Is the closest diagonal state to a given state always the dephased original state?

Question

This question is about the following optimization problem:

Given some density matrix $\rho\in\mathbb C^{n\times n}$ find the diagonal state which is closest to it in trace norm. More precisely, find $$ \underset{s\in\Delta^{n-1}}{\rm argmin}\|\rho-{\rm diag}(s)\|_1 $$ with $\Delta^{n-1}\subset\mathbb C^n$ the standard simplex of probability vectors.

A naive first guess is that this minimum is attained on $s=d(\rho)$ where $d(A):=(A_{ii})_i$ just collects the diagonal entries, i.e., $d$ is (equivalent to) the dephasing channel. This is motivated by the following two facts:

1. If the trace norm is replaced by the Frobenius norm, then $s=d(\rho)$ because minimizing $\|\rho-{\rm diag}(s)\|_2$ amounts to minimizing the individual entries. Therefore, having the diagonal vanish (which is the only thing one can "control" here) is optimal.
2. For $n=2$ our guess is provably true:

Result. Given any density matrix $\rho\in\mathbb C^{2\times 2}$ one has $$ \underset{s\in\Delta^{n-1}}{\rm argmin}\|\rho-{\rm diag}(s)\|_1=\begin{pmatrix} \rho_{11}\\ \rho_{22} \end{pmatrix}. $$

Proof. The key is the qubit formula $ \|A\|_1^2={\rm tr}(A^\dagger A)+2|\det(A)| $, i.e., this is valid for all $A\in\mathbb C^{2\times 2}$. Because $\rho-{\rm diag}(s)$ is a traceless Hermitian matrix—the off-diagonals of which are the same as for $\rho$—our minimization problem simplifies as follows: \begin{align*} \min_{s\in\Delta^{n-1}}\|\rho-{\rm diag}(s)\|_1&=\min_{b}\Big\| \begin{pmatrix} b&\rho_{12}\\\rho_{12}^*&-b \end{pmatrix} \Big\|_1\\ &=\min_{b}\sqrt{ \Big\| \begin{pmatrix} b&\rho_{12}\\\rho_{12}^*&-b \end{pmatrix} \Big\|_2^2+2\Big|\det \begin{pmatrix} b&\rho_{12}\\\rho_{12}^*&-b \end{pmatrix} \Big| }\\ &=\min_{b}\sqrt{ 2|\rho_{12}|^2+2b^2+2\big|-b^2-|\rho_{12}|^2\big| }\\ &=\min_{b}\sqrt{4|\rho_{12}|^2+4b^2}\\ &=\sqrt{4|\rho_{12}|^2}=2|\rho_{12}| \end{align*} and the minimum is attained for $b=0$, i.e., $s=d(\rho)$. $\square$

This begs the question: does this result hold up for higher dimensions, as well?

---

(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

Answer 1

It turns out that this is not true, and I want to thank Rubén Ibarrondo (UPV/EHU) for suggesting this problem & pointing out Section II.B of this paper (arXiv) to me; therein, the authors consider the pure qutrit state $\psi=(\frac23,\frac23,\frac13)^T$ which is a counterexample because

• the trace-norm distance to dephased state $D(|\psi\rangle\langle \psi|)={\rm diag}(\frac49,\frac49,\frac19)$ equals $\frac{4}{9} (\sqrt{3}+1)\simeq 1.214$
• on the other hand, the argmin is $s=(\frac12,\frac12,0)$, and the trace distance in question is \begin{align*} \big\|\,|\psi\rangle\langle\psi|-{\rm diag}(s)\big\|_1&= \Big\| \begin{pmatrix} -\frac1{18} &\frac49 &\frac29 \\ \frac49 &-\frac1{18} &\frac29 \\ \frac29 &\frac29 &\frac19 \end{pmatrix} \Big\|_1 \\ &=\frac{1}{6} (\sqrt{17}+3)\simeq 1.187<1.214 \end{align*}

---

Another example one may consider is $$ \rho=\begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & \frac{1}{6} \end{pmatrix} $$ and note that $$ \|\rho-{\rm diag}(d(\rho))\|_1=\Big\| \begin{pmatrix} 0 & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & 0 & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & 0 \end{pmatrix} \Big\|_1\simeq 0.685515\,. $$ Now numerical search suggests that the argmin in question is $$ \begin{pmatrix} \frac7{12}\\\frac13\\\frac1{12} \end{pmatrix}(\neq d(\rho)) $$ for which the corresponding trace norm difference equals \begin{align*} \Big\| \begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & \frac{1}{6} \end{pmatrix} - \begin{pmatrix} \frac{7}{12} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{12} \end{pmatrix}\Big\|_1&=\Big\|\begin{pmatrix} \frac{1}{12} & -\frac{1}{4} & -\frac{1}{6} \\ -\frac{1}{4} & 0 & -\frac{1}{12} \\ -\frac{1}{6} & -\frac{1}{12} & -\frac{1}{12} \end{pmatrix}\Big\|_1 \\ &=\Big|-\frac{1}{3}\Big|+\frac{2+\sqrt{3}}{12}+\frac{2-\sqrt{3}}{12}=\frac23. \end{align*} While rigorously proving that this is in fact the argmin is difficult—so this could, theoretically, be a numerical artifact—this does show without a doubt that $s=d(\rho)$ cannot be the argmin in question (as, obviously, $\frac23<0.685515$). In fact, this seems to be "generic" behaviour, i.e., generating qutrit states at random seems to lead to counterexamples quite quickly.