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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.

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2 Answers 2

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  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 ${\rm range}(U)={\rm supp}(\rho)$. Note that this also means that $\sqrt\rho UU^\dagger\sqrt\rho=\rho$, i.e. $U^\dagger$ acts as an isometry at least on the support of $\rho$. In fact, we could simply assume $UU^\dagger=I$ without losing anything, because the action of $U^\dagger$ outside of the support of $\rho$ is never relevant to this discussion. Hence why the result can be perhaps more simply stated saying that $\psi_\rho=\sqrt\rho V^\dagger$ for some isometry $V$.

    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 $$|\psi_\rho\rangle = \sum_i \sqrt{p_i} |u_i, v_i\rangle$$ 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.

Toy example

Purifications of $\rho$ Let's consider a concrete toy example. Say $\rho=I/2$ be a maximally mixed qubit state. Its purifications are then states of the form $\Psi=\frac12 V^\dagger$ for isometries $V$. Examples of such purifications might be $$|\Psi_1\rangle= \frac1{\sqrt2} (|00\rangle+|11\rangle), \quad\text{using}\quad V_1=I, \\ |\Psi_2\rangle = \frac1{\sqrt2}(|01\rangle+|10\rangle), \quad\text{using}\quad V_2=X, \\ |\Psi_3\rangle = \frac{|0\rangle(|0\rangle+|1\rangle+|2\rangle) +|1\rangle(|0\rangle+\omega_3|1\rangle+\omega_3^2|2\rangle}{\sqrt6}, \, V_3=\frac1{\sqrt3}\begin{pmatrix}1&1\\1&\omega_3\\1&\omega_3^2\end{pmatrix}, $$ where $\omega_3\equiv\exp(2\pi i/3)$. Note that the support of all these $V$ is the same as that of $\rho$. And you can always assume this without loss of generality.

Purifications of $\sigma$ Let's say $\sigma=\operatorname{diag}(1/3,2/3)$. Using the same isometies as above would then give the purifications $$|\Phi_1\rangle= \frac1{\sqrt3} (|00\rangle+\sqrt2|11\rangle), \quad\text{using}\quad V_1=I, \\ |\Phi_2\rangle = \frac1{\sqrt3}(|01\rangle+\sqrt2|10\rangle), \quad\text{using}\quad V_2=X, \\ |\Phi_3\rangle = \frac{|0\rangle(|0\rangle+|1\rangle+|2\rangle) +\sqrt2|1\rangle(|0\rangle+\omega_3|1\rangle+\omega_3^2|2\rangle}{3}, \, V_3=\frac1{\sqrt3}\begin{pmatrix}1&1\\1&\omega_3\\1&\omega_3^2\end{pmatrix}. $$

Inner products of purifications — Let's try something like $$\langle\Psi_1|\Phi_2\rangle = \operatorname{tr}[(\sqrt\rho V_1^\dagger)^\dagger\sqrt\sigma V_2^\dagger] = \frac{1}{\sqrt2}\operatorname{tr}[\sqrt\sigma X]=0.$$ Well, that obviously isn't what we expect, because we know the fidelity between $\rho$ and $\sigma$ isn't zero. If we instead try using $V_1$ for both states, we get instead $$\langle\Psi_1|\Phi_1\rangle = \operatorname{tr}[\sqrt\rho\sqrt\sigma] = \frac{1+\sqrt2}{\sqrt6}.$$ And this is the maximum because it's trivial to check that in this case $\sqrt\rho\sqrt\sigma=|\sqrt\rho\sqrt\sigma|$. Note that we could have equivalently used the other purifications, as $\langle\Psi_1|\Phi_1\rangle=\langle\Psi_2|\Phi_2\rangle=\langle\Psi_3|\Phi_3\rangle$. This happens because, as discussed before, the only thing that matters to the overlaps is the product $V_1^\dagger V_2$ for the two isometries.

Slightly less toyish examples

Example with less trivial states — The example above is kinda trivial because $\rho$ and $\sigma$ commute (and in fact, $\rho$ commutes with everything). For a less trivial analysis consider $$ \rho' \equiv p \mathbb{P}_0 + (1-p) \mathbb{P}_1,\,\, p\neq1/2,\qquad \sigma' \equiv q \mathbb{P}_+ + (1-q)\mathbb{P}_-,\,\, q\neq1/2, $$ using the notation $\mathbb{P}_\psi\equiv |\psi\rangle\!\langle\psi|$. These ensure $[\rho',\sigma']\neq0$. Defining purifications in analogy of the calculations above in terms of the isometries $V_i$, we have for example $$\langle\Psi_1|\Phi_1\rangle = \operatorname{tr}[\sqrt{\rho'}\sqrt{\sigma'}]= \frac{(\sqrt p+\sqrt{1-p})(\sqrt q+\sqrt{1-q})}{2},$$ but this time this isn't the max overlap. After some calculation, you indeed find that $$\operatorname{tr}|\sqrt{\rho'}\sqrt{\sigma'}| =\frac{\sqrt{1+\sqrt{u}}+\sqrt{1-\sqrt{u}}}{2}, \quad u\equiv 1-16 pq(1-p)(1-q),$$ and you can verify numerically that this is always larger than the other result. I couldn't find any simple expression for the purifications/isometries that produce this fidelity.

Pure states — If $\rho=\mathbb{P}_\psi$ and $\sigma=\mathbb{P}_\phi$ are pure states, we of course know that the fidelity is just $|\langle\psi|\phi\rangle|$. It might be tempting to think that we get this following the argument above by simply having the unitary as $U=I$, and thus $F=\operatorname{tr}[\sqrt\rho\sqrt\sigma]$. This is however not so: $$\operatorname{tr}[\sqrt\rho\sqrt\sigma]=|\langle\psi|\phi\rangle|^2,$$ and we should note that the square here is a problem, because the formula $\operatorname{tr}|\sqrt\rho\sqrt\sigma|$ is for the "square root" fidelity. You can also see that this is wrong because the result is trivially reachable reasoning in terms of purifications: the "purifications" of $\mathbb{P}_\psi$ and $\mathbb{P}_\phi$ are just $|\psi\rangle$ and $|\phi\rangle$, and thus their overlap is just $|\langle\psi|\phi\rangle|$. The apparent discrepancy happens because for pure states $|\sqrt\rho\sqrt\sigma|\neq \sqrt\rho\sqrt\sigma$. Instead, one has $$|\sqrt\rho\sqrt\sigma|=|\mathbb{P}_\psi\mathbb{P}_\phi| = |\langle\psi|\phi\rangle| \sqrt{\mathbb{P}_\psi} = |\langle\psi|\phi\rangle| \mathbb{P}_\psi,$$ whose trace is then clearly just $|\langle\psi|\phi\rangle|$, as expected.

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  • $\begingroup$ The standard matrix inequality quoted only works when $VU^\dagger$ is unitary, but here, it's only a composition of two partial isometries. $\endgroup$ Commented Jun 2, 2021 at 23:55
  • $\begingroup$ @MaudPieTheRocktorate that's true, but you can generalise that maximisation result to work on any operator with $\|U\|_{\rm op}\le 1$. I edited quantumcomputing.stackexchange.com/a/6141/55 to add the proof for the generalised inequality $\endgroup$
    – glS
    Commented Jun 5, 2021 at 15:18
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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 $A\simeq B$.

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