# $M(\rho)=\operatorname{Tr}_2[U(\rho\otimes\rho_2)U^{\dagger}]$ 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/… May 3, 2020 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? May 3, 2020 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? May 4, 2020 at 0:30
• @JohnWatrous could you give an example of that (if I understand you correctly and you are saying that that is possible)?
– glS
May 4, 2020 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 $U_1$. May 4, 2020 at 23:01

To add to the other nice answers as well as John Watrous' great counterexample: Interestingly one can characterize when "$$\operatorname{Tr}_2(U(\rho\otimes\omega)U^{\dagger})$$ is a unitary channel" is guaranteed to imply "$$U$$ is of the form $$U_1\otimes U_2$$":

Theorem. Given any $$n\geq 2$$, $$m\in\mathbb N$$ and any density matrix $$\omega\in\mathbb C^{m\times m}$$ the following statements are equivalent.

1. $${\rm rank}(\omega)=m$$
2. If $$\operatorname{Tr}_2(U((\cdot)\otimes\omega)U^{\dagger})=W(\cdot)W^\dagger\tag{1}$$ for any unitaries $$U\in\mathbb C^{mn\times mn}$$, $$W\in\mathbb C^{n\times n}$$, then $$U=W\otimes\tilde W$$ for some unitary $$\tilde W\in\mathbb C^{m\times m}$$.

Proof. 2. $$\Rightarrow$$ 1. By way of contradiction assume $$\omega$$ is not full rank, that is, $$m\geq 2$$ and $$r:={\rm rank}(\omega). Now what we have to do is construct unitaries $$U,W$$ such that (1) holds but $$U$$ is not of product form (hence 2. cannot hold for all $$U,W$$ meaning it must be false, as desired). The idea will be to generalize John Watrous' counterexample from the above comments. More precisely, we diagonalize $$\omega=\sum_{k=1}^r \gamma_k|g_k\rangle\langle g_k|$$ for some $$\gamma_k>0$$, $$\sum_k\gamma_k=1$$ and some orthonormal system $$\{g_k\}_{k=1}^r$$ in $$\mathbb C^m$$ which can be extended to an orthonormal basis $$\{g_k\}_{k=1}^m$$, and we define $$U:=W\otimes(\sum_{k=1}^r|g_k\rangle\langle g_k|)+V\otimes (\sum_{k=r+1}^m|g_k\rangle\langle g_k|)$$ for some unitary $$V\in\mathbb C^{n\times n}$$ which is linearly independent of $$W$$ (this is always possible because $$n\geq 2$$). From this it is easy to see that $$U$$ cannot be of the desired product form.

$${}$$1. $$\Rightarrow$$ 2. The rough idea here is that if $$\omega$$ is full rank then "all information of $$U$$ gets used" so there are no detached entries of $$U$$ which could prevent the desired product structure. The mathematical tool we will need are the Kraus operators of the partial trace and of the extension map $$(\cdot)\otimes\omega$$. Defining $$E_k:={\bf1}\otimes|g_k\rangle$$ (i.e. $$E_k|x\rangle=|x\rangle\otimes|g_k\rangle$$) for all $$k$$ it holds that $${\rm Tr}_2=\sum_{k=1}^mE_k^\dagger(\cdot)E_k\quad\text{ and }\quad(\cdot)\otimes\omega=\sum_{k=1}^m (\sqrt{\gamma_k}E_k)(\cdot)(\sqrt{\gamma_k}E_k)^\dagger\,.\tag{2}$$ Substituting (2) into (1) shows that $$\{\sqrt{\gamma_k}E_j^\dagger UE_k\}_{j,k=1}^m$$ is a set of Kraus operators of $$\operatorname{Tr}_2(U((\cdot)\otimes\omega)U^{\dagger})$$. But we already had a set of Kraus operators of this channel: $$\{W\}$$. Thus by unitary equivalence of different Kraus representations (see also Corollary 2.23 in Watrous' book) there exists $$z\in\mathbb C^{m^2}\simeq\mathbb C^{m\times m}$$ with $$\|z\|=1$$ such that $$\sqrt{\gamma_k}E_j^\dagger UE_k=z_{jk}W$$ for all $$j,k=1,\ldots,m$$. By assumption all $$\gamma_k>0$$ so this is equivalent to $$E_j^\dagger UE_k=\gamma_k^{-1/2}z_{jk}W$$ for all $$j,k$$. The final piece of the puzzle is the identity $$\sum_{j=1}^mE_jE_j^\dagger={\bf1}$$ (see, e.g., Lemma C.3 in this paper -- arXiv version) because it lets us compute \begin{align*} U&=\Big(\sum_{j=1}^mE_jE_j^\dagger\Big)U\Big(\sum_{k=1}^mE_kE_k^\dagger\Big)\\ &=\sum_{j,k=1}^mE_j(E_j^\dagger UE_k)E_k^\dagger\\ &=\sum_{j,k=1}^mE_j^\dagger UE_k\otimes |g_j\rangle\langle g_k|\\ &=\sum_{j,k=1}^m\gamma_k^{-1/2}z_{jk}W\otimes |g_j\rangle\langle g_k|\\ &=W\otimes \Big(\sum_{j,k=1}^m\gamma_k^{-1/2}z_{jk}|g_j\rangle\langle g_k|\Big)\,, \end{align*} that is, $$U=W\otimes\tilde W$$ for some $$\tilde W\in\mathbb C^{m\times m}$$. Because $$U,W$$ are unitary $$\tilde W$$ has to be unitary as well. This concludes the proof. $$\square$$

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. May 4, 2020 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, 2020 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 $p_k = \delta_{k0}$. May 4, 2020 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 $U$. May 4, 2020 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, 2020 at 17:27