Suppose $\Phi$ is a CPTP map with Kraus operators $\phi_n$, so that $\hat{\Phi} := Σ_n (\phi_n ⊗ \phi_n^*)$ is the matrix representation (here $*$ being entry-wise complex conjugate).

Is there an upper bound for $\|\hat{\Phi}\|_{\rm op}$ (where $\|\cdot\|_{\rm op}$ is "operator norm" / max-singular value)? And how should I interpret this quantity?

I need this for purely technical reasons (i.e. to bound some other quantity), but I'd still like to know what it means. I know it must be at least $\leq\sqrt{d}$, because e.g. it happens for the 1-qubit completely amplitude damping channel (defined here, with $\gamma=1$). I also generated a 2-qubit channel from the single-qubit amplitude damping (with Kraus operators $\phi_i\otimes\phi_j$, which to be fair I'm assuming it is indeed a channel) and the same holds. Furthermore, I'm tempted to understand this operator norm as a measure of how much the purity of a state (or I guess in a sense it's coherence) can be increased with the channel: for the example I cited of the amplitude damping channel, such case takes any state to the $|0\rangle$ state, on the other hand, if I try a totally depolarizing channel (such that $\Phi(\cdot)=\boldsymbol{1}/d$ ), then I get $\|\hat{\Phi}\|=1/2$ (which anyhow, is not $1/\sqrt{2}$, as I would've expected). I just haven't been able to prove or understand this generally.

  • $\begingroup$ "Matrix representation" is rather ambigous, it could also refer to the Choi matrix. $\endgroup$ Commented Dec 30, 2021 at 19:17

4 Answers 4



Below, we prove that

$$ \|\hat\Phi\|_{\rm op}=\sup_{\rho\in D(\mathcal{X})}\sqrt\frac{\gamma(\Phi(\rho))}{\gamma(\rho)} $$

where $D(\mathcal{X})$ denotes the set of density matrices on the Hilbert space $\mathcal{X}$ associated with the input of the quantum channel $\Phi$ and $\gamma(\rho)=\mathrm{tr}(\rho^2)$ is the purity of $\rho$.

This justifies the interpretation of $\|\Phi\|_{\rm op}$ as the square root of the largest multiplicative increase in purity due to the action of the channel $\Phi$. As a corollary, we obtain $\|\Phi\|_{\rm op}\in[1,\sqrt{d}]$ where $d$ is the dimension of the input Hilbert space $\mathcal{X}$.

Channel operator norm

Operator norm $\|.\|_{\rm op}$ is defined$^1$ as

$$ \|A\|_{\rm op}=\sup_{\|v\|_2\le 1}\|Av\|_2=\sup_{v\ne 0}\frac{\|Av\|_2}{\|v\|_2}.\tag1 $$

where $\|v\|^2_2=\sum_k |v_k|^2$ is the vector $\ell_2$ norm on the domain and codomain of $A$.

If $\hat\Phi$ is the matrix representing the quantum channel $\Phi:L(\mathcal{X})\to L(\mathcal{Y})$, then for all $X\in L(\mathcal{X})$

$$ \mathrm{vec}({\Phi(X)}) = \hat\Phi\,\mathrm{vec}(X)\tag2 $$

where $\mathrm{vec}$ is the vectorization map. Moreover, the $\ell_2$ norm of the vectorization of $X$ coincides with the Frobenius norm $\|.\|_F$ of $X$

$$ \|\mathrm{vec}(X)\|_2=\sqrt{\sum_{i,j=1}^n|x_{ij}|^2}=\|X\|_F\tag3 $$

where $n=\dim\mathcal{X}$. Thus, from $(1)$ we have

$$ \|\hat\Phi\|_{\rm op}=\sup_{\|X\|_F\le 1}\|\Phi(X)\|_F=\sup_{X\ne 0}\frac{\|\Phi(X)\|_F}{\|X\|_F}.\tag4 $$

Matrix achieving supremum is positive semidefinite

We will prove that the supremum in $(4)$ is achieved for a positive semidefinite $X$. First note that $\|\Phi(.)\|_F$ is continuous and $\{X\,|\,\|X\|_F\le 1\}$ is compact, so the supremum is achieved for some operator $M$. By linearity of $\Phi$ and absolute homogeneity$^2$ of the norm, $\|M\|_F=1$. We can write $M$ as the sum of its Hermitian and anti-Hermitian parts

$$ M=\frac{M+M^\dagger}{2}+\frac{M-M^\dagger}{2}.\tag5 $$

Setting $\alpha:=\frac12\|M+M^\dagger\|_F$ and $\beta:=\frac12\|M-M^\dagger\|_F$ we can rewrite $(5)$ as

$$ M=\alpha H+i\beta G\tag6 $$

where $H=\frac{M+M^\dagger}{2\alpha}$ and $G=\frac{M-M^\dagger}{2i\beta}$ are Hermitian matrices with unit Frobenius norm. Moreover, for any Hermitian $A,B$ we have $\|A+iB\|_F^2=\|A\|_F^2+\|B\|_F^2$, so $\alpha^2+\beta^2=1$.

Now, $\Phi$ sends Hermitian matrices to Hermitian matrices, so

$$ \|\Phi(M)\|_F^2=\alpha^2\|\Phi(H)\|_F^2+\beta^2\|\Phi(G)\|_F^2.\tag7 $$

But $\|\Phi(.)\|_F^2$ is convex$^3$, so $\|\Phi(M)\|_F^2\le\max\{\|\Phi(H)\|_F^2, \|\Phi(G)\|_F^2\}$. Therefore, we can take $M$ to be Hermitian.

Writing $M=R-S$ for some positive semidefinite $R$ and $S$ with orthogonal supports, we have $|M|=R+S$ and

$$ \begin{align} \|\Phi(M)\|_F^2&=\|\Phi(R)-\Phi(S)\|_F^2\\ &=\|\Phi(R)\|_F^2+\|\Phi(S)\|_F^2-2\langle\Phi(R),\Phi(S)\rangle_{HS}\\ &\le\|\Phi(R)\|_F^2+\|\Phi(S)\|_F^2+2\langle\Phi(R),\Phi(S)\rangle_{HS}\\ &=\|\Phi(R)+\Phi(S)\|_F^2\\ &=\|\Phi(|M|)\|_F^2 \end{align}\tag8 $$

where we used the fact that the Hilbert-Schmidt inner product of two positive semidefinite operators is a non-negative real number. Therefore, the supremum in $(4)$ is achieved for a positive semidefinite $X$ and we can rewrite $(4)$ as

$$ \|\hat\Phi\|_{\rm op}=\sup_{\rho\in D(\mathcal{X})}\frac{\|\Phi(\rho)\|_F}{\|\rho\|_F}\tag9 $$

where $D(\mathcal{X})$ denotes the set of density matrices on Hilbert space $\mathcal{X}$.

Relation to purity

Finally, Frobenius norm $\|\rho\|_F$ of a density matrix $\rho$ is the square root of its purity $\gamma(\rho)$. Therefore, $(9)$ becomes

$$ \|\hat\Phi\|_{\rm op}=\sup_{\rho\in D(\mathcal{X})}\sqrt\frac{\gamma(\Phi(\rho))}{\gamma(\rho)}.\tag{10} $$

This confirms the interpretation of $\|\hat\Phi\|_{\rm op}$ as the square root of the maximum multiplicative increase in purity due to the action of the channel $\Phi$.

In $d$-dimensional Hilbert space we have $\gamma(\rho)\in [\frac{1}{d},1]$, so $\|\Phi\|_{\rm op}\le\sqrt{d}$. Also, by Schauder's theorem, $\Phi$ has a fixed point, so $\|\Phi\|_{\rm op}\in[1,\sqrt{d}]$.

$^1$ Strictly speaking, operator norm may be induced by any pair of vector norms on the domain and codomain. Since operator norm $\|A\|_{\rm op}$ induced by the $\ell_2$ norm $\|v\|^2_2=\sum_k |v_k|^2$ is equal to $A$'s largest singular value, we assume that this is the operator norm the question is concerned with.

$^2$ Absolute homogeneity means that $\|sX\|_F=|s|\|X\|_F$ for all $s\in\mathbb{C}$. As pointed out by @Markus Heinrich in the comments, we may alternatively appeal to the convexity of $\|\Phi(.)\|_F$ and $\{X\,|\,\|X\|_F\le 1\}$.

$^3$ Because $\Phi$ is linear, $\|.\|_F$ is convex and $f(x)=x^2$ is convex and non-decreasing on $[0,+\infty)$.

  • 3
    $\begingroup$ Great answer! I hope you don't mind if I offer some improvements: First, you need that $\|\Phi(\cdot)\|_F$ is convex such that the supremum is attained at the boundary. Next, you have that $\|\Phi(M)\|_F^2 \leq \max\{ \|\Phi(H)\|_F^2, \|\Phi(G)\|^2\}$ which slightly simplifies your argument. Finally, since $\|M\|_F = \| |M| \|_F$, I think one can take the supremum over psd matrices with unit Frobenius norm, i.e. pure states. $\endgroup$ Commented Jan 4, 2022 at 12:21
  • $\begingroup$ I cannot edit my comment anymore, but the last point is of course nonsense. The matrices are not normalized in trace norm, hence it is not enough the optimize over pure states only. This would have led to a wrong upper bound on the spectral norm, too. $\endgroup$ Commented Jan 4, 2022 at 13:30
  • $\begingroup$ @MarkusHeinrich Thank you for reviewing the proof and offering improvements! Re first suggestion: My reasoning there was too implicit. I clarified that I obtain $\|M\|_F=1$ by linearity of $\Phi$ and absolute homogeneity of the norm. I also included your suggestion to use convexity in the footnotes. Re second suggestion: Nice simplification. I have incorporated it in the post. Re third suggestion: No worries :-) $\endgroup$ Commented Jan 5, 2022 at 8:34

To complement the two other answers, which give the bounds $1\le\|\Phi\|\le\sqrt{d}$: Both bounds are tight.

For the lower bound, this is saturated by the identity channel.

For the upper bound, this is e.g. the channel which maps all states to $\lvert0\rangle\langle0\rvert$: The output has norm $1$, while the maximally mixed state as an input has norm $1/\sqrt{d}$.


The spectral radius of (the natural representation of) positive trace-preserving maps, and thus that of channels as well, is equal by $1$. This is proved in Watrous' book, prop 4.26, page 230 of the online edition.

This implies a lower bound on the operator norm: $\|K(\Phi)\|\ge1$. One further thing proved in the book is that $\|K(\Phi)\|=1$ if and only if $\Phi$ is unital, that is, $\Phi(I)=I$. Here $\Phi$ is a generic positive trace-preserving map $\mathrm{Lin}(\mathcal X)\to\mathrm{Lin}(\mathcal X)$, and $K(\Phi)\in\mathrm{Lin}(\mathcal X\otimes\mathcal X)$ is its natural representation (which I think is what you are referring to as "matrix representation").

  • $\begingroup$ Thanks, I think spectral radius is upper-bounded by max-singular value? $\endgroup$
    – Pedro
    Commented Dec 30, 2021 at 14:26
  • $\begingroup$ @Pedro yes, indeed. I also forgot to mention that one can prove that $\|K(\Phi)\|=1$ if and only if $\Phi$ is unital. It doesn't fully answer the question, but can be useful information. $\endgroup$
    – glS
    Commented Dec 30, 2021 at 14:36

Found it indeed to be $\|\hat{\Phi}\|_{\mathrm{op}}\leq\sqrt{d}$, see for example here:

  • 1
    $\begingroup$ Key point: This is tight, e.g. by taking the map which maps everything to a pure state. $\endgroup$ Commented Dec 30, 2021 at 19:21

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