Prove that different purifications of a state can be mapped into one another via local unitaries

Question

Let $\rho \in \mathfrak{D}(A)$ be a density matrix. Show that $\left|\psi^{A B}\right\rangle \in A B$ and $\left|\phi^{A C}\right\rangle \in A C$ (assuming $\left.|B| \leqslant|C|\right)$ are two purifications of $\rho \in \mathfrak{D}(A)$ if and only if there exists an isometry matrix $V: B \rightarrow C$ such that $$ \left|\phi^{A C}\right\rangle=I^{A} \otimes V^{B \rightarrow C}\left|\psi^{A B}\right\rangle $$

$\mathbf{attempt}$:

I first prove $\rightarrow$ side of this theorem. So I assume that $\left|\psi^{A B}\right\rangle \in A B$ and $\left|\phi^{A C}\right\rangle \in A C$ (assuming $\left.|B| \leqslant|C|\right)$ are two purifications of $\rho \in \mathfrak{D}(A)$. We can write $\left|\psi^{A B}\right\rangle $ and $\left|\phi^{A C}\right\rangle $ as follows

\begin{equation} \begin{aligned} &|\psi\rangle^{A B}=\sum_{x=1}^{|A|}|x\rangle^{A}\left(\sum_{y=1}^{|B|} m_{x y}|y\rangle^{B}\right)\\ &|\psi\rangle^{A C}=\sum_{z=1}^{|A|}|z\rangle^{A}\left(\sum_{w=1}^{|C|} m^{\prime}_{z w}|w\rangle^{C}\right) \end{aligned} \end{equation}

So then we can right them as follows

\begin{equation} \begin{aligned} &|\psi\rangle^{A B}=I \otimes M\left|\phi_{+}^{A \tilde{A}}\right\rangle\\ &|\psi\rangle^{A C}=I \otimes M^{\prime}\left|\phi_{+}^{A \tilde{A}}\right\rangle\\ \end{aligned} \end{equation}

Which $M: H^{\tilde{A}} \rightarrow H^{B}$ and $M^{\prime}: H^{\tilde{A}} \rightarrow H^{C}$ and

\begin{equation} \begin{aligned} &M|x\rangle^{\tilde{A}}:=\sum_{y=1}^{|B|} m_{xy}|y\rangle^{B} \quad and \quad M^{\prime}|z\rangle^{\tilde{A}}:=\sum_{w=1}^{|c|} m^{\prime}_{zw}|y\rangle^{C} \\ &\left|\phi_{+}^{\tilde{A} A}\right\rangle:=\sum_{x=1}^{|A|} |xx\rangle^{\tilde{A} A}\\ \end{aligned} \end{equation}

Now we want $|\psi\rangle^{A B}$ and $|\psi\rangle^{A c}$ to be purification of $\rho \in \mathfrak{D}(A)$. So according to the definition, we should have

\begin{equation} \begin{aligned} &\psi^{A}=MM^{*}=M^{\prime}(M^{\prime})^{*}=\rho \\ \end{aligned} \end{equation}

And we assume $M^{\prime}=VM$, So So we can write

\begin{equation} \begin{aligned} &M^{\prime}(M^{\prime})^{*}=VMM^{*}V^{*}=V\rho V^{*} \end{aligned} \end{equation}

Now, what should I do? Is my procedure correct?

$\mathbf{Note}$:

We know that $\left|\phi^{A B}\right\rangle=I^{A} \otimes M \left|\Phi^{A \tilde{A}}\right\rangle$ is called a purification of $\rho$ if reduced density matrix $\psi^A$

\begin{equation} \begin{aligned} &\psi^A := M M^* \in Pos(A) \end{aligned} \end{equation}

is equal to our density matrix $\rho$. And

\begin{equation} \begin{aligned} &\left|\phi_{+}^{\tilde{A} A}\right\rangle:=\sum_{x=1}^{|A|} |xx\rangle^{\tilde{A} A} \end{aligned} \end{equation}

at the end I should mention that $\tilde{A}$ is the same as $A$.

Answer 1

I guess that part of the confusion is that you defined the matrix

$$ M = [m_{x,y}] ,\quad 1 \leq x \leq |A|, 1 \leq y \leq |B| $$

which is actually the transpose of the common matrix representation of a linear operator. This later on means that composition of the linear operators $ V M $ is not the usual matrix multiplication. All this to say, that you should have defined

$$ M|x\rangle^A = \sum_{y=1}^{|B|}m_{yx} |y\rangle^B $$

However, with the above definition we have that $ \hspace{0.2em}|\phi \rangle^{AB} = (I^A \otimes M) \big|\Phi_+^{AA} \big\rangle $ is a purification of $ \rho $ if $$ \rho = \text{Tr}_B\Big[|\phi^{AB} \rangle \langle \phi^{AB}| \Big] = \big( M^{\dagger} M \big)^T $$

After this clarification, let us first prove the reverse direction of the theorem, meaning if $ V^{B\rightarrow C} $ is an isometry and

$$ |\phi^{AC} \rangle = (I^A \otimes V^{B\rightarrow C}) |\phi^{AB} \rangle = (I^A \otimes V^{B\rightarrow C} M ) \big|\Phi_+^{AA} \big\rangle $$

then the reduced density matrices of $ |\phi^{AC} \rangle, |\phi^{AB} \rangle $ are equal. Indeed: $$ \text{Tr}_C\Big[|\phi^{AC} \rangle \langle \phi^{AC}| \Big] = \text{Tr}_C\Big[ (I^A \otimes V^{B\rightarrow C} M ) \big|\Phi_+^{AA} \big\rangle \Big] = \Big( \big(VM \big)^{\dagger} VM \Big)^T = \Big( M^{\dagger} V^{\dagger}V M \Big)^T = \Big( M^{\dagger} M \Big)^T = \text{Tr}_B\Big[|\phi^{AB} \rangle \langle \phi^{AB}| \Big] $$ since $ V $ is an isometry so $ V^{\dagger} V = I_B $.

For the other direction, if $ \hspace{0.2em}|\phi \rangle^{AB} = (I^A \otimes M) \big|\Phi_+^{AA} \big\rangle $ and $ \hspace{0.2em}|\phi \rangle^{AC} = (I^A \otimes M') \big|\Phi_+^{AA} \big\rangle $ are two purifications of $ \rho $, it must hold that $$ \rho^T = M^{\dagger} M = M'^{\dagger} M' $$ So if $ \rho^T = \sum_{j=1}^{r} \lambda_j |x_j \rangle \langle x_j| $ is the eigendecomposition of $ \rho^T $, by the singular value theorem it must hold that

$$ M = \sum_{j=1}^{r} \sqrt{\lambda_j} \cdot |y_j \rangle \langle x_j| $$ $$ M' = \sum_{j=1}^{r} \sqrt{\lambda_j} \cdot |z_j \rangle \langle x_j| $$

for two orthonormal sets $\{ |y_j \rangle \} \in H^B $ and $ \{ |z_j \rangle \} \in H^C $.

Now we can define $ V: H^B \rightarrow H^C $ with $ V |y_j \rangle = |z_j \rangle $ and extend this matrix, if needed, to an isometry (we can always do this). This means that

$$ M' = V M \implies |\phi^{AC} \rangle = (I^A \otimes V) |\phi^{AB} \rangle$$

Answer 2

You seem to have got, pretty successfully (I won't claim to have checked all the fine details), to the point of showing that you need $$ MM^\star=M'(M')^\star=\rho. $$ However, you then assume $M'=VM$. You cannot do this as what you're trying to prove is that the only option is for $M'=VM$.

What you could do is assume a singular value decomposition of both $M$ and $M'$. For example, $$ M=UDV, $$ where $D$ is diagonal (with non-negative entries) and $U$ and $V$ are unitaries. Similarly, $$ M'=U'D'V'. $$ We calculate $MM^\star=UD^2U^\star$, so $D^2$ must correspond to the eigenvalues of $\rho$, the $U$ transforms the computational basis to the eigenbasis of $\rho$.

Compare this to the same calculation for $M'$. We see that $D=D'$ and $U'=e^{i\theta}U$ (I suppose you could get super fussy about degeneracies/multiplicities in the eigenvalues. These won't affect the final outcome because they'll commute with $D^2$, and hence we can absorb into the $V$ instead).

This then proves that the only difference between $M$ and $M'$ is a unitary $e^{i\theta}V^\star V'$.

Answer 3

This is a special case of the following more general statement:

Let $A,B$ be matrices such that $AA^\dagger=BB^\dagger$. Then $A=BU$ for some unitary $U$.

That this is true follows easily looking at the singular value decomposition of the matrices: $AA^\dagger=BB^\dagger$ implies that $A$ and $B$ have the same singular values and same left singular vectors, therefore their SVDs have the form $$A=\sum_k s_k |u_k\rangle\!\langle v_k|, \qquad B = \sum_k s_k |u_k\rangle\!\langle w_k|.$$ We then get the conclusion by simply choosing $U\equiv\sum_k |w_k\rangle\!\langle v_k|$.

To see why this statement is relevant to the original statement about states, notice that if $(\psi_{ij})_{ij}$ is the matrix of coefficients of a bipartite state $|\psi\rangle$, then $\operatorname{Tr}_2(|\psi\rangle\!\langle\psi|)=\psi\psi^\dagger$. Therefore $\operatorname{Tr}_2(|\psi\rangle\!\langle\psi|)=\operatorname{Tr}_2(|\phi\rangle\!\langle\phi|)$ implies $\psi=\phi U$, which is equivalent to $|\psi\rangle=(I\otimes U^T)|\phi\rangle$.