I have a $d$-dimensional state $\rho$. Is there any way to find the (possibly not unique) trace distance to the nearest pure state: $$ \min_{|\psi\rangle} \,\,\lVert \rho - |\psi\rangle\langle \psi| \rVert_1 $$

where $\lVert A\rVert_1 = \mathrm{tr}\,\sqrt{A^\dagger A}$. I am not concerned with actually identifying $|\psi\rangle$. An answer that finds the maximum fidelity with a pure state is equally good for my purposes.

For instance, with $d=2$ we can use the Bloch sphere picture to solve this problem: if we represent $\rho$ and $|\psi\rangle \langle \psi|$ as \begin{align} \rho &= \frac{I + \vec{r}\cdot \vec{\sigma}}{2}\\ |\psi\rangle \langle \psi| &= \frac{I + \vec{s}\cdot \vec{\sigma}}{2} \quad, \quad \lVert \vec{s}\rVert=1 \end{align}

where $\vec{\sigma} = (\sigma_1, \sigma_2, \sigma_3)$ is the vector of Pauli operators, then the closest point $\vec{s}$ on the surface of the sphere (pure state) to a point internal to the sphere (mixed state) will lie along the ray formed by $(0,0,0)$ and $\vec{r}$, so the desired pure state is just $$ \vec{s}= \frac{1}{\lVert \vec{r} \rVert} \vec{r} $$

and the corresponding distance is $d(\rho, |\psi\rangle\langle\psi|) = 1 - \lVert \vec{r} \rVert$. However I don't really know how to extend this kind of approach for $d>2$ because I can't really wrap my head around the geometry of higher dimensional state spaces.


3 Answers 3


Recall that for any Hermitian operator $A$ and any unit vector $|\psi\rangle$ the real number $\langle \psi|A|\psi\rangle$, known as the Rayleigh quotient, is bounded by the largest eigenvalue $\lambda_{max}$ of $A$

$$ \langle \psi|A|\psi\rangle \le \lambda_{max}. $$

Moreover, the maximum is achieved when $|\psi\rangle$ is the unit norm eigenvector of $A$ associated with $\lambda_{max}$. When $A=\rho$ is a density matrix and $|\psi\rangle$ a pure state then the Rayleigh quotient equals the fidelity $$F(\rho, |\psi\rangle\langle\psi|) = \langle \psi|\rho|\psi\rangle.$$

Therefore, the maximum fidelity of $\rho$ with a pure state is

$$ \max_{|\psi\rangle} F(\rho, |\psi\rangle\langle\psi|) = \max_{|\psi\rangle}\, \langle\psi|\rho|\psi\rangle = \lambda_{max} $$

where $\lambda_{max}$ is the largest eigenvalue of $\rho$. (Moreover, $F(\rho, |\psi\rangle\langle\psi|)$ is maximized when $|\psi\rangle$ is the eigenvector of $\rho$ associated with $\lambda_{max}$.)


The fidelity case was already worked in the other answer. Here is an idea for the trace distance one.

The trace distance between $\rho$ and some $|\psi\rangle\!\langle\psi|$ is $$\|\rho - |\psi\rangle\!\langle\psi|\|_1 = \operatorname{Tr}\lvert \,\rho - |\psi\rangle\!\langle\psi|\,\rvert, $$ which is equal to the sum of the singular values of $\rho-|\psi\rangle\!\langle\psi|$. Write the eigendecomposition of $\rho$ as $$\rho = \sum_k p_k |u_k\rangle\!\langle u_k|,$$ for some orthonormal basis $\{|u_k\rangle\}_k$ and $p_k\ge0$ with $\sum_k p_k=1$.

Notice that if we choose $|\psi\rangle=|u_1\rangle$ (I fix the first eigenvector here without loss of generality), then $$\rho - |\psi\rangle\!\langle\psi| = (p_1-1)|u_1\rangle\!\langle u_1| + \sum_{k>1} p_k |u_k\rangle\!\langle u_k|.$$ Because the singular values of a Hermitian matrix are the moduli of the eigenvalues, we see that $$ \|\rho - |\psi\rangle\!\langle\psi| \|_1 = |p_1-1| + \sum_{k>1} p_k = 2(1 - p_1). $$ Minimising this expression with respect to $|\psi\rangle$, restricting the possible choices to the eigenvectors of $\rho$, thus leads to the minimum trace distance being $2(1- \max_k p_k)$, achieved with $|\psi\rangle$ the eigenstate of $\rho$ which largest probability. This is the same conclusion obtained for the fidelity.

I should stress that I didn't actually prove that there cannot be some $|\psi\rangle$ that is not an eigenvector of $\rho$ for which we can get an even smaller trace distance. That's unlikely to be the case, but a bit more thought might be required to show it formally.

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    $\begingroup$ See my answer for a formal argument. (The first statement is very general and quite useful. Thanks to K. Audenaert for teaching me it.) $\endgroup$ May 7, 2021 at 10:04

The trace norm is unitarily invariant. For any unitarily invariant matrix norm, the norm-distance $\|A-B\|$ of two matrices becomes extremal when they are diagonal in the same basis. (Specifically, for the minimal value, the eigenvalues need to be ordered the same way, and for the maximum, the opposite way. This is a consequence of Lidskii's theorem, see e.g. Bhatia, Eq. (IV.62).)

In your case, you have that $B=\vert\phi\rangle\langle\phi\vert$ has only one eigenvalue $1$. For a minimal norm distance, you want $B$ is diagonal in the same basis as $\rho$ with the same eigenvalue order.

Thus, it follows that for the minimum trace distance, $\vert\phi\rangle$ must be the leading eigenvector of $\rho$, and consequently $\|\rho-\vert\phi\rangle\langle\phi\vert\|_1=1-\lambda_1$, with $\lambda_1$ the eigenvalue.

  • $\begingroup$ that's a nice one. I wonder if with something like this one could prove that the result is the same in an arbitrary (unitarily invariant, monotonic) distance $\endgroup$
    – glS
    May 8, 2021 at 9:57
  • $\begingroup$ @glS Yes, I"m not using anywhere that this is the trace distance. Any distance arising from a unitarily invariant norm will do. $\endgroup$ May 8, 2021 at 20:38

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