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.

  • $\begingroup$ crossposted to pse: physics.stackexchange.com/questions/548847/… $\endgroup$ May 3, 2020 at 17:20
  • $\begingroup$ 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? $\endgroup$ May 3, 2020 at 17:20
  • 1
    $\begingroup$ 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? $\endgroup$ May 4, 2020 at 0:30
  • $\begingroup$ @JohnWatrous could you give an example of that (if I understand you correctly and you are saying that that is possible)? $\endgroup$
    – glS
    May 4, 2020 at 16:34
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    $\begingroup$ 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 $U_1$. $\endgroup$ May 4, 2020 at 23:01

2 Answers 2


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 $A$.

  • $\begingroup$ Yes, I assumed $\rho_2 = |0\rangle\langle0|$ for simplicity. $\endgroup$
    – Danylo Y
    May 4, 2020 at 16:41
  • $\begingroup$ 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) $\endgroup$
    – glS
    May 4, 2020 at 16:44
  • $\begingroup$ 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 $p_k = \delta_{k0}$. $\endgroup$
    – Danylo Y
    May 4, 2020 at 17:20
  • $\begingroup$ 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 $U$. $\endgroup$
    – Danylo Y
    May 4, 2020 at 17:25
  • $\begingroup$ 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 $U$? $\endgroup$
    – glS
    May 4, 2020 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 $M$).


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