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I sometimes see the "Fubini-Study distance" between two (pure) states $|\psi\rangle,|\phi\rangle$ written as $$ d(\psi,\phi)_1=\arccos(|\langle\psi|\phi\rangle|), $$ for example in the Wikipedia page. Other sources (e.g. this paper in pag. 16), use the definition $$ d(\psi,\phi)_2=\sqrt{1-|\langle\psi|\phi\rangle|}. $$

What is the difference between these two definitions? Is one preferred over the other?

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    $\begingroup$ The first one is giving you a path that connects two points so distance in the sense of geometry. The second is more of an analysis flavor where you talk about various norms on vector spaces. The difference of two vectors gives the intuition of the path being a straight line in the above vector space. It does not talk directly about giving any geometric path in the manifold. $\endgroup$
    – AHusain
    Jan 24, 2020 at 17:05
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    $\begingroup$ @AHusain but what confuses me is that they are both called "Fubini-Study distance". Then when someone refers to the "Fubini-Study metric/distance" without explicitly giving the expression, which of the two expression should we think they are referring to? Also, if they are both distances, they both should come with a geometric interpretation, no? $\endgroup$
    – glS
    Jan 24, 2020 at 17:11
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    $\begingroup$ I'd only call the first one a distance. The difficulty comes from the coincidences of R^n. So many possible notions coincide leading to different definitions with the same name. $\endgroup$
    – AHusain
    Jan 24, 2020 at 17:55
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    $\begingroup$ The first distance is the length of the geodesic (corresponding to the Fubiny-Study metric) on the complex projective space $CP^{n-1}$ connecting the two rays (defined by the two vectors). It can be computed, for instance, by solving the Hamilton-Jacobi equation. $\endgroup$ Jan 27, 2020 at 13:33
  • $\begingroup$ @DavidBarMoshe if you could expand on that, it would make for a great answer I think $\endgroup$
    – glS
    Jan 28, 2020 at 15:01

2 Answers 2

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Recall the law of cosines for two unit vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb R^2$: $$ \|\mathbf{u}-\mathbf{v}\|^2 = 2-2\cos\theta, $$ where $\theta$ is the angle between the vectors. Similarly, you'll recall the definition of the inner product, $$ \langle \mathbf u|\mathbf {v}\rangle = \cos\theta. $$ So, $$ \|\mathbf{u}-\mathbf{v}\| = \sqrt{2}\sqrt{1- \langle\mathbf{u}|\mathbf{v}\rangle}. $$ For these unit vectors in $\mathbb R^2$, $\theta$ is the arc length on the unit circle intersecting these vectors. It is a distance. In your notation, $d(\mathbf{u},\mathbf{v})_1$ is the arc length and $d(\mathbf{u},\mathbf{v})_2$ is the Euclidean distance. Before we generalize to quantum states, you should note that nothing said so far would suggest one is preferred over the other.

For complex vectors we can do something analogous. Define the "Hermitian angle" of two complex vectors $\psi$ and $\phi$ in $\mathbb C^n$ to be $$ \cos\theta = |\langle\psi|\phi\rangle|. $$ Now consider $$ \|\psi -\phi\|^2 = 2-2\Re[\langle\psi|\phi\rangle] \geq 2-2|\langle\psi|\phi\rangle|. $$ So you can see that $d(\psi,\phi)_1$ and $d(\psi,\phi)_2$ are analogous, but not quite, the arc length and distance between the quantum states.

Both are distances and both are referred to by the name "Fubini-Study". This sort of overloading of definitions is unfortunate but happens all the time. For example, quantum fidelity could mean one of two things as well, and they differ quadratically.

When trying to generalize things from simple cases, there are many choices and so we end up with lots of different distances. (Just wait until you find out how many their are for matrices!) Just as there is no objective way to choose between arc length and Euclidean distance, there is no objective way to choose between the Fubini-Study distances.

Why does a particular researcher use one over the other? The answer varies, but it could be convenience—sometimes you choose the definition for which you can prove things about. Other times, a scenario is set up so a particular definition has a physical or operational meaning—you might be able to connect it to something measurable or it could correspond to the success of some protocol. But without some assumptions in the set-up of the problem, there is no preferred distance.

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  • $\begingroup$ Hello. I just ended on this question. Why for the Bure distance: $d(\psi,\phi)_2$ is not $\sqrt{2-2Re[\langle \psi | \phi \rangle]}$ then instead of $\sqrt{2-2|\langle \psi | \phi \rangle|}$ ? I agree with your calculation, then I don't get why the modulus is taken in the original definition ? $\endgroup$ Aug 17, 2020 at 8:30
  • $\begingroup$ My associated topic: quantumcomputing.stackexchange.com/questions/13348/… $\endgroup$ Aug 17, 2020 at 8:30
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I'll try to address the problem from the Riemannian geometry point of view. In this approach, the distances are identified as length of geodesics of Riemannian metrics on spaces of quantum states. The advantage of this approach lies in the fact that the Riemannian distances automatically satisfy the metric axioms of positivity, symmetry and the triangle inequality and sometimes additional properties of interest in quantum information theory such as monotonicity.

The distance $d_1$ is the geodesic distance of the Fubini-Study metric on the pure state space $\mathbb{C}P^{N-1}$ connecting the rays $\psi$ and $\phi$. It is the only distance on $\mathbb{C}P^{N-1}$ invariant under unitary evolution.

The elements of the pure state space are rank $1$ projectors onto rays of a projective Hilbert space. This space is a subspace of the following spaces:

  1. The cone $\mathcal{P}^N$ of positive definite $N$ dimensional complex matrices.
  2. The quantum state space of density matrices $\mathcal{M}^N$ consisting of the intersection of $\mathcal{P}^N$ with the space of unit trace matrices. (Using the notation of Bengtsson and Życzkowski , pages 192 and 200). The spaces $\mathcal{P}^N$ and $\mathcal{M}^N$ can be equipped with a metric called the Bures metric (Or the Bures-Wasserstein metric). $$ds_B^2 = \text{tr}(GdX)$$ Where, in the first case, $X=R$, a positive definite complex matrix, and in the second case $X=\rho$, a density matrix The matrix valued one form $G$ is given implicitly in: $$dX = XG+GX$$ The geodesic distance function associated to this metric on $\mathcal{P}^N$ is given by ( Bengtsson and Życzkowski equation 9.31): $$D^{\mathcal{P}^N }_{\text{Bures}}(R_1, R_2) = \text{tr}(R_1)+ \text{tr}(R_2)-2 \sqrt{F}( R_1, R_2)$$ Where $\sqrt{F}$ is the Uhlmann fidelity: $$\sqrt{F}( R_1, R_2) = \text{tr}\left(\sqrt{\sqrt{R_1}R_2\sqrt{R_1})}\right)$$ While on the space $\mathcal{M}^N$ of density matrices, the same metric gives rise to the geodesic distance function (equation 9.32): $$D^{\mathcal{M}^N }_{\text{Bures}}(\rho_1, \rho_2) = \arccos(\sqrt{F}( \rho_1, \rho_2))$$ As can be easily seen, when the initial and final points are pure states, the first formula reduces to $d_2$ (multiplied by a factor of $2$), while the second formula reduces to $d_1$.

Thus, if the projectors representing the initial and final pure states are regarded as merely positive matrices and we don't mind that the intermediate points on the geodesic are only positive matrices with not necessarily a unit trace, then we get the second distance $d_2$. On the other hand, if we insist that the intermediate points are also quantum states we get the first distance $d_1$.

The equality of the Bure's distance for density matrices when restricted to pure states and the Fubini-Study distance proves that at the metric level, the restriction of the Bures metric to pure states is the Fubini-Study metric. Another metric possessing this property is the Wigner-Yanase metric. Both Bures and Wigner-Yanase metrics are known to be monotone.

The Fubini-Study metric has a very well known expressions in coordinates and otherwise, for example in Bengtsson and Życzkowski equation 4.51. The Bures metric in the other hand can be given in general only implicitly. It is not easy to write an explicit formula except when $N=2$ (Bengtsson and Życzkowski equation 9.50). The distance function of Bures was known long ago, but it was only much later when Uhlmann proved that the distance function originates from a Riemannian metric.

Uhlmann's proof is indirect due to the implicitness of the Bures metric, but the simpler case of the Fubini-Study metric the computation of the distance function from the metric is rather simple, which I'll include in this answer: On a Riemannian manifold $(M, g)$, a geodesic is a curve: $[0, 1] \rightarrow M$ starting at $x(0) = x_i$ and ending at $x(1) = x_f$ minimizing the functional (arc length): $$I[x]= \int_0^1 g\left(\frac{dx}{dt}, \frac{dx}{dt}\right) dt$$ When a unique solution $x_c(t)$ exists, then the arc length (regarded as a function of the initial and final points): $$S(x_i, x_f) = I[x_c],$$ satisfies the Hamilton-Jacobi equation: $$ g(\nabla_{x_i}S, \nabla_{x_i}S) = 1$$ The Hamilton-Jacobi equation has exact solutions in rare cases when the system is integrable , such as in the case of the harmonic oscillator and the Kepler problem; in many multidimensional cases the solution is obtained by means of separation of variables. The Fubini-Study geodesic motion on $\mathbb{C}P^{N-1}$, is integrable and the Hamilton-Jacobi equation is exactly solvable. Substituting $|\phi\rangle =|\psi\rangle + |d\psi\rangle $ in $d_1$, we get: $$ds_{FS} = \arccos(1+|\langle \psi| d\psi\rangle|) = \sqrt{\langle d\psi|d\psi\rangle-\langle d\psi|\psi\rangle\langle \psi|d\psi\rangle }$$ One way to accomplish that is to exploit the homogeneity of $\mathbb{C}P^{N-1} = SU(N)/S(U(N-1) \times U(1))$. In calculating the geodesic distance between two rays $\psi_i$ and $\psi_f$, we can perform a special unitary transformation to bring $\psi_i$ to: $$|\psi_i\rangle = [1, 0, …, 0]^t$$ We still have an $S(U(N-1) \times U(1))$, freedom which can be used to bring any other unit vector to the form: $$|\psi\rangle = [\cos(\theta), \sin(\theta), 0, …, 0]^t$$ Substituting, the last expression in the Fubini-Study metric we obtain: $$ds^2_{FS} =d\theta^2$$ Thus, the Hamilton-Jacobi equation takes the form: $$\left(\frac{dS_{FS}}{d\theta}\right)^2 = 1$$ Whose, solution is: $$S_{FS} = \theta + \text{Const.}$$ Fixing the constant by requiring the distance to be from the initial point ($\theta = 0$) to vanish: Thus: $$ S_{FS} = \theta = \arccos|\langle \psi|\psi_i\rangle|$$

$\mathbb{C}P^{N-1}$ as a Riemannian symmetric space. In this respect, the angle $\theta$ on $\mathbb{C}P^{N-1}$ is called a radial coordinate ( a generalization of the radial coordinate in spherical coordinates of the Euclidean space). The number of radial coordinates is called the rank of the symmetric space. (This means that $\mathbb{C}P^{N-1}$ is symmetric space of rank one). The space of density matrices $\mathcal{M}^N$ possesses subspaces which are symmetric spaces of higher rank. Thus, these spaces have multiple radial coordinates and other distance functions.

The Hamilton-Jacobi equation on $\mathbb{C}P^{N-1}$ can also be directly solved using the local Kählerian coordinates given in in Bengtsson and Życzkowski equation 4.51.

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