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

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edited Sep 12 at 5:49

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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|)$$D(|\psi\rangle\langle \psi|)={\rm diag}(\frac49,\frac49,\frac19)$ equals $\frac{4}{9} (\sqrt{3}+1)\simeq 1.214$
on the other hand, the argmin is $s=(\frac12,\frac12,0)$, and the trace distance in question is $$ \big\|\,|\psi\rangle\langle\psi|-{\rm diag}(s)\big\|_1=\frac{1}{6} (\sqrt{17}+3)\simeq 1.187<1.214 $$\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.

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 $D(|\psi\rangle\langle \psi|)$ 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 $$ \big\|\,|\psi\rangle\langle\psi|-{\rm diag}(s)\big\|_1=\frac{1}{6} (\sqrt{17}+3)\simeq 1.187<1.214 $$

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.

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.

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edited Sep 11 at 12:17

Frederik vom Ende

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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 / examplethis paper (arXiv) to me! Considerme; 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 $D(|\psi\rangle\langle \psi|)$ 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 $$ \big\|\,|\psi\rangle\langle\psi|-{\rm diag}(s)\big\|_1=\frac{1}{6} (\sqrt{17}+3)\simeq 1.187<1.214 $$

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.

It turns out that this is not true, and I want to thank Rubén Ibarrondo (UPV/EHU) for suggesting this problem / example to me! Consider $$ \rho=\begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & \frac{1}{6} \end{pmatrix} $$ 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, $\frac23<0.685515$).

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 $D(|\psi\rangle\langle \psi|)$ 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 $$ \big\|\,|\psi\rangle\langle\psi|-{\rm diag}(s)\big\|_1=\frac{1}{6} (\sqrt{17}+3)\simeq 1.187<1.214 $$

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.

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created Sep 11 at 10:48

Frederik vom Ende

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1
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47

It turns out that this is not true, and I want to thank Rubén Ibarrondo (UPV/EHU) for suggesting this problem / example to me! Consider $$ \rho=\begin{pmatrix} \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \\ \frac{1}{4} & \frac{1}{3} & \frac{1}{12} \\ \frac{1}{6} & \frac{1}{12} & \frac{1}{6} \end{pmatrix} $$ 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, $\frac23<0.685515$).

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