If two operators are related to each other by a unitary operator, are their eigenvalues the same? If we consider 2 linearly independent basis as follow:
$$\{ |\psi_1 \rangle  , |\psi_2 \rangle ... |\psi_n \rangle\}$$
$$\{ |\phi_1 \rangle  , |\phi_2 \rangle ... |\phi_n \rangle\}$$
And they are related by a unitary tranform such that:
$$U|\psi_i\rangle = |\phi_n\rangle$$
If O is an oprator in basis $\{ |\psi_1 \rangle  , |\psi_2 \rangle ... |\psi_n \rangle\}$ and O' is operator in basis $\{ |\phi_1 \rangle  , |\phi_2 \rangle ... |\phi_n \rangle\}$, and we could write the spectral decomposition as follows:
$$O = \sum_n \lambda_n |\psi_n \rangle \langle\psi_n|$$
$$O' = \sum_n \mu_n |\phi_n \rangle \langle\phi_n|$$
Would it be true to say that the eigenvalues $\mu_n$ and $\lambda_n$ are equal to one another?
 A: You probably mean that, since $\langle \psi_n|U^\dagger =\langle \phi_n|$,
$$ O'=UOU^\dagger = \sum_n \lambda_n U|\psi_n \rangle \langle\psi_n| U^\dagger=  \sum_n \lambda_n |\phi_n \rangle \langle\phi_n|=   
 O'  ,$$   unitarily equivalent.
So, if $O$ and $O'$ are unitarily equivalent, the corresponding transformation of their eigenstate decompositions forces $\lambda_n=\mu_n$.
If they are not unitarily equivalent, the eigenvalues won't coincide: any operator can be written in any unitarily equivalent basis, and the basis by itself cannot tell it what it is or isn't!
Try a 2x2 matrix example. Note the trace and determinant of two unitarily equivalent matrices coincide.
A: It is given that
$$U|\psi_n\rangle = |\phi_n\rangle$$
where $U$ is a unitary matrix and $|\psi_n\rangle$ are eigenkets of $\hat{O}$ and $|\phi_n\rangle$ are eigenkets of $\hat{O'}$ with eigenvalues $\lambda_n$ and $\mu_n$, respectively.
Equivalently,
$$|\psi_n\rangle = U^\dagger|\phi_n\rangle$$
Since $\hat{O'}|\phi_n\rangle = \mu_n|\phi_n\rangle$ and $\hat{O}|\psi_n\rangle = \lambda_n |\psi_n\rangle$, and since $\hat{O'} = (U\hat{O}U^\dagger)$, then
$$ \mu_n|\phi_n\rangle = \hat{O'}|\phi_n\rangle = \hat{O'} U|\psi_n\rangle = (U\hat{O}U^\dagger)U|\psi_n\rangle = U\hat{O}|\psi_n\rangle = U\lambda_n|\psi_n\rangle = \lambda_n U|\psi_n\rangle = \lambda_n |\phi_n\rangle$$
Thus, $\lambda_n = \mu_n$ and they share the same eigenvalues.
