I am interested in the relation of the eigenvalues of two Hermitian operators $A$ and $B$ that are related via $$A = \sum_j c_j U_j B U^\dagger_j.$$

Is there anything useful I can say about the spectrum? It looks suspiciously similar to the case of a regular unitary transformation, i.e. where the eigenvalues are preserved. I wonder if something similar can be said here. I don't necessarily need to know the full spectrum, already just knowing that the spectrum is the same but stretched would be helpful.

I was trying the following: I can write $B = \sum_n \lambda_n |n \times n|$ where $\{\lambda_n\}$ is the spectrum of $B$. Plugging that into the definition above yields $$A = \sum_n \lambda_n \sum_j c_j U_j |n \times n| U^\dagger_j =: \sum_n \lambda_n \sum_j c_j |\tilde{n}_j \times \tilde{n}_j|$$ where $\tilde{n}_j = U_j |n\rangle$.

It almost looks like there should be a way to relate to the spectrum of $A$, but here I am stuck. Any ideas how to proceed?

Some specifications that may help reducing the complexity: The $U_j$ have the form $U_j = \exp(-i \phi_j \sum_\ell P_\ell)$ where $P_\ell$ are (non-commuting) Pauli-words and $B$ is one of those Pauli words. So to specify the original relation: $$ A = \sum_j c_j e^{i \phi_j \sum_\ell P_\ell} P_k e^{-i \phi_j \sum_\ell P_\ell} $$ Pauli word here means that it is a tensor product of Pauli-operators, e.g. $P_0 = X_1 \otimes Y_2$. The case that all $c_j\equiv 1$ may also be interesting.

  • $\begingroup$ I suspect there's not muchthat you can say. After all,if I picked the set of unitaries right, this could be twirling, and basically just return an identity matrix. The only thing that would get preserved is the trace (i.e. sum of the eigenvalues). $\endgroup$
    – DaftWullie
    Dec 6, 2022 at 12:14
  • $\begingroup$ Thank you or the quick replies. I specified the conditions a bit further, maybe that can help? $\endgroup$
    – Korbinian
    Dec 6, 2022 at 12:50
  • $\begingroup$ related: quantumcomputing.stackexchange.com/q/13013/55 $\endgroup$
    – glS
    Dec 9, 2022 at 10:17

1 Answer 1


I assume that the $c_j$ form a convex combination such that $X\mapsto \sum_j c_j U_j X U_j^\dagger$ is a unital quantum channel.

I think the only thing that one can say about this problem in its generality is that the $B$ majorizes $A$. This means that if $\lambda_i(A)$ and $\lambda_i(B)$ are the eigenvalues of $A$ and $B$, respectively, and sorted in descending order, then $$ \sum_{i=1}^k \lambda_i(B) \geq \sum_{i=1}^k \lambda_i(A) \,, $$ for all $k=1,\dots,d$, where $d$ is the dimension of underlying Hilbert space.

Note that for certain channels, more can be said. For instance, the completely depolarizing channel is of this form and maps any Hermitian operator to a multiple of the identity (we get a flat spectrum).

If you want to know more about majorization, there's a section in Watrous' book about it.


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