# $M(\rho)=\operatorname{Tr}_2\left(\ U\ \rho\otimes\rho_2\ U^{\dagger}\right)$is unitary $\iff\ U=U_1\otimes U_2$, a product of $2$ unitary operators?

Let $$\rho : V_1 \to V_1$$ and $$\rho_2 : V_2 \to V_2$$, where $$V_1$$ and $$V_2$$ are Hilbert spaces.

Suppose that $$U:V_1\otimes V_2 \to V_1\otimes V_2$$ is a unitary operator.

Define a map $$M : L(V_1, V_1) \to L(V_1, V_1)$$ as \begin{align*} M(\rho) := \operatorname{Tr}_2 \left(\ U\ \rho\otimes\rho_2 \ U^{\dagger}\ \right) \end{align*} where $$\rho_2 \in L(V_2, V_2)$$ is a fixed density operator, and $$\operatorname{Tr}_2$$ is the partial trace of vector space $$V_2$$.

Then, trivially $$M$$ is a unitary operator, if $$U = U_1 \otimes U_2$$ for some unitary operators $$U_1 \in L(V_1,V_1)$$ and $$U_2 \in L(V_2,V_2)$$.

Is the converse also true? If $$U$$ cannot be expressed as a tensor product of $$2$$ unitary operators, then is $$M$$ non-unitary ?

I am lost how to prove this statement. Any hints or references are appreciated.

• crossposted to pse: physics.stackexchange.com/questions/548847/… – Norbert Schuch May 3 '20 at 17:20
• Let me iterate my comment from physics.SE: " I bet this can be proven using the uniqueness ("up to ...") of the Stinespring dilation." (And given you say "Any hints [...] are appreciated" -- would this qualify as an answer then? – Norbert Schuch May 3 '20 at 17:20
• It is possible to find a density operator $\rho_2$ and a unitary operator $U$ that does not happen to be expressible as $U = U_1\otimes U_2$ for which $M$ as you define it is a unitary channel. Is this the correct interpretation of the question? – John Watrous May 4 '20 at 0:30
• @JohnWatrous can you give an example of that (if I understand you correctly and you are saying that that is possible)? – glS May 4 '20 at 16:34
• Take $\rho_2 = |0\rangle\langle 0|$ and $U = U_1\otimes |0\rangle\langle 0| + V \otimes |1\rangle\langle 1|$ for any unitary $V$ that is linearly independent of $U_1$. – John Watrous May 4 '20 at 23:01

This is probably not the answer to what you've meant, but it's still relevant.

Assume that $$\rho_2 = |0\rangle\langle0|$$ $$-$$ it's known that quantum channels have such representation.

If $$U = U_1 \otimes U_2$$ then $$M(\rho) = U_1\rho U_1^\dagger.$$ This $$M$$ is "unitary" if we consider the space $$L(V_1, V_1)$$ as a vector space of matrices with Hilbert-Schmidt inner product given by $$(A,B) = \text{Tr}(B^\dagger A)$$. Indeed, we have $$\text{Tr}(M(B)^\dagger M(A)) = \text{Tr}\big((U_1BU_1^\dagger)^\dagger(U_1AU_1^\dagger) \big)= \text{Tr}(B^\dagger A),$$ so the inner product remains the same.

Now suppose $$M$$ is unitary in this sense. Consider any pure state $$\theta$$ (density matrix of it, e.g. $$|1\rangle\langle1|$$). We must have $$\text{Tr}(M(\theta)^\dagger M(\theta)) = \text{Tr}(\theta^\dagger \theta) = 1.$$ But $$\text{Tr}(M(\theta))=1$$. Let $$\lambda_i$$ be eigenvalues of $$M(\theta)$$, so $$0\leq \lambda_i \leq 1$$ and $$\sum_i \lambda_i = 1$$. The above equality gives us that $$\sum_i \lambda_i^2 = 1$$. From this it's easy to deduce that for some index $$k$$ it must be $$\lambda_k=1$$ and $$\lambda_i = 0$$ for $$i\neq k$$. That is, $$M(\theta)$$ also must be a pure state. So, $$M$$ maps pure states to pure states.

Notice that partial trace $$\text{Tr}_2(s)$$ is pure for a density matrix $$s$$ only if the state $$s$$ is a product state: $$s = \text{Tr}_2(s) \otimes \text{Tr}_1(s)$$ (here $$\text{Tr}_1(s)$$ is not necessary pure).

So we can write $$U\ \theta \otimes\rho_2 \ U^{\dagger}\ = M(\theta) \otimes N(\theta),$$ where $$N(\theta) = \text{Tr}_1(U\ \theta \otimes\rho_2 \ U^{\dagger})$$ is a complementary channel.

Now take two pure states $$\theta_1, \theta_2$$. We have that

$$M(\theta_1\theta_2) = \text{Tr}_2(U\ \theta_1\theta_2 \otimes\rho_2 \ U^{\dagger}) = \text{Tr}_2(U\ \theta_1 \otimes\rho_2 \ U^{\dagger} \cdot U\ \theta_2 \otimes\rho_2 \ U^{\dagger}) =$$ $$= \text{Tr}_2( M(\theta_1) \otimes N(\theta_1) \cdot M(\theta_2) \otimes N(\theta_2)) = M(\theta_1)M(\theta_2).$$ So, for any pure states $$\theta_1, \theta_2$$ we have that $$M(\theta_1\theta_2) = M(\theta_1)M(\theta_2).$$ By linearity it can be proved that for any matrices $$A,B \in L(V_1, V_1)$$: $$M(AB) = M(A)M(B).$$ It also can be shown that $$M(I)=I$$ and $$M(A^\dagger) = M(A)^\dagger$$. So $$M$$ is a unital $$*$$-homomorphism and this is a known fact that such homomorphism from matrix algebra to itself always corresponds to a unitary conjugation, i.e. it must be $$M(A) = U_1 A U_1^\dagger$$ for some unitary $$U_1$$ and any matrix $$A$$.

• Yes, I assumed $\rho_2 = |0\rangle\langle0|$ for simplicity. – Danylo Y May 4 '20 at 16:41
• right, I missed the remark at the beginning. But do you think the result holds for non-pure $\rho_2$? Because to me it looks like it can't. If $\rho_2=\sum p_k|p_k\rangle\!\langle p_k|$ then $M(\rho)=\sum_k p_k \operatorname{Tr}[U(\rho\otimes|p_k\rangle\!\langle p_k|)U^\dagger]$, which can only be pure (as must be the case if the action on $\rho$ is unitary) if $p_k=\delta_{k0}$ (or similar) – glS May 4 '20 at 16:44
• Well, if $U = U_1 \otimes U_2$ then it doesn't matter if $\rho_2$ is pure or not. In your expression all $\operatorname{Tr}[U(\rho\otimes|p_k\rangle\!\langle p_k|)U^\dagger]$ can be the same (as in the case $U = U_1 \otimes U_2$), so you can't deduce that $p_k = \delta_{k0}$. – Danylo Y May 4 '20 at 17:20
• It's a know fact that any quantum channel has the form $\operatorname{Tr}_2[U(\rho\otimes|0\rangle\langle 0|)U^\dagger]$. So, if we are given a channel with some non-pure $\rho_2$, then this same channel has the form were $\rho_2=|0\rangle\langle 0|$ but with some different $U$. – Danylo Y May 4 '20 at 17:25
• I know its "a known fact", but I haven't actually ever seen a proof of it. I see that my argument doesn't work for $U$ separable though. Do you also know of a counterexample for non-separable $U$? – glS May 4 '20 at 17:27

$$\newcommand{\calU}{\mathcal{U}}\newcommand{\calV}{\mathcal{V}}\newcommand{\red}[1]{{\color{red}#1}}\newcommand{\purple}[1]{{\color{purple}#1}}\newcommand{\green}[1]{{\color{green}#1}}$$You can prove this as follows:

Suppose $$\sigma=|k\rangle\!\langle k|$$.

Expliciting the expression of $$M(\rho)$$ in its matrix components we get $$M(\rho)_{ij} = \sum_{a,n,m} \calU_{i a}^{n k}(\calU^*)_{j a}^{m k} \rho_{nm}.$$ This gives you the Kraus representation $$M(\rho)=\sum_a A_a^{(k)}\rho A_a^{(k)\dagger}$$ with $$(A_a^{(k)})_{i,n}\equiv \calU_{ia}^{nk}$$.

Our hypothesis is that, for some unitary $$\calV$$, we have $$M(\rho)=\calV\rho\calV^\dagger$$ for all $$\rho$$. This would then imply $$\calV \rho\calV^\dagger = \sum_a A_a^{(k)}\rho A_a^{(k)\dagger}\quad\forall\rho,$$ This, in turn, implies that $$A_a^{(k)}=C_{a}^{(k)}\calV$$ with $$C$$ such that $$\sum_a |C_a^{(k)}|^2=1$$. This follows from the fact that if $$\sum_a A_a\rho A_a^\dagger=\sum_a B_a \rho B_a^\dagger$$ for all $$\rho$$ then for some unitary $$C$$ we have $$A_a=\sum_b C_{ab}B_b$$ (which in turn is a direct application of the SVD decomposition). If $$B_b=\delta_{b0}\calV$$ we get the result.

We thus proved that $$(A_a^{(k)})_{i,n}=\calU_{ia}^{nk}=C_a^{(k)}\calV_{i}^n$$. This is essentially the conclusion: it means that $$\calU=\calV\otimes \tilde C$$ with $$\tilde C$$ a unitary whose first column (or row, depending on the convention we are using) equals $$(C_a^{(k)})_a$$ (it can be any such unitary, as $$\calU$$ is not fully defined by the definition of $$M$$).