Is there an upper-bound on the operator norm (max-singular value) of the matrix representation of a quantum channel?

Question

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.

Answer 1

Summary

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 $[0,+\infty)$.

Answer 2

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 $1/\sqrt{d}$.

Answer 3

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").

Answer 4

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

• https://arxiv.org/abs/math-ph/0601063 (Theorem II.1 with $p=2$)
• https://arxiv.org/abs/1404.6025 (Proposition 5 item (ii), refers back to the article above)