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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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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 $r:={\rm rank}(\omega)<m$. 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. $\square$

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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 $A$.

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  • $\begingroup$ 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

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