# Closest quantum state with a fixed marginal: Analytical solution?

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;
cvx_end


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!

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

Compute

\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 $$(1)$$.

• 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. Dec 22, 2021 at 9:45
• 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. Dec 22, 2021 at 14:09
• 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. Dec 22, 2021 at 18:29