1
$\begingroup$

I am trying to prove the following

enter image description here

For this I should use the following fact:

Let $ρ_A = \sum_{i=1}^r p_i|e_i⟩⟨e_i|$, where $p_i$ are the nonzero eigenvalues of $ρ_A$ and $|e_i⟩$ corresponding orthonormal eigenvectors Then,any purification $|\psi_{AB}⟩$ of $ρ_A$ has a Schmidt decomposition of

the form $|\psi_{AB}⟩ = \sum_{i=1}^r s_i|e_i⟩ ⊗ |f_i⟩$

(i.e with the same $|e_i⟩$ as in $ρ_A$)

Using this fact I can write indeed

$|\psi_{AB}⟩ = \sum_{i=1}^r s_i|e_i⟩ ⊗ |f_i⟩$

and

$|\phi_{AC}⟩ = \sum_{i=1}^r \tilde s_i|e_i⟩ ⊗ |\tilde f_i⟩$

(I am not sure if $ \tilde s_i=s_i $)

Then I thought I could define a linear map: $V_{B \to C}:H_B\to H_C: v \mapsto Dv$

where C is the change of basis matrix. If I can prove that this is an isometry I am practically done. So I have to prove that: $ V_{B \to C}^\dagger V_{B \to C}=I_{H_B}$

This is where I am having trouble. Is a the linear map given by the change-of-basis matrix between complex (finite dimensional) Hilbert spaces given by a hermitian matrix? If so , how do I prove it ? If not, it looks like this definition or the approach at all does not work, how do I prove the isometry then?

$\endgroup$

1 Answer 1

0
$\begingroup$

As also discussed e.g. in this other answer, the set of purifications of a state $\rho$ with eigendecomposition $\rho=\sum_i p_i \mathbf e_i\mathbf e_i^\dagger$ can be characterised as the set of bipartite vectors of the form $$\boldsymbol\Psi = \sum_k \sqrt{p_k} \mathbf e_k\otimes \mathbf u_k\tag1$$ for any orthonormal set of vectors $\mathbf u_k$ living in some ancillary space.

This follows e.g. from the fact that $\rho=\sqrt\rho\sqrt\rho=AA^\dagger$ implies that $A$ has the same singular values as $\sqrt\rho$, and the same associated left principal components. In other words, the only freedom is in the choice of right principal components. This connects to purifications because $\boldsymbol\Psi$ being a purification of $\rho$ is the same as saying $\rho=M_{\boldsymbol\Psi}M_{\boldsymbol\Psi}^\dagger$, with $M_{\boldsymbol\Psi}$ the matrix obtained "devectorising" $\boldsymbol\Psi$, that is, $M_{\boldsymbol\Psi}\equiv \sum_k\sqrt{p_k}\mathbf e_k\mathbf u_k^T$.

In synthesis, $\boldsymbol\Psi$ and $\boldsymbol\Phi$ being purifications of $\rho$ means they read $$\boldsymbol\Psi=\sum_k\sqrt{p_k} \mathbf e_k\otimes \mathbf u_k, \qquad \boldsymbol\Phi=\sum_k\sqrt{p_k} \mathbf e_k\otimes \mathbf v_k,\tag2$$ for some pair of orthonormal sets $\{\mathbf u_k\}_k$ and $\{\mathbf v_k\}_k$. Note that $\mathbf u_k$ and $\mathbf v_k$ need not, and do not, in general belong to spaces of the same dimension. We can therefore always connect these purifications as $\boldsymbol\Psi=(I\otimes V)\boldsymbol\Phi$ with $V\equiv \sum_k \mathbf u_k \mathbf v_k^\dagger$. If furthermore $\Psi$ uses a larger purification space, that is, $\mathbf u_k$ live in a larger space than $\mathbf v_k$, then $V$ is an isometry.

For example, as discussed in the answer linked above, three purifications for the state $$\rho = \begin{pmatrix}2/3&0\\0&1/3\end{pmatrix}$$ are (I'll omit the $\otimes$ symbols for brevity): $$\boldsymbol\Psi_1 \equiv \sqrt{\frac23}\mathbf e_0\mathbf e_0 + \sqrt{\frac13}\mathbf e_1\mathbf e_1,\\ \boldsymbol\Psi_2 \equiv \sqrt{\frac13}\, \mathbf e_0\mathbf e_0 + \sqrt{\frac13}\, \left(\frac{\mathbf e_0+\mathbf e_1}{\sqrt2}\right)\mathbf e_1 + \sqrt{\frac13}\, \left(\frac{\mathbf e_0-\mathbf e_1}{\sqrt2}\right)\mathbf e_2, \\ \boldsymbol\Psi_3 \equiv \sqrt{\frac13}\, \left(\frac{\sqrt2 \mathbf e_0+\mathbf e_1}{\sqrt3}\right)\mathbf e_0 + \sqrt{\frac13}\, \left(\frac{\sqrt2 \mathbf e_0+\omega_3\mathbf e_1}{\sqrt3}\right)\mathbf e_1 + \sqrt{\frac13}\, \left(\frac{\sqrt2 \mathbf e_0+\omega_3^2\mathbf e_1}{\sqrt3}\right)\mathbf e_2. $$ Although these look different from (2), they can be rearranged to look like that, by simply collecting suitable factors in the first space. Explicitly, $\boldsymbol\Psi_1$ is already in the correct form, while the other ones can be rewritten as $$\boldsymbol\Psi_2 = \sqrt{\frac23} \mathbf e_0\left( \frac{\sqrt2 \mathbf e_0 + \mathbf e_1+\mathbf e_2}{2} \right) + \sqrt{\frac13} \mathbf e_1\left( \frac{\mathbf e_1 - \mathbf e_2}{\sqrt2} \right), \\ \boldsymbol\Psi_3 = \sqrt{\frac23} \mathbf e_0\left( \frac{\mathbf e_0 + \mathbf e_1+\mathbf e_2}{\sqrt3} \right) + \sqrt{\frac13} \mathbf e_1\left( \frac{\mathbf e_0 + \omega_3 \mathbf e_1 + \omega_3^2\mathbf e_2}{\sqrt3} \right).$$ You should be able to see easily from this how you can go between $\boldsymbol\Psi_2$ and $\boldsymbol\Psi_3$ (and the other way around) via a unitary, while you need an isometry (that is not a unitary) to go from $\boldsymbol\Psi_1$ to one of $\boldsymbol\Psi_2$ and $\boldsymbol\Psi_3$. Generally speaking, the dimension of the required ancillary space corresponds to the number of (nontrivially different) elements in the pure state decomposition of the density matrix.

Note that a map of the form $V=\sum_k \mathbf u_k\mathbf v_k^\dagger$ is always an isometry when $\{\mathbf u_k\}$, $\{\mathbf v_k\}$ are orthonormal, because a matrix is an isometry iff it has orthonormal columns, as discussed e.g. in this other answer.

A very similar question was also asked and answered some time ago on qc.SE here.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .