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

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{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}

So then we can right them as follows

\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}

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

\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}

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

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

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

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

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

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

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

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

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

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 $$e^{i\theta}V^\star V'$$.

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 $$|\psi\rangle=(I\otimes U^T)|\phi\rangle$$.