Posted on 3th July 2023
The first of hopefully a few posts about actual Mathematics, the context of which will follow. In Watrous's QIT book, Theorem 4.25 states the following:
Let \( \Phi:\mathbb M_n\rightarrow\mathbb M_n \) be a unital quantum channel with Kraus representation \( \Phi(x) = \sum_a A_a x A_a^* \). Then \( \Phi(x)=x \) if and only if \( A_a x = x A_a \) for each \( a \).
(Here I translate slightly the language). Whenever I see a statement about quantum channels, I ask myself if I can translate it into a more Operator Algebraic statement making more use of the Stinespring representation theorem, for example. After much messing about, I realised that the key property is that \( \Phi \) is trace preserving, and that this, together with multiplicative domain theory, can be used to give a simple proof. (Well, maybe not simple, but one adding some context.)
Let's start by recalling some basic results, which I always forget somehow. Let \( A,B \) be \( C^* \)-algebras and \( \Phi:A\rightarrow B \) be a completely positive map. The Stinespring representation theorem tells us that if \( B\subseteq \mathcal B(H) \) is a faithful representation, then there is a Hilbert space \( K \), a \( * \)-representation \( \pi:A\rightarrow\mathcal B(K) \), and a bounded linear map \( V:H\rightarrow K \) such that \( \Phi(a) = V^*\pi(a)V \) for each \( a\in A \). We call \( (\pi,K,V) \) a Stinespring dilation of \( \Phi \), which is said to be minimal if \( \{ \pi(a)V\xi: \xi\in H, a\in A\} \) has dense linear span in \( K \). A minimal dilation is unique up to (obvious) unitary equivalence. (It can sometimes be helpful to know that when \( (\pi,K,V) \) is minimal, there is a \( * \)-homomorphism \( \rho:\Phi(A)' \rightarrow \pi(A)' \) with \( \rho(x)V = Vx \) for each \( x\in\Phi(A)'\subseteq\mathcal B(H) \). Then \( \rho \) is automatially normal.)
For the following, see Brozwn+Ozawa Proposition 1.5.7.
Theorem: Let \( \Phi:A\rightarrow B \) be a contractive completely positive map. Then: 1. (The Schwarz Inequality) We have that \( \Phi(a)^*\Phi(a) \leq a^*a \) for each \( a\in A \); 2. (Bimodule structure) For \( a\in A \), if \( \Phi(a^*a) = \Phi(a)^*\Phi(a) \) then \( \Phi(ba)=\Phi(b)\Phi(a) \) for each \( b\in A \); and if \( \Phi(aa^*) = \Phi(a)\Phi(a)^* \) then \( \Phi(ab)=\Phi(a)\Phi(b) \) for each \( b\in A \); 3. The space \( A_{\Phi}=\{a\in A:\Phi(a^*a) = \Phi(a)^*\Phi(a), \Phi(aa^*)=\Phi(a)\Phi(a)^*\} \) is a \( C^* \)-subalgebra of \( A \), and \( \Phi \) restricted to \( A_\Phi \) is a \( * \)-homomorphism.
Proof: Let \( B\subseteq\mathcal B(H) \) be faithful, and let \( (\pi,K,V) \) be a Stinespring dilation of \( \Phi \). As \( \Phi \) is a contraction, \( \|V\|\leq 1 \) and so \( VV^*\leq 1 \). For \( a\in A \) we have \[ \begin{split} \Phi(a^*a)-\Phi(a)^*\Phi(a) &= V^*\pi(a)^*\pi(a)V^* - V^*\pi(a)^*VV^*\pi(a)V \\ &= V^* \pi(a)^* (1-VV^*) \pi(a)V \geq 0, \end{split} \] giving the Schwarz Inequality.
Furthermore, we notice that \( \Phi(a^*a) = \Phi(a)^*\Phi(a) \) if and only if \( (1-VV^*)^{1/2}\pi(a)V=0 \). If this holds, then for \( b\in A \), \[ \Phi(ba) - \Phi(b)\Phi(a) = V^*\pi(b)(1-VV^*)\pi(a)V = 0. \] Similarly, \( \Phi(aa^*)=\Phi(a)\Phi(a)^* \) is equivalent to \( (1-VV^*)^{1/2}\pi(a)^*V=0 \), equivalently, \( V^* \pi(a) (1-VV^*)^{1/2} = 0 \). This implies that \( \Phi(a)\Phi(b) = \Phi(ab) \) for each \( b\in A \).
We now consider (3). As \( \Phi \) is positive, it is a \( * \)-map, and so \( a\in A_\Phi \) implies that \( a^*\in A_\Phi \). If both \( \Phi(a^*a) = \Phi(a)^*\Phi(a) \) and \( \Phi(aa^*)=\Phi(a)\Phi(a)^* \) then by (2) we know that \( \Phi(ba)=\Phi(b)\Phi(a) \) and \( \Phi(ab)=\Phi(a)\Phi(b) \) for each \( b\in A \). Hence, given \( a,b\in A_\Phi \) we have \[ \Phi((ab)^*ab) = \Phi(b^*a^*a) \Phi(b) = \Phi(b^*a^*) \Phi(a) \Phi(b) = \Phi((ab)^*) \Phi(ab), \] using that \( b\in A_\Phi \), then that \( a\in A_\Phi \), and then that \( b\in A_\Phi \). Similarly, \( \Phi((ab)(ab)^*) = \Phi(ab)\Phi((ab)^*) \) and so \( ab\in A_\Phi \). As \( \Phi \) is continuous, it is easy to see that \( A_\Phi \) is closed, hence a \( C^* \)-subalgebra of \( A \). We have already shown that \( \Phi \) restricted to \( A_\Phi \) is a \( * \)-homomorphism.
We can now prove our abstract commutation result.
Proposition: Let \( \Phi:A\rightarrow A \) be completely positive, and suppose there is a faithful state \( \varphi\in A^* \) with \( \varphi\circ\Phi = \varphi \). The collection of fixed points of \( \Phi \) (those \( a\in A \) with \( \Phi(a)=a \)) forms a \( C^* \)-subalgebra of \( A_\Phi \).
Proof: Given \( a\in A \) with \( \Phi(a)=a \), also \( \Phi(a^*)=a^* \), by the Schwarz Inequality, \( a^*a \leq \Phi(a^*a) \), and so \( \Phi(a^*a)-a^*a\geq 0 \), but also \( \varphi( \Phi(a^*a)-a^*a ) = 0 \), and so as \( \varphi \) is faithful, \( \Phi(a^*a) = a^*a \). As \( a^* \) is also fixed by \( \Phi \), also \( \Phi(aa^*) = aa^* \). Hence \( a\in A_\Phi \). Given \( a,b \) fixed, also \( a^* \) is fixed, and \( \Phi(ab) = \Phi(a)\Phi(b) \) because \( a,b\in A_\Phi \), so \( \Phi(ab)=ab \). We conclude that the fixed points form a \( C^* \)-subalgebra of \( A_\Phi \).
A Cesaro mean arguments shows that if \( \Phi \) is unital, there is always some invariant state. However, without there being a faithful invariant state, the collection of fixed points can fail to be an algebra. For example, let \( A=\mathbb C^3 \) with the pointwise product, let \( \Phi \) be the linear map with matrix form \[ \begin{pmatrix} 1/4 & 1/4 & 1/2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \] This is positive, hence completely positive (as \( A \) is abelian), but for example, the vector \( (1,-1,2)^T \) is fixed while its square \( (1,1,4)^T \) is not.
Proposition: Let \( \Phi:A\rightarrow A \) be completely positive and unital, and suppose there is a faithful state \( \varphi\in A^* \) with \( \varphi\circ\Phi = \varphi \). Let \( (\pi,K,V) \) be a minimal dilation of \( \Phi \). Then \( a\in A \) is fixed by \( \Phi \) if and only if \( \pi(a)V = Va \).
Proof: If \( \pi(a)V=Va \) then \( \Phi(a) = V^*\pi(a)V=V^*Va=a \), as \( V^*V=1 \) as \( \Phi \) is unital. Conversely, by the above, we know that if \( \Phi(a)=a \) then \( a\in A_{\Phi} \) so \( \Phi(ba)=\Phi(b)\Phi(a) = \Phi(b)a \) for each \( b\in A \). For \( \xi,\eta\in H \), \[ (\pi(b) V \eta | \pi(a) V \xi) = (\eta | \Phi(b^*a) \xi) = (\eta | \Phi(b^*)a\xi) = (\pi(b)V\eta | Va\xi). \] (Here I write my inner-products linear on the right.) By minimality, vectors of the form \( \pi(b)V\eta \) are linearly dense, and so we conclude that \( \pi(a)V\xi = Va\xi \), that is, \( \pi(a)V = Va \) as claimed.
If \( (\pi',K',V') \) is another dilation, not assumed minimal, then there is always a unitary \( u:K' \rightarrow K' \) which intertwines \( \pi \) and \( \pi' \), and satisfies \( uV = V' \). (Use that \( (\pi,K,V) \) is minimal, and check that \( u : \pi(a)V\xi \mapsto \pi'(a)V'\xi \) extends by linearity to a well-defined isometry.) Then \( \pi'(a)V' = u \pi(a)V = uVa = V'a \) if \( a \) is fixed (and \( \Phi \) satisfies our hypotheses).
Now come back to a quantum channel \( \Phi:\mathbb M_n \rightarrow \mathbb M_n \) (so \( \Phi \) is completely positive and trace-preserving). As the trace is invariant (and can be normlised to a state!) our proposition applies. To make a link with Kraus operators, we need to recall what the Stinespring representation looks like in this case. Any \( * \)-representation \( \pi:\mathbb M_n\rightarrow\mathcal B(K) \) has the form \( \pi(a)=a\otimes 1 \) where \( K\cong\mathbb C^n \otimes H \) for some auxiliary Hilbert space \( H \). Let \( (e_i) \) be an orthonormal basis for \( H \), so \( V:\mathbb C^n\rightarrow K = \mathbb C^n\otimes H \) has the form \( V\xi = \sum_i A_i\xi \otimes e_i \) for some matrices \( (A_i) \subseteq \mathbb M_n \) with \( 1 = V^*V = \sum_i A_i^*A_i \). Then \( \Phi(a) = V^*(a\otimes 1)V = \sum_i A_i^* a A_i \) gives the Kraus form (here I adopt a convention that privilages UCP maps). That \( \Phi \) is trace-preserving means that \( \sum_i A_i A_i^* = 1 \) as well. (Notice that any minimal dilation will have \( K \) finite-dimensional, so \( H \) will also be finite-dimensional, and all sums will be finite.)
Theorem: Let \( \Phi:\mathbb M_n\rightarrow\mathbb M_n \) be a unital quantum channel with Kraus representation \( \Phi(x) = \sum_a A_a x A_a^* \). Then \( \Phi(x)=x \) if and only if \( A_a x = x A_a \) for each \( a \).
Proof: From the proposition above and the discussion after, \( \Phi(x)=x \) if and only if \( \pi(x)V=Vx \) for any Stinespring dilation \( (\pi,K,V) \). Such a dilation is given by \( \pi(x)=x\otimes 1 \) on \( K=\mathbb C^n\otimes K' \) and \( V\xi = \sum_a A_a^*(\xi)\otimes e_a \) where \( (e_a) \) is an orthonormal basis of \( K' \). Then \( \pi(x)V=Vx \) if and only if \( A_a^* x = x A_a^* \) for each \( a \), which is equivalent to \( A_a x^* = x^* A_a \) for each \( a \). As \( \Phi(x)=x \) if and only if \( \Phi(x^*)=x^* \), the claim follows.
So we have our more abstract proof. Update: Of course, there is nothing new under the sun, and by chance I came across the paper
Arias, A.; Gheondea, A.; Gudder, S., Fixed points of quantum operations. J. Math. Phys. 43, No. 12, 5872-5881 (2002). Zbl 1060.81009
which gives essentially the same argument, and a lot more: well worth a read.