# Intuitive role of the polar decomposition in proof of Uhlmann's theorem for fidelity

I have read the Wikipedia article which relates the polar decomposition to a complex number being split into its modulus and phase but this analogy isn't very intuitive to me.

In Nielsen and Chuang, the polar decomposition is especially important in proving results around fidelity. If I have states $$\rho \in H_{R}$$ and $$\sigma \in H_{Q}$$, Uhlmann's theorem shows that the best purification that preserves fidelity $$F(\rho, \sigma)$$ is some state in $$R\otimes Q$$ and gives us a method to find it. Set state $$m = \sum_i\vert i_R\rangle\vert i_Q\rangle$$. Now

$$\vert\psi\rangle = 1\otimes \sqrt{\rho}\vert m\rangle$$

$$\vert\phi\rangle = 1\otimes \sqrt{\sigma}V^{\dagger}\vert m\rangle$$

gives the purification that preserves fidelity. Here, the polar decomposition of $$\sqrt{\rho}\sqrt{\sigma}$$ is $$|\sqrt{\rho}\sqrt{\sigma}|V$$.

While I can follow the proof in Nielsen and Chuang, I have no intuition about what the polar decomposition is doing here and why this particular purification preserves fidelity.

I'm looking for intuition only, so proofs are unimportant. Could someone help with simple example states and how they transform under different purifications and why the polar decomposition-based purification works best?

• I was already going to write an answer over on physics, but then I realized I had no idea what you were looking for. In the proof, $\sqrt{\rho}\sqrt{\sigma}V^\dagger$ shows up, and you want it to be as large as possible, so it should be the absolute value of $\sqrt{\rho}\sqrt{\sigma}$. This is achieved by the $V$ in the polar decomposition. --- But this is the proof. What are you looking for instead? (EDIT: I see you added a request for examples (which I didn't remember from physics) -- which examples did you try?) – Norbert Schuch Nov 10 '18 at 11:22
• @NorbertSchuch I think the point is whether there is any intuition as to why should the polar decomposition be important here. It is also something I personally wondered more than once. I don't know if that's the same with the OP, but I for example would love some physical intuition regarding the role of the singular vectors of $\sqrt\rho\sqrt\sigma$. This because understanding what the polar decomposition is doing is essentially equivalent to understanding what the SVD is doing. – glS Nov 11 '18 at 21:20
• Indeed, I believe a partial answer to the question would be to say that the polar decomposition is not really the important point here. The unitary $V$ is essentially just a mathematical gimmick to select the singular values of $\sqrt\rho\sqrt\sigma$. But the question then becomes why are the singular values of $\sqrt\rho\sqrt\sigma$ so well-fit to measure the indistinguishability or $\rho$ and $\sigma$? I don't know of any "nice" expression of these singular values in terms of properties of $\rho$ and $\sigma$. – glS Nov 11 '18 at 21:23
• @glS My problem is that what I consider intuitive is still a mathematical explanation -- maybe that's what you (or the OP) find intuitive, maybe not. --- To me, the point here is to align the bases of the purifying system so that $\sqrt{\rho}\sqrt{\sigma}$ (essentially the overlap) has positive eigenvalues, thus maximize the overlap. Is this intuitive? No idea. But that's what the polar decomposition does -- it decomposes an operator into a positive part -- a distortion/stretching/boost/..., if you wish -- times a basis rotation. – Norbert Schuch Nov 11 '18 at 21:56
• @NorbertSchuch obviously intuition is somewhat subjective, but I think anything elucidating in any way why the theorem holds can be useful. For example, it would be great if there was a way to understand physically what the singular values of $\sqrt\rho\sqrt\sigma$ represent, that could explain why they are a useful measure of indistinguishability. – glS Nov 12 '18 at 1:55

(I will give the argument with formulas for now, hopefully I find time for some pictures later.)

Let $$|m\rangle$$ be the (unnormalized) maximally entangled state. Then, a purification of $$\rho$$ is given by $$|\rho\rangle_{AB}=(\sqrt{\rho}_A\otimes1\!\!1_B)|m\rangle\ ,$$ and correspondingly for $$\sigma$$ -- this can be seen most easily by first tracing the $$B$$ system from $$|m\rangle$$, which gives $$1\!\!1$$, which implies that the partial trace of $$|\rho\rangle_{AB}$$ is $$\sqrt{\rho}1\!\!1\sqrt{\rho}=\rho$$.

Now there is a unitary degree of freedom in the purification -- $$B$$ can apply any unitary and it is the same purification (as $$B$$ is being traced anyway). Let's use it for the purification of $$\sigma$$: $$|\sigma\rangle = (\sqrt{\sigma}\otimes U)|m\rangle\ .$$

Now what is the overlap of the two states? It is $$\langle\rho|\sigma\rangle = \langle m| \sqrt{\rho}\sqrt{\sigma}\otimes U|m\rangle\ .$$ It is not straightforward to work out that this equals $$\langle\rho|\sigma\rangle = \mathrm{tr}[\sqrt{\rho}\sqrt{\sigma}U^T]\ .$$ Now we want this quantity to be maximal. When is it maximal? It is maximal if $$\sqrt{\rho}\sqrt{\sigma}U^T = |\sqrt{\rho}\sqrt{\sigma}|$$, i.e., $$\bar U$$ is given by the polar decomposition of $$\sqrt{\rho}\sqrt{\sigma}$$. (Here -- and in the question -- $$|X|$$ denotes the positive part of the polar decomposition of $$X$$, which is $$\sqrt{X^\dagger X}$$, different from the absolute value function defined on the eigenvalues.)

(The last statement holds since $$|\mathrm{tr}[XU]|\le \|X\|_1\|U\|_\infty = \|X\|_1$$ -- this is Hölder's inequality -- where $$\|X\|_1=\mathrm{tr}|X|$$ is the sum of the singular values. Alternatively (following @glS) one can use the SVD of $$X$$, $$X=\sum \sigma_i|s_i\rangle\langle q_i|$$: $$|\mathrm{tr}[XU]|=|\sum \sigma_i\langle q_i|U|s_i\rangle| \le \sum \sigma_i|\langle q_i|U|s_i\rangle| \le \sum \sigma_i\ ,$$ where we have used that $$|\langle q_i|U|s_i\rangle|\le1$$, since $$||q_i\rangle|=1$$ and $$|U|s_i\rangle|=||s_i\rangle|=1$$.)

• This really helped! Thank you. I'm leaving it open for a bit since you and gIs are discussing the graphical notation but I think for me, this answer already gave me the intuition I was missing. – user1936752 Nov 12 '18 at 23:15
• @user1936752 Happy to hear that! Which part was more relevant - the first or the second? – Norbert Schuch Nov 12 '18 at 23:23
• The first part. I'm trying to work out the bit where the inner product over |m> is replaced with the trace but aside from that, it's much clearer – user1936752 Nov 12 '18 at 23:27
• @user1936752 Next, transpose U. Then, write the trace formula and compare. – Norbert Schuch Nov 13 '18 at 11:32
• @user1936752 by the way, that inequality is also the object of this other question on physics.SE. @ Norbert you mentioned graphical notation.. did you mean as a way to show the equality between expval and trace (like I did in the linked answer), or can you actually visualize the last bit of this argument in diagrammatic notation? – glS Nov 14 '18 at 19:25

Purifications play an important role in the theory of density matrices (or more generally quantum states) because they provide a geometric tool in the explanation and description algebraic relations.

(I'll be following here Bengtsson and Życzkowski's reasoning in derivation of the fidelity formula (section 9.4)).

A positive $$N \times N$$ matrix $$\rho$$ can be (nonuniquely) written as: $$\rho = A^{\dagger} A$$ Where $$A$$ is an $$N \times M$$ matrix. The matrix $$A$$ can be viewed as a vector in $$\mathbb{C}^N \otimes \mathbb{C}^M$$, i.e. a purification. Of course, the purification is not unique and any matrix of the form: $$A' = A U$$ where $$U$$ is an $$M\times M$$ unitary matrix, is an equally good purification.

Now, in physics, distance functions between points in various spaces can be frequently obtained as the solutions of variational principles, or optimization problems. This is the method adopted by Bengtsson and Życzkowski to define the fidelity function:

Since the multiplication by a unitary matrix $$A' = A U$$ doesn't change the matrix norm, the various purifications of a density matrix can be viewed as points on a sphere, given purifications of two density matrices: $$\rho_1 = A_1^{\dagger} A_1$$ and $$\rho_2 = A_2^{\dagger} A_2$$, a reasonable distance function can be taken as the spherical geodesic distance between their purification vectors: $$\cos d = \frac{1}{2} \operatorname{Tr}(A_1A_2^{\dagger} + A_2A_1^{\dagger})$$ Where $$d$$ is the angle between the two directions $$A_1$$, $$A_2$$. Now, since the above formula depends on the purification, it is reasonable to define the distance between the density matrices (the fidelity) as the maximum distance as we run over all the purifications: $$F(\rho_1, \rho_2) = \max_{U, V} \frac{1}{2} \operatorname{Tr}(A_1 U V^{\dagger} A_2^{\dagger} + A_2 V U^{\dagger} A_1^{\dagger})$$

The solution of the above optimization problem which depends only on the density matrices and not on the specific purifications(as explicitly done by Bengtsson and Życzkowski) is exactly the fidelity function.

Uhlmann extensively uses purifications in his work; one of the most important objects defined by him using purifications is the Uhlmann's phase which is the generalization of Berry's geometric phase for mixed states (and which has numerous applications in quantum computing). Please see the following review of his.

• a bit late but: when you mention the spherical geodesic distance between purifications, this is essentially a metric in the projective space, right? I mean, $A$ are written as matrices here but they represent ket states upon "vectorization", as far as I can tell – glS Nov 8 '19 at 12:12
• @glS Bengtsson and Życzkowski used two geometric ideas in their construction of the Bures distance: The first was in the definition of the distance denoted in the text by $\cos d …$. This formula is inspired by the geodesic distance of the sphere; if $A_1$ and $A_2$ were unit vectors, then the formula would have been a geodesic distance on the sphere. But nevertheless, when $A_1$ and $A_2$ are matrices, this function still defines a distance satisfying the axioms of a metric space: en.wikipedia.org/wiki/Metric_space. (they leave the responsibility of proving this fact to the reader). – David Bar Moshe Nov 10 '19 at 8:28
• Secondly, they are not satisfied with this distance, because $A_1$ and $A_2$ are not uniquely defined by the given density matrices. They want to run over all the matrices $A_1$ and $A_2$ , corresponding to the two density matrices to find the minimal distance of this type (which will necessarily be a function of the density matrices and not any specific purification). Thus, the variational principle is to be performed on $U$ and $V$ in the last equation written in the text. – David Bar Moshe Nov 10 '19 at 8:30
• Therefore, running over the purifications just means running over the unitary groups on which $U$ and $V$ live, that is the group $U(N) \times U(N)$. Since $U$ and $V$, occur only in the combination $UV^{\dagger}$ and its Hermitian conjugate, this problem has a gauge freedom and many identical minima will be found. This is essentially what Bengtsson and Życzkowski do on the next page in their proof of Uhlmann's fidelity theorem. – David Bar Moshe Nov 10 '19 at 8:31

## Short version

Consider the two following observations:

1. Given a state $$\rho$$, the problem of finding purifications of $$\rho$$ is equivalent to that of finding matrices $$A$$ such that $$\rho=AA^\dagger$$. The purifications of $$\rho$$ are then the vectorisations of these $$A$$ (i.e. a vector $$\Psi$$ is a purification of $$\rho$$ iff its matrix of coefficients $$\Psi_{ij}$$ satisfies $$\Psi\Psi^\dagger=\rho$$). In this notation, the overlap between two purifications $$A,B$$ is written as $$\operatorname{Tr}(A^\dagger B$$).

2. We want to find the value of $$\operatorname{Tr}|\sqrt\rho\sqrt\sigma|$$. As shown in this other answer, this is equivalent to finding the unitary $$U$$ maximising $$\lvert\operatorname{Tr}(U\sqrt\rho\sqrt\sigma)\rvert$$, and the unitary achieving this maximum turns out to equal $$V^\dagger$$ where $$V$$ is the unitary in the polar decomposition of $$\sqrt\rho\sqrt\sigma$$.

Observation (2) above tells us why finding the fidelity reduces to a maximisation problem, and how the polar decomposition (or more generally, the singular values) enters into it, while observation (1) tells us why we can understand the terms in this maximisation as overlaps of purifications of the states.

Consider then the $$U$$ such that $$F(\rho,\sigma)=\operatorname{Tr}(U\sqrt\rho\sqrt\sigma)$$, and define $$A\equiv U\sqrt\rho$$ and $$B=\sqrt\sigma$$. Then $$F(\rho,\sigma)=\operatorname{Tr}(AB^\dagger)$$, and thus $$F(\rho,\sigma)$$ equals the overlap between the purification of $$\rho$$ corresponding to $$A$$ and the purification of $$\sigma$$ corresponding to $$B$$.

## Detailed version

The starting observation is that there are two "natural" ways to write down $$\sqrt\rho\sqrt\sigma$$ (or any other product of two normal operators): using their spectral decompositions, and using the singular value decomposition of their product: $$\sqrt\rho\sqrt\sigma=\sum_{jk} \sqrt{\lambda_j\mu_k}\lvert\lambda_j\rangle\!\langle\lambda_j\rvert \mu_k\rangle\!\langle \mu_k\rvert = \sum_m s_m \lvert s_m^L\rangle\!\langle s_m^R\rvert,$$ where $$\rho=\sum_j\lambda_j\lvert\lambda_j\rangle\!\langle\lambda_j\rvert$$ and $$\sigma=\sum_k\mu_k\lvert\mu_k\rangle\!\langle\mu_k\rvert$$ are the spectral decompositions of $$\rho$$ and $$\sigma$$, and I denoted with $$s_m$$ the singular values of $$\sqrt\rho\sqrt\sigma$$, and with $$\lvert s_m^{L(R)}\rangle$$ the left (right) singular vectors of $$\sqrt\rho\sqrt\sigma$$. Note that using these definitions, we have $$\lvert\sqrt\rho\sqrt\sigma\rvert=\sum_m s_m \lvert s_m^R\rangle\!\langle s_m^R\rvert$$ (if using the definition $$\lvert A\rvert\equiv\sqrt{A^\dagger A}$$, otherwise replace $$R$$ with $$L$$ if you want to define $$\lvert A\rvert\equiv\sqrt{AA^\dagger}$$), and thus $$\operatorname{tr}\lvert\sqrt\rho\sqrt\sigma\rvert=\sum_m s_m$$.

Let us now denote with $$\lvert\psi_\rho\rangle$$ and $$\lvert\psi_\sigma\rangle$$ a pair of purifications of $$\rho$$ and $$\sigma$$. These can be written in general as $$\lvert\psi_\rho\rangle=\sum_k \sqrt{\lambda_k}\lvert\lambda_k\rangle\otimes\lvert u_k\rangle, \\ \lvert\psi_\sigma\rangle=\sum_k \sqrt{\mu_k}\lvert\mu_k\rangle\otimes\lvert v_k\rangle,$$ for arbitrary orthonormal bases $$\{u_k\}_k, \{v_k\}_k$$. We can then write their overlap as $$\langle\psi_\rho\rvert\psi_\sigma\rangle = \sum_{jk}\sqrt{\lambda_j\mu_k} \langle\lambda_j\rvert\mu_k\rangle\langle u_k\rvert v_k\rangle = \sum_{jk}\langle\lambda_j\rvert\sqrt\rho\sqrt\sigma\lvert\mu_k\rangle \langle u_j\rvert v_k\rangle,$$ where I exploited the fact that $$\sqrt\rho\lvert\lambda_j\rangle=\sqrt{\lambda_j}\lvert\lambda_j\rangle$$ and $$\sqrt\sigma\lvert\mu_k\rangle=\sqrt{\mu_k}\lvert\mu_k\rangle$$. Using the singular value decomposition of $$\sqrt\rho\sqrt\sigma$$ we thus get

\begin{align}\langle\psi_\rho\rvert\psi_\sigma\rangle &= \sum_{jkm} s_m \langle\lambda_j\rvert s_m^L\rangle\!\langle s_m^R\rvert\mu_k\rangle \langle u_j\rvert v_k\rangle \\ &= \sum_m s_m \langle s_m^R\rvert \Bigg(\underbrace{\sum_k \lvert \mu_k\rangle\!\langle \bar{v}_k\rvert }_{U_1}\Bigg) \Bigg(\underbrace{\sum_j \lvert\bar{u}_j\rangle\!\langle \lambda_j\rvert}_{U_2}\Bigg) \lvert s_m^L\rangle, \end{align} where I denoted with $$\lvert \bar{u}_j\rangle$$ the complex conjugate vector of $$\lvert u_j\rangle$$, so that $$\langle u_j\rvert v_k\rangle=\langle \bar{v}_k\rvert \bar{u}_j\rangle$$.

Uhlmann's theorem is almost straightforward from this. The triangle inequality gives $$\lvert\langle\psi_\rho\rvert\psi_\sigma\rangle\rvert\le \sum_m s_m$$ because matrix elements of unitary matrices are always less than $$1$$ in modulus, and the inequality is saturated when $$U_1 U_2=\sum_m \lvert s_m^R\rangle\!\langle s_m^L\rvert\equiv \mathcal U_{PD}^\dagger$$. In terms of the purification vectors $$\lvert u_j\rangle,\lvert v_k\rangle$$, this happens when $$\langle u_j\rvert v_k\rangle = \langle \mu_k \rvert \Bigg(\sum_m \lvert s_m^R\rangle \!\langle s_m^L\rvert \Bigg) \lvert \lambda_j\rangle = \langle \mu_k\rvert \mathcal U_{PD}^\dagger\rvert \lambda_j\rangle.$$ Note that here $$\mathcal U_{PD}$$ is the unitary matrix that you get out of the polar decomposition, that is, the $$V$$ in your post. We can thus conclude that the purifications that saturate the inequality are those of the form \begin{align} \lvert u_j\rangle = V\lvert\bar{\lambda}_j\rangle, \qquad \lvert v_k\rangle = V\mathcal U_{PD}^*\lvert\bar{\mu}_k\rangle, \end{align} or, equivalently, \begin{align} \lvert u_j\rangle = V\lvert j\rangle, \qquad \lvert v_k\rangle = V\lvert j\rangle \langle \mu_k\rvert\mathcal U_{PD}^\dagger\lvert\lambda_j\rangle, \end{align} for any unitary $$V$$.

So, in conclusion, what does this tell us? That the vectors $$\lvert u_j\rangle,\lvert v_k\rangle$$ that make the purifications the most aligned are determined by the overlap between the eigenvectors of $$\sigma$$, and the eigenvectors of $$\rho$$ rotated through unitary that maps the right singular vectors of $$\sqrt\rho\sqrt\sigma$$ into the left ones.

Why is this the case? I have no idea, mostly because I don't know of any easy way to relate the SVD of a product of two operators with their eigenvectors.