of operators. If the eigenvectors are superpositions..

yurius about 7 yearsI think if $A$ has no multiple eigenvectors then your your conjecture is true.