# Is the trace norm monotone with respect to quantum operations?

The trace norm is defined to be $$\| K \| = \mathrm{tr}\sqrt{K^\dagger K}.$$ Is it true that we have $$\| \mathcal E(K) \|\leq \|K\|,$$ for any quantum operation $$\mathcal{E}: A\otimes B \to A\otimes B$$? If it is not true, then is it at least true for hermitian $$K$$?

I'm asking this because trace norm defines a distance measure on the space of density operators $$d(\rho, \sigma) = || \rho - \sigma ||$$, and I think it should be true that pushing two states through a quantum operation should not increase the distance between them, as it destroys distinctions that used to be visible.

Theorem 9.2 on p.406 in Nielsen & Chuang shows that this is true when $$\mathcal E$$ is trace-preserving and $$K = \rho - \sigma$$ where $$\rho, \sigma$$ are density operators, but I'm asking if the more general case is true, where $$\mathcal E$$ is not required to be trace-preserving, and $$K$$ is not required to be a difference of two density operators.

• This is true. See theorem 9.2 on p.406 in NIelsen & Chuang. A key step is the use of Jordan-Hahn decomposition wherein a Hermitian operator is written as a difference of two positive semidefinite operators with orthogonal support. Commented Jun 30, 2021 at 3:26
• @AdamZalcman Thanks for that link. I edited the question. Commented Jun 30, 2021 at 3:40
• If $\mathcal{E}$ is not a TP channel, and $K$ is just 'any old' positive semidefinite operator, then it's not true in general. Take $\mathcal{E}(K) = 2K$; this is CP but not TP. Then obviously the inequality does not hold.
– JSdJ
Commented Jun 30, 2021 at 9:05
• But in the question you first state 'for any quantum operation' which is TP (or trace decreasing, up to your definition), and then in the final paragraph of your question you ask the case where $\mathcal{E}$ is not required to be trace-preserving. Those two options do not align!
– JSdJ
Commented Jun 30, 2021 at 9:08
• @JSdJ I specified that $\mathcal E$ is a quantum operator, so it is $\leq 1$. The $\mathcal E$ you gave does not satisfy that. Commented Jun 30, 2021 at 9:08

The first step is to notice that $$\mathcal{E}$$ is CP $$\iff \mathcal{E} \otimes \mathcal{I}$$ is positive (for any ancillary space on which we act trivially with $$\mathcal{I}$$). Therefore, we can prove a simpler result, namely, that every positive map, let's call it $$\mathcal{F} := \mathcal{E} \otimes \mathcal{I}$$, is contractive (that is, it decreases the $$1$$-norm).
The contractivity of positive maps follows immediately from the following sequence of observations. If $$\Delta \geq 0$$ then $$\left\Vert \Delta \right\Vert_{1}^{} = \operatorname{Tr}\left[ \Delta \right]$$, that is, for positive semidefinite operators, the trace norm is the trace of the operator. Moreover, since $$\mathcal{F}$$ is a positive map, $$\Delta \geq 0 \implies \mathcal{F}(\Delta) \geq 0$$, that is, it maps positive semidefinite operators to positive semidefinite operators. Therefore, $$\left\Vert \mathcal{F}(\Delta) \right\Vert_{1}^{} = \operatorname{Tr}\left[ \mathcal{F}(\Delta) \right] \leq \operatorname{Tr}\left[ \Delta \right] = \left\Vert \Delta \right\Vert_{1}^{}$$ where in the last inequality I have used the fact that $$\mathcal{F}$$ is trace non-increasing.
One can, in fact, prove a more general claim: A trace non-increasing map $$\mathcal{F}$$ is positive if and only if for any hermitian operator $$\Delta$$ acting on $$\mathcal{H}$$, $$\left\Vert \mathcal{F}(\Delta) \right\Vert_{1}^{} \leq \left\Vert \Delta \right\Vert_{1}^{}$$. I'll leave this as an exercise with the following hint, start by writing a Hermitian operator $$\Delta = \Delta_{+} - \Delta_{-}$$ (which follows by separating the positive and negative eigenvalues in the spectral decomposition) and apply the same sequence of observations.