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That may appear as a dumb question, but:

Does Hilbert space have curvature, or is it a flat space? How and why?

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    $\begingroup$ Related: physics.stackexchange.com/q/134172 $\endgroup$ – QuantumBrick Jun 28 '16 at 16:24
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    $\begingroup$ @ACuriousMind As a vector space, indeed it does not. But Hilbert space is also geometric, because besides the vector structure it does have a scalar product. If I take three wave functions, f g and h, and I put them at the vertex of a triangle in an Hilbert space, and I then use the integral scalar product on the functions I highly doubt that the sum of the triangle's angles will be 180 degrees. Hence it has a curvature. Or.. not? $\endgroup$ – Les Adieux Jun 28 '16 at 17:40
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This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to consider the distance function $$d_\textrm{proj}(\psi,\phi) =\min_\alpha || \psi - e^{i\alpha} \phi ||$$ Or equivalently, $d_\textrm{proj}(\psi,\phi) = \sqrt{2 -2 |\langle \psi,\phi\rangle|}$. This is for example natural when we are looking at the ground state of a Hamiltonian: this is naturally only defined up to a phase. We might then ask how much our ground state changes when we tune a parameter in our Hamiltonian. The above projective distance function is then the natural quantity to look at (and it becomes singular at a quantum phase transition).

Our distance function of course defines a metric, similar to what you might be used to from general relativity. It is in fact the natural metric associated to $\mathbb CP^n$, called the Fubini-Study metric. Moreover this has non-trivial curvature. (In the case of the Bloch sphere, $n = 1$, it coincides with the spherical metric on $\mathbb CP^1 \cong S^2$ but for $n>1$ it is much more non-trivial with e.g. the metric depending on the direction you go.) This metric of course also defines a distance function (defined by taking the shortest geodesic between two states), which turns out to be $$ d_\textrm{FS}(\psi,\phi) = \arccos \left( \left|\langle \psi \right| \phi \rangle| \right) $$ Note that this is basically the angle between the two states. This is the Fubini-Study distance between $\psi$ and $\phi$. By expanding out the $\arccos$ it is clear that for $\psi$ close to $\phi$ it agrees with $d_\textrm{proj}$ (as it must, since by definition both have the same (local) metric). But they disagree for finite distances, and the fact that $d_\textrm{FS} \neq d_\textrm{proj}$ shows that the metric space defined by $d_\textrm{proj}$ is not an inner metric space/length space. (I.e. its distances are not given by geodesics. Usually in physics we work with inner metric spaces so it might be surprising to see a case pop up where this is not the case. However if we work with $d_\textrm{FS}$ then it is indeed an inner metric space.)

The Fubini-Study distance is operationally very meaningful. It arises naturally as follows:

  1. First there is the general result in statistics that the natural ``distance'' between two probability distributions $(p_1,p_2,\cdots)$ and $(q_1,q_2\cdots)$ is given by $\arccos\left( \sum_i \sqrt{p_i q_i} \right)$. This is the so-called Fisher distance. (It is in fact very conceptual as explained here by Alioscia Hamma in his lecture on the quantum informational perspective of quantum phase transitions (at Perimeter Institute 2013).)
  2. Of course we know that any wavefunction $\psi$ defines a probability distribution if we are given an observable $A$. More exactly $p_n = |\langle \psi| A_n\rangle|$ where $A_n$ are the eigenfunctions of $A$. Similarly for $\phi$. So for a given $A$ we then have the natural distance $\arccos\left( \sum_i \sqrt{ \left|\langle \psi| A_i\rangle \langle \phi| A_i\rangle \right|} \right)$.
  3. It is then natural to define the operational distance between $\phi$ and $\psi$ by maximizing the previous expression over all possible observables $A$. It is not hard to show that this then gives $d_\textrm{FS}(\psi,\phi)$.

The Fubini-Study metric/distance has many nice mathematical properties (e.g. it makes our projective Hilbert space into a so-called Kahler manifold, is related to fun stuff like Hopf fibrations, et cetera). But it is also a natural language for some physical concepts. One example is as I hinted at above: suppose we have a space of parameters $\Lambda$ and for each choice we have a Hamiltonian. I.e. we have a map of $\Lambda$ into the space of Hamiltonians. Then through the ground state this effectively defines a map $\Lambda \to P\mathcal H$ (for each parameter we have a state in our projective Hilbert space). Hence we can pull back our Fubini-Study metric to define a metric on our manifold $\Lambda$. It turns out that then looking at the curvature of this contains the complete physical information of the phase transitions of our model.

Another situation where it comes up is concerning entanglement. Given a state $\psi$, we can ask what is the closest separable state nearby. This Fubini-Study distance is then a measure of entanglement (it in fact contains the same information as the largest Schmidt value of our state--in case that means something to you.). More generally it is still quite an open question as to what extent we can relate entanglement and geometry, but it seems like a very promising path. For more information in this direction, here is the freely accessible chapter by Uhlmann and Crell on the geometry of state space from the book Entanglement and Decoherence.

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  • $\begingroup$ Nice answer! Two remarks 1) I guess you are restricting to elements $\psi$ with $\|\psi\| = 1$ (otherwise you'd have to factor that out from the distance) 2) are you sure the curvature is not constant? I would think that the space is homogeneous, every point can be transformed into any other by a rotation, which seems to be an isometry for this metric. $\endgroup$ – doetoe Jun 29 '16 at 13:31
  • $\begingroup$ @doetoe (1) Good point, I am indeed assuming normalized vectors. (2) You are completely right. I wanted to contrast the case $n >1$ to the simple case of $n=1$ where the scalar curvature is constant and completely characterizes the metric. It is still true that for $n>1$ the scalar curvature is constant, and I should have instead said that now curvature is not isotropic, as captured by the sectional curvature. $\endgroup$ – Ruben Verresen Jun 29 '16 at 14:36
  • $\begingroup$ The geometric insight seems indeed promising! I wonder though what is the meaning of curvature here. So the points 1-3 talk about a metric on a space, not a tangent metric like in GR. The concept of curvature is linked to a connection. Giving a metric on a manifold is giving a connection, and we may calculate its curvature, so that makes sense. But the connection differentiates tangent vectors, which use do they have? $\endgroup$ – Lorenz Mayer Oct 18 '18 at 14:01
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Hilbert spaces are vectorspaces by definition. If you interpret a vector space as a manifold (which you can do) then it's a flat manifold.

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  • $\begingroup$ It would be interesting to see this fleshed out. e.g. what is the metric and how would we calculate the curvature from it? $\endgroup$ – John Rennie Jun 28 '16 at 16:55
  • $\begingroup$ I read a paper which attempts to define the curvature as a Gaussian curvature using Heisenberg-Weyl representation. It states that "The metric can be defined by considering a distance function as $D^2(|\psi_i\rangle,|\psi_j\rangle)=inf_{(\delta,\gamma)}|||\psi_i\rangle e^{\mathrm{i}\delta}-|\psi_j\rangle e^{\mathrm{i}\gamma}||$ and then considering a coherent space representation.". I am looking for the link of the paper. $\endgroup$ – vbj Jun 28 '16 at 17:11
  • $\begingroup$ Here's the link of the paper, worldscientific.com/doi/pdf/10.1142/S0217732393001148 $\endgroup$ – vbj Jun 28 '16 at 17:17
  • $\begingroup$ Since nobody wants to pay for it, I hacked it. $\endgroup$ – Les Adieux Jun 28 '16 at 17:42
  • $\begingroup$ How would you interpret a curved Hilbert space? It's a mathematical tool for quantum states. I don't see the geometric extension of the idea. $\endgroup$ – Peter R Jun 28 '16 at 20:17
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As commentators have indicated Hilbert space is a vector space. A manifold is a space with an atlas-chart construction with maps on overlapping regions that define connection coefficients and ultimately curvature.

It is certainly possible to think of a finite dimensional complex vector space that is a locally flat region in an otherwise curved space. This sort of holomorphic manifold construction is well established. If one wanted to get on their Rocinante and propose some general quantum physics in a finite Hilbert space of states, they are free to do so. The finite complex vector space might be proposed as some basis set of states that are local in some general curved space. For all I know somebody has probably already tried this.

It is almost certain you will not be able to do this with an infinite dimensional Hilbert space. The infinite dimensions will make convergence issues difficult. In particular if you had some atlas-chart construction with transition functions on the overlap you would have difficulties with defining connection coefficients in a consistent manner. It is for reasons similar to defining a distance in a general sense in Hilbert space.

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Hilbert space is a complex normed vector space equipped with an inner product, where this inner product comes from the norm on the space (the norm of a vector on the hilbert space is the square root of the inner product of the vector with itself), but for a curved space like the minkowski space we will use a minkowski metric that differs from the usual euclidean one, plus we will also have a Minkowski inner product which is different than that deployed in a Hilbert space.

Topological spaces->Metric spaces->Normed spaces->Banach spaces->Hilbert spaces.

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    $\begingroup$ Why do you say Minkowski space is curved? $\endgroup$ – John Rennie Jun 28 '16 at 16:58
  • $\begingroup$ @JohnRennie Maybe Elmo meant non-Euclidean for Minkowski. $\endgroup$ – K7PEH Jun 29 '16 at 15:10

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