# Prove that a change-of-basis map is an isometry between complex hilbert spaces, to prove uniqueness of purifications

I am trying to prove the following

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?

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.