How to write a two qubit state as "diagonal" in the basis of Pauli matrices?

Question

Any two qubit density matrix can be written as

$$ \rho = \frac{1}{4} \sum_{n,m = 0}^{3} M_{nm} (\sigma_n \otimes \sigma_m), $$

where $\sigma_\mu$'s are the identity and Pauli matrices.

Is it possible to do a local unitary transformation on the system, like $U_1 \otimes U_2$, which will transform $\rho$ into $\rho'$ which looks like this:

$$\rho' = \frac{1}{4} \left( \sigma_0 \otimes \sigma_0 + M^{'}_{11} \sigma_1 \otimes \sigma_1 + M^{'}_{22} \sigma_2 \otimes \sigma_2 + M^{'}_{33} \sigma_3 \otimes \sigma_3 \right)$$

i.e. the matrix whose elements are $\{M_{nm}\}$ is diagonal?

Answer 1

This is indeed possible. Consider the SVD of $M$, written as $$M_{nm}=\sum_k s_k \langle n |u_k\rangle \overline{\langle m| v_k\rangle} \equiv \sum_k s_k u_{nk} \bar v_{mk}$$ for some pair of orthonormal bases $\{|u_k\rangle\}$ and $\{|v_k\rangle\}$ and positive reals $s_k>0$. More concisely, this amounts to $M=\sum_k s_k |u_k\rangle\!\langle v_k|$. Then $$ 4\rho = \sum_{knm} s_k u_{nk} \bar v_{mk}(\sigma_n\otimes \sigma_m) = \sum_k s_k \left( \sum_n u_{nk} \sigma_n \right) \otimes \left( \sum_m \bar v_{mk} \sigma_m \right). $$ Now observe that any unitary $U$ acts via the adjoint on Pauli matrices (or any other orthonormal basis of operators) as a unitary operator, meaning $$U\sigma_i U^\dagger = \sum_j v_{ji} \sigma_j$$ for some unitary $\tilde U$ with components $v_{ji}$. With this, we see immediately that we can write $$4\rho = \sum_k s_k (U\sigma_k U^\dagger)\otimes(V\sigma_k V^\dagger)$$ for some pair of unitaries $U,V$, and thus we conclude $$4(U\otimes V)^\dagger\rho (U\otimes V)=\sum_k s_k (\sigma_k\otimes\sigma_k).$$

Some (maybe too many?) details about the above relation for $U\sigma_i U^\dagger$ can be found here and here. I'm sure I've seen more direct treatments of it on some SE site, but I can't find the post right now.

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Some more details:

1. After writing the SVD of $M$, I implicitly assumed that the numbers $u_{nk}$ are the elements of some unitary matrix $U\equiv (u_{nk})_{nk}$. This is not always the case. We can however assume that is without loss of generality, by simply allowing $s_k=0$ for some $k$. This way we always have $4$ elements in the sum, and $U$ is always unitary.

2. Any special unitary $U\in\mathbf{SU}(2)$ can be written as a sum of Pauli operators as $$U = \alpha_0 I + i \sum_{j=1}^3 \alpha_j \sigma_j,$$ with $\alpha_i\in\mathbb{R}$ and $\sum_{i=0}^4\alpha_i^2=1$. From this decomposition, you can compute explicitly $$U \sigma_i U^\dagger = \left(\alpha_0^2-\sum_{j=1}^3\alpha_j^2\right)\sigma_i + 2 \alpha_i (\boldsymbol\alpha\cdot\boldsymbol\sigma) +2i\alpha_0 (\boldsymbol\sigma\times\boldsymbol\alpha)_i. $$