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I am wondering if there exist CPTP maps $T$ such that the purity of a quantum state $\rho$ can increase, i.e.

$$ \text{tr} ( T ( \rho )^2 ) \geq \text{tr} ( \rho ^2). $$

If so, what are the conditions on $T$ and/or $\rho$ for this to be possible?

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3 Answers 3

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Yes, some quantum channels can increase purity. For example the preparation channel $$ T(X) = \mathrm{Tr}[X] |\psi\rangle \langle \psi| $$ that can be thought of as throwing away your system and preparing $|\psi\rangle$ will always produce a pure state. Therefore $\mathrm{Tr}[T(\rho)^2] = 1$ for any $\rho$.

A channel $T$ is called unital if $T(I) = I$ (i.e. preserves the identity matrix). If $T$ is a unital channel then it is possible to show that $\mathrm{Tr}[T(X)^2] \leq \mathrm{Tr}[X^2]$. Thus any channel that increases purity must be non-unital. It turns out non-unitality is necessary and sufficient see Theorem II.4 in Contractivity of positive and trace preserving maps under Lp norms.

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  • $\begingroup$ how do you see that unital channels can't increase purity? (or equivalently, where is it shown?) $\endgroup$
    – glS
    Mar 17 at 18:58
  • $\begingroup$ Added a citation. $\endgroup$
    – Rammus
    Mar 17 at 19:15
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    $\begingroup$ interesting, thanks! So summarising (and changing a bit) the argument there: in the HS inner product you have $\langle T(\rho),T(\rho)\rangle=\langle \rho,T^\dagger\circ T(\rho)\rangle\le \|\rho\|^2_2 \|T^\dagger\circ T\|_\infty$, and if $T$ is unital then $T^\dagger$ is a channel, thus so is $T^\dagger\circ T$, and a channel has operatorial norm smaller than one (and $\|\rho\|_2^2\equiv{\rm tr}(\rho^2)$) $\endgroup$
    – glS
    Mar 17 at 19:28
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    $\begingroup$ @glS A channel does not necessarily have operator/spectral norm smaller than one (e.g. replacement channel!). However, the channels with unit spectral norm are exactly the unital channels (Watrous, Thm. 4.27) $\endgroup$ Mar 18 at 9:33
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    $\begingroup$ @MarkusHeinrich ah, thanks for catching that! $\endgroup$
    – glS
    Mar 18 at 9:45
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Here's the simplest argument I could come up with for the statements Rammus made about unital channels. We actually don't need complete positivity, just positivity is enough.

Suppose that $T$ is a positive, unital, and trace-preserving map and let $\rho$ be a density operator input to $T$. Consider a spectral decomposition of $\rho$. $$ \rho = \sum_{k = 1}^{n} p_k \vert\psi_k\rangle\langle\psi_k\vert $$

Define $\sigma_k = T(\vert\psi_k\rangle\langle\psi_k\vert)$ for $k = 1,\ldots,n$. We can use our assumptions on $T$ as follows.

  • Because $T$ is positive, $\sigma_1,\ldots,\sigma_n$ are positive semidefinite.
  • Because $T$ is unital, $\sigma_1 + \cdots + \sigma_n = T(\mathbb{I}) = \mathbb{I}$.
  • Because $T$ is trace-preserving, $\sigma_1,\ldots,\sigma_n$ all have unit trace.

We can now obtain the desired inequality by Cauchy-Schwarz.

\begin{align*} \operatorname{Tr}(T(\rho)^2) & =\sum_{j,k=1}^n p_j p_k \operatorname{Tr}(\sigma_j\sigma_k)\\ & \leq\sqrt{\sum_{j,k=1}^n p_j^2 \operatorname{Tr}(\sigma_j\sigma_k)} \sqrt{\sum_{j,k=1}^n p_k^2 \operatorname{Tr}(\sigma_j\sigma_k)}\\ & = \sum_{k=1}^n p_k^2\\ & = \operatorname{Tr}(\rho^2) \end{align*} For the inequality, we're making use of the fact that $\operatorname{Tr}(\sigma_j\sigma_k)$ is always a nonnegative real number, which follows from the fact that $\sigma_1,\ldots\sigma_n$ are positive semidefinite, and for the equality that follows we're using both the equality $\sigma_1+\cdots+\sigma_n = \mathbb{I}$ and the fact that $\sigma_1,\ldots,\sigma_n$ all have unit trace.

On the other hand, if $T$ isn't unital, then the inequality above will be violated for the completely mixed state $\rho = \mathbb{I}/n$. This is because $\sigma = T(\rho)$ is a density operator that isn't completely mixed, so $$ 0 < \operatorname{Tr}((\sigma - \rho)^2) = \operatorname{Tr}(\sigma^2) - \frac{1}{n} = \operatorname{Tr}(T(\rho)^2) - \operatorname{Tr}(\rho^2). $$ In short, the completely mixed state is uniquely qualified as the most impure a state can be, so you necessarily get more pure if you move away from it.

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  • $\begingroup$ re last sentence: you're probably not quite saying this, but the statement makes me wonder; is it true that for any positive TP map, $T(I/d)$ is always the most mixed state in the image of $T$? Ie that ${\rm tr}(\Phi(\rho)^2)\ge {\rm tr}(\Phi(I/d)^2)$ for all $\rho$? though this might warrant a separate post if it's not obvious $\endgroup$
    – glS
    Mar 19 at 14:34
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    $\begingroup$ No, that's not true — take for instance a classical-quantum qubit channel defined as $\Phi(\vert 0\rangle\langle 0\vert) = \mathbb{I}/2$ and $\Phi(\vert 1\rangle\langle 1 \vert) = \vert 1\rangle\langle 1 \vert$. $\endgroup$ Mar 19 at 15:58
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Another common example of a CPTP channel which can increase purity (excluded from that link of canonical channels) is amplitude damping, with Kraus operators (parameterised by probability $\lambda$) $$ K_1 = \begin{pmatrix} 1 & \\ & \sqrt{1-\lambda} \end{pmatrix}, \;\;\;\;\;\;\;\; K_2 = \begin{pmatrix} & \sqrt{\lambda} \\ & \end{pmatrix}, $$ which act on a general one-qubit state $\rho$ as $$ \rho \rightarrow K_1 \rho K_1^\dagger + K_2 \rho K_2^{\dagger}. $$ This channel represents a simple dissipative process toward a minimum observable state, above assumed as $|0\rangle$. It is easy to see that as $\lambda$ approaches $1$ and ergo the probability of the dissipative transition $|1\rangle \rightarrow |0\rangle$ becomes certain, the maps become $$ K_1 \rightarrow |0\rangle\langle 0|, \;\;\;\; K_2 \rightarrow |1\rangle\langle 0|. $$ The maps take both $|0\rangle$ and $|1\rangle$ states and output a completely pure $|0\rangle$ state, even if the input $\rho$ is maximally mixed.

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  • $\begingroup$ For $\lambda=1$ this recovers the replacement channel of the answer and for $\lambda <1$ this will always decrease the purity of $\rho=|1\rangle \langle 1|$. So this probably shouldn't be called a "purity-increasing channel". $\endgroup$
    – forky40
    Mar 18 at 18:24
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    $\begingroup$ For $\lambda < 1$, it remains a channel capable of increasing purity as per the OP's question, which I pointed out explicitly for the maximally mixed state. I'll change the wording you mention. Indeed the channels are equivalent for specific respective parameterisations ($\lambda = 1$, $|\psi\rangle = |0\rangle$), though I'm unsure why that's notable $\endgroup$
    – Anti Earth
    Mar 18 at 18:49

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