What does equality of partial traces, ${\rm Tr}_1\rho={\rm Tr}_1\sigma$, say about a pair of states $\rho,\sigma$?

Question

Let $\rho,\sigma$ be a pair of bipartite quantum states such that ${\rm Tr}_1\rho={\rm Tr}_1\sigma$. What does this tell us about $\rho,\sigma$? More precisely, is there a way to write more explicitly the relation between these two states?

In the special case in which they are pure, this is doable in a relatively straightforward manner via Schmidt decomposition: write $$|\psi\rangle = \sum_k \sqrt{p_k}(|u_k\rangle\otimes |v_k\rangle), \qquad |\psi'\rangle = \sum_k \sqrt{p_k'}(|u_k'\rangle\otimes |v_k'\rangle),$$ and then equality of partial traces implies that (up to some reordering of the elements in the sums) $p_k=p_k'$ and $|v_k\rangle=|v_k'\rangle$, while there is no additional constraint on $|u_k'\rangle$. So this provides a relatively nice way to see how the equality of partial traces condition reflects on the structure of the states, and we can write that in general two such states are related by $|\psi\rangle=(U\otimes I)|\psi'\rangle$ for some unitary $U$.

Is there any similar type of "nice" characterisation doable for more general mixed states?

Answer 1

If $\rho = (U \otimes I)\sigma(U^\dagger \otimes I)$ then ${\rm Tr}_1\rho={\rm Tr}_1\sigma$. But the reverse implication isn't true.

For example, for a maximally entangled pure state $\rho=|\psi\rangle\langle\psi|$ we have ${\rm Tr}_1\rho = I/d_2$, where $d_2$ is dimension of the second subsystem (assuming $d_2\le d_1$). But for any convex combination $\sigma$ of maximally entangled states we also have ${\rm Tr}_1\sigma = I/d_2$. So that $\rho$ and $\sigma$ can have arbitrary sets of eigenvalues, while ${\rm Tr}_1\rho={\rm Tr}_1\sigma$ is true.

I don't think there is some simple general characterization. The best I can imagine is the following.

Let $\{U_i\}_{i=1}^{d_1^2}$ be a unitary basis on the first subsystem. Then $\rho$ can be reconstructed from the matrices $A_i = {\rm Tr}_1 (U_i \otimes I \cdot \rho)$ by the formula $$ \rho = \frac{1}{d_1}\sum_{i=1}^{d_1^2} U_i^\dagger\otimes A_i. $$

If we drop the condition that $\rho$ is a state then $A_i$ can be any complex matrices. In other words, we have a bijection between general complex matrices $\rho$ and tuples $\{A_i\}_{i=1}^{d_1^2}$. And If $U_1 = I$ then ${\rm Tr}_1\rho = A_1$, i.e. it's just the first "coefficient". So that, for $\sigma = \frac{1}{d_1}(I\otimes A_1 + \sum_{i=2}^{d_1^2} U_i^\dagger\otimes A_i')$ we have ${\rm Tr}_1\sigma= {\rm Tr}_1\rho$ for any matrices $A_i', i>1$. But the condition that $\sigma$ is a state imposes restrictions on $A_i'$, which is hard to express easily.