Let $\rho_{AB}$ be a bipartite state and let $\sigma_{B}$ be another state. What state $\tilde{\rho}_{AB}$ is closest to $\rho_{AB}$ and satisfies $\tilde{\rho}_B = \sigma_B$? We can define closeness in many ways but I pick fidelity here arbitrarily. Hence, the optimization problem is

\begin{align} &\max_{\tilde{\rho}_{AB}} F(\tilde{\rho}_{AB}, \rho_{AB})\\ &s.t. \tilde{\rho}_B = \sigma_B \end{align}

This can be written as a semidefinite program and solved numerically. MATLAB code below

rho = RandomDensityMatrix(4,4);
sigma = RandomDensityMatrix(2,2);

cvx_begin sdp
    variable rho_tilde(4, 4) hermitian;
    maximize Fidelity(rho_tilde, rho)
    rho_tilde >= 0;
    trace(rho_tilde) == 1;
    PartialTrace(rho_tilde, 1, [2,2]) == sigma;

The code requires both cvxquad and QETLAB to be installed if you wish to run it. My question is if there is some analytical form for $\tilde{\rho}$? I tried the following attempt

$$\tilde{\rho} = (I_A\otimes \sigma_B^{1/2})(I_A\otimes\rho_B^{-1/2})\rho_{AB}(I_A\otimes\rho_B^{-1/2})(I_A\otimes \sigma_B^{1/2})$$

followed by a normalization of $\tilde{\rho}$ to get unit trace but my numerics showed that this is not the correct solution!


1 Answer 1


Below is a partial analytical solution to a variant of the problem using the trace distance.

Solution approach

We find an analytical expression for a lower bound on the trace distance $D(\rho_{AB}, \tilde{\rho}_{AB})$ and - under certain additional conditions described below - quantum states that achieve the bound. This gives us a partial solution of the variant

$$ \min_{\tilde{\rho}_{AB}} D(\tilde{\rho}_{AB},\rho_{AB})\\ s.t. \tilde{\rho}_B = \sigma_B\tag1 $$

of the problem where we use trace distance $D(\rho,\sigma)=\frac12\mathrm{tr}|\rho-\sigma|$, rather than the fidelity, to measure closeness of quantum states.

Lower bound on trace distance


$$ \begin{align} D(\tilde{\rho}_{AB},\rho_{AB}) &\ge D(\mathrm{tr}_A\tilde{\rho}_{AB},\mathrm{tr}_A\rho_{AB})\\ &=D(\sigma_B,\mathrm{tr}_A\rho_{AB})\\ &=\frac12\mathrm{tr}|\sigma_B-\mathrm{tr}_A\rho_{AB}| \end{align}\tag2 $$

where the inequality follows from the fact that partial trace is a completely positive trace-preserving (CPTP) map and CPTP maps are contractive, see e.g. theorem $9.2$ on page $406$ and equation $(9.45)$ on page $407$ in Nielsen & Chuang.

States achieving the lower bound

Let $\tau_A$ be any quantum state, so that $\mathrm{tr}\,\tau_A=1$. Define

$$ X:=\rho_{AB}+\tau_A\otimes(\sigma_B-\mathrm{tr}_A\rho_{AB}).\tag3 $$

Note that $X$ is unit trace, but not necessarily positive semidefinite (thanks for spotting this, @DaftWullie!) and thus not necessarily a quantum state. However,

$$ \begin{align} D(X,\rho_{AB}) &= \frac12\mathrm{tr}|\rho_{AB}+\tau_A\otimes(\sigma_B-\mathrm{tr}_A\rho_{AB})-\rho_{AB}|\\ &=\frac12\mathrm{tr}|\tau_A\otimes(\sigma_B-\mathrm{tr}_A\rho_{AB})|\\ &=\frac12\mathrm{tr}|\sigma_B-\mathrm{tr}_A\rho_{AB}| \end{align}\tag4 $$

so every $X$ defined in $(3)$ achieves the lower bound in $(2)$. Thus, if $X$ is positive semidefinite then the state $\tilde\rho_{AB}:=X$ is a solution to $(1)$.

Positive semidefiniteness

A candidate solution $X$ is positive semidefinite, and thus a solution to $(1)$, for example if

$$ \frac{1}{d_A}\|\sigma_B-\mathrm{tr}_A\rho_{AB}\|_2\le\lambda_{min}\tag5 $$

where $\|.\|_2$ is the spectral norm, $\lambda_{min}$ is the least eigenvalue of $\rho_{AB}$ and $d_A$ is the dimension of the Hilbert space of subsystem $A$.

Another case occurs when $\rho_{AB}=\rho_A\otimes\rho_B$ is a product state. Then by setting $\tau_A:=\rho_A$ we find that $\tilde\rho_{AB}=\rho_A\otimes\sigma_B$ which is positive semidefinite and thus a solution to $(1)$.

  • 2
    $\begingroup$ Are you certain that (3) is always a valid state (i.e. positive semi-definite)? Consider $\rho_{AB}=|00\rangle\langle 00|$, $\sigma_B=|1\rangle\langle 1|$ and $\tau_A=|1\rangle\langle 1|$. Perhaps there always exists a valid $\tau_A$, but I guess this requires a bit more work. $\endgroup$
    – DaftWullie
    Commented Dec 22, 2021 at 9:45
  • 3
    $\begingroup$ Actually, I don't think there's even always a valid $\tau$. Consider $\rho_{AB}$ to be the pure state $|00\rangle+|11\rangle$ and $\sigma_B=|0\rangle\langle 0|$. Then $\langle 01|\tilde\rho_{AB}|01\rangle=-\langle 0|\tau|0\rangle/2$. So the only possible solution is $\tau=|1\rangle\langle 1|$. But try that specific case, and you also have negative eigenvalues. $\endgroup$
    – DaftWullie
    Commented Dec 22, 2021 at 14:09
  • 1
    $\begingroup$ You're right. Thank you, @DaftWullie! For now, I reworded the answer to indicate that this is a partial solution. It's not clear to me that a general analytical solution exists. $\endgroup$ Commented Dec 22, 2021 at 18:29

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