How to prove generalized Uhlmann's theorem?

Question

I think the Uhlmann theorem should be in general of this form:

Let $\rho$ and $\sigma$ be density operators acting on $A$, with Schmidt degrees at most $r$, and let $B$ be another Hilbert space with dimension at least $r$, so that $\rho, \sigma$ can both be purified over $B$.

Let $\left|\psi_{\rho}\right\rangle, \left|\psi_{\sigma}\right\rangle$ denote arbitrary purifications of them over $A \otimes B$, then $$ F(\rho, \sigma)=\max _{\left|\psi_{\rho}\right\rangle, \left|\psi_{\sigma}\right\rangle}\left|\left\langle\psi_{\rho} \mid \psi_{\sigma}\right\rangle\right|^{2} $$ Therefore, in general, the fidelity is the maximum overlap between purifications.

In all the proofs I found (such as the one in Nielson and Chuang), the Uhlmann theorem is proved only for the case where $A\simeq B$. How to prove this general case?

Edit: In Fidelity for Mixed Quantum States (Richard Jozsa, 1993), Uhlmann theorem is proved in this general case (as can be seen from Definition 1 and Theorem 2). However, I find the proof dubious, and would like another reference on this.

Answer 1

1. Let $|\psi_\rho\rangle\in\mathcal S(\mathcal X\otimes\mathcal Y)$ be a purification of $\rho\in\mathrm D(\mathcal X)$. This means that $$\operatorname{Tr}_{\mathcal Y}|\psi_\rho\rangle\!\langle\psi_\rho|=\rho \Longleftrightarrow \psi_\rho\psi_\rho^\dagger=\rho \Longleftrightarrow \psi_\rho=\sqrt\rho U,$$ where $\psi_\rho\in\mathrm{Lin}(\mathcal Y,\mathcal X)$ is the linear operator with matrix elements $(\psi_\rho)_{ij}=\langle i,j|\psi_\rho\rangle$, and in the last step we took the polar decomposition of $\psi_\rho$, so that $U\in\mathrm U(\mathcal Y,\mathcal X)$ is any partial isometry such that ${\rm range}(U)={\rm supp}(\rho)$.

   You can equivalently write the above saying that the purifications of $\rho$ have the form $$|\psi_\rho\rangle = \operatorname{vec}(\sqrt\rho U)$$ for any $\mathcal Y$ such that $\dim(\mathcal Y)\ge\operatorname{rank}(\rho)$, and any partial isometry $U\in\mathrm U(\mathcal Y,\mathcal X)$ with range equal to that of $\rho$.

   I personally think the easier way to put this is to say that, if $\rho$ has eigendecomposition $\rho=\sum_i p_i|u_i\rangle\!\langle u_i|$, then its purifications are all and only the vectors of the form $$|\psi_\rho\rangle = \sum_i \sqrt{p_i} |u_i\rangle\!\langle v_i|$$ for any orthonormal set of vectors $\{|v_j\rangle\}_j$.

2. Note that for this to be possible, you need $\operatorname{rank}(\psi_\rho)=\operatorname{rank}(\rho)$, which is only possible if $\dim(\mathcal Y)\ge\operatorname{rank}(\rho)$. This is the only assumption that needs to be made on $\mathcal Y$.

3. Now, back to Uhlmann. Clearly, to take an inner product between purifications, you need them to be defined in spaces of the same dimensions. All finite-dimensional spaces are isomorphic, so it's not really meaningful to ask anything about the structure of such spaces aside from their dimensions. We can thus take some $|\psi_\rho\rangle$ and $|\psi_\sigma\rangle$ defined over spaces of the same dimensions, and then notice that $$\langle \psi_\rho,\psi_\sigma\rangle = \operatorname{Tr}(VU^\dagger \sqrt\rho\sqrt\sigma),$$ for some $U,V\in\mathrm U(\mathcal Y,\mathcal X)$, for some $\mathcal Y$ with $\dim(\mathcal Y)\ge \min(\operatorname{rank}(\rho),\operatorname{rank}(\sigma))$. The conclusion follows from here using standard matrix inequalities, discussed e.g. here. More specifically, we get $$\lvert \langle \psi_\rho,\psi_\sigma\rangle\rvert \le \|VU^\dagger\|_{\rm op}{\rm Tr}|\sqrt\rho\sqrt\sigma| \le {\rm Tr}|\sqrt\rho\sqrt\sigma|,$$ where the maximum is achieved when $VU^\dagger$ are unitary. Note that here $U,V$ are defined as surjective partial isometries, and thus $U^\dagger$ is an isometry. Their product is therefore also an isometry (and also a unitary, if the purification spaces of $\rho$ and $\sigma$ coincide, which we can however always assume without loss of generality) iff $U$ and $V$ have identical support.

Answer 2

To define the inner product between $|\psi_\rho\rangle$ and $|\psi_\sigma\rangle$, the two vectors must exist in the same Hilbert space. If one purification exists in a Hilbert space smaller than the other, the smaller one can be trivially extended to the Hilbert space of the larger one by taking a tensor product with any other pure state in the extra dimensions of the larger Hilbert space.

Now, why must the purified space be isomorphic to the unpurified space? If it wasn't, the overlap would not be maximized. In general, a purified state for $\rho$ acting on $\mathbb{C}^n$ is written as $$|\psi_\rho\rangle=\sum_{i=1}^n\rho^{1/2}|e_i\rangle\otimes|f_i\rangle,$$ where $|e_i\rangle$ form an orthonormal basis of $\mathbb{C}^n$. In general, the $n$ orthonormal states $|f_i\rangle$ may exist in any vector space $\mathbb{C}^m$ with $m\geq n$, but they'll always only span an $n$-dimensional subspace of any larger vector space. We might purify another state $\sigma$ using another set of $n$ orthonormal states $|g_i\rangle$ that form another $n$-dimensional subspace of $\mathbb{C}^m$: $$|\psi_\sigma\rangle=\sum_{i=1}^n\sigma^{1/2}|e_i\rangle\otimes|g_i\rangle.$$ The maximal overlap between $|\psi_\rho\rangle$ and $|\psi_\sigma\rangle$ is found when $$\mathrm{Span}\left(\{|f_i\rangle\}\right)=\mathrm{Span}\left(\{|g_i\rangle\}\right),$$ so we might as well choose $\mathbb{C}^m=\mathbb{C}^n$, so the optimal overlap always has $A\simeq B$.