# Can a CPTP map increase the purity of a state?

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?

• – glS
Mar 17 at 18:49
• When in doubt about a property of channels, here's a list of canonical channels to check: physics.stackexchange.com/questions/291810/… Mar 17 at 19:20
• $T=\text{const}$ may be CPTP. Mar 18 at 14:18
• @AdamZalcman What do you mean by that? The only constant linear map is the zero map, which is certainly not trace-preserving Mar 20 at 13:59
• @AdamZalcman Ah, you mean a reset channel; that makes more sense, thanks for clarifying! Mar 20 at 14:12

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.

• how do you see that unital channels can't increase purity? (or equivalently, where is it shown?)
– glS
Mar 17 at 18:58
• Added a citation. Mar 17 at 19:15
• 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 $\|\rho\|_2^2\equiv{\rm tr}(\rho^2)$)
– glS
Mar 17 at 19:28
• @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) Mar 18 at 9:33
• @MarkusHeinrich ah, thanks for catching that!
– glS
Mar 18 at 9:45

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.

• 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
– glS
Mar 19 at 14:34
• 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 $\Phi(\vert 1\rangle\langle 1 \vert) = \vert 1\rangle\langle 1 \vert$. Mar 19 at 15:58

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.

• 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". Mar 18 at 18:24
• 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 Mar 18 at 18:49