# QM only possible in flat space?

1. May 1, 2012

### TrickyDicky

I was trying to think of QM in the context of different possible spatial curvatures since the standard cosmology (FRW model) admits at least in principle that 3-space can have positive, negative or no curvature, even if the flat space is favored by the CMB WMAP-COBE observations etc, when I noticed that the axioms of QM demand the Schrodinger equation to be linear and therefore apparently only Euclidean space would be acceptable in QM, is this a correct conclusion?
Is this one of the reasons (or at least a further constraint on space geometry) most physicists (especially in the particle-high energy physics subfield) favored a flat space even before the WMAP etc ?

2. May 1, 2012

### fzero

The Laplacian in curved space still a linear operator, so we can discuss its eigenfunctions and eigenvalues. In fact, the spherical harmonics can be thought of as the wavefunctions for a particle which is confined to the surface of a sphere, which is in fact a curved space. So we do not have to require flat space to have a well-defined quantum mechanics.

More generally, you can discuss QFT in a curved background spacetime. What has not been solved is how to properly treat the backreaction of quantum matter on the geometry of spacetime in a way consistent with general relativity. This is the problem of quantum gravity.

Last edited: May 1, 2012
3. May 1, 2012

### tom.stoer

There are well-known examples for QM on curved manifolds; this does by no means affect the lineareity of the Hamiltonian. A well-known example is an underlying metric space (M,g) with manifold M and metric g on which the so-called Laplace-Beltrami Operator ΔM as a generalization of Δ can be defined.

$$\Delta_\mathcal{M}\,f = \frac{1}{\sqrt{\text{det}g}} \, \partial_i \, (\sqrt{\text{det}g}\,g^{ik}\,\partial_k\,f)$$

Well-known example which can be solved analytically are the spheres Sn or Lie-group manifolds.

For S2 the Laplace-Beltrami operator Δ is nothing else but the angular part of the 3-dim Δ, therefore the eigenfunctions of Δ are the well-known Ylm(Ω). This example can be constructed by starting with a free particle in R³ an constraining its motion to S² with radius r via δ(xi xi - r²).

4. May 1, 2012

### tom.stoer

Not really; please refer to my post for the correct definition

5. May 1, 2012

### TrickyDicky

But with the time dependent Schrodinger equation how would this linear equation give us the nonlinear evolution in time if the space is curved?

6. May 1, 2012

### tom.stoer

What do you mean by non-linear evolution?

Look at the S² example: the time-evolution for angular momentum L and Ylm(Ω) is linear (even though the underlying S² is curved).

Last edited: May 1, 2012
7. May 1, 2012

### TrickyDicky

doesn't curvature introduce an element of non-linearity in the space where the wave function evolves in time?

8. May 1, 2012

### tom.stoer

the space (as a manifold) is curved, but the evolution of states, wave functions etc. defined over this space remains linear

9. May 1, 2012

### TrickyDicky

Hmmm, how so? I don't understand it. Do you mean the equations of the evolution of say a wave packet is indifferent to the geometry of the space it lives in, it is always linear?

10. May 1, 2012

### TrickyDicky

I'm referring to an intrinsic curvature of the 3-space, in this example there is a dimensional reduction, I mean that it is still possible to consider it as R³ constrained to S² as you explained.

11. May 2, 2012

### tom.stoer

It is not indifferent, it "feels" the underlying geometry, nevertheless the Hilbert space operators remain linear, the superposition principle holds. etc.

Look at the S² with non-vanishing curvature. If you keep only the Ω-piece of the flat Δ you get the Δ as Laplace-Beltrami operator; it takes into account the curvature due to the non-trivial metric g (if you reduce R³ to S² g is nothing else but the induced metric).

Now look at the Schroedinger equation for a wave function Y(Ω). It reads

i∂0Y(Ω) = -ΔY(Ω)

You can look at it from two different perspectives:

1) Usually you derive it via factorization ψ(r,Ω) = R(r) Y(Ω) for rotational invariant problems. Then you solve for Ylm(Ω), i.e. you get the usual spherical harmonics Ylm(Ω) as angular part of ψ(r,Ω). In the next step you solve for the R(r) piece which depends on a potential term V(r). In that way you can e.g. derive the eigenfunctions for the hydrogen atom.

2) If you constrain the particle motion from R³ to S² via δ(x²-r²) with constant r you get the same operator Δ. In addition the r-terms vanish (the particle is confined to fixed r). Alternatively you can start with the postulate that momentum and kinetic energy on an arbitary manifold (M,g) must be defined via gradM and ΔM. This is nothing else but a generalization to the usual, xi and ∂i for flat space. One finds that the dimensional reduction (here: R³ → S²) and the starting point gradM and ΔM are equivalent (perhaps up to ordering ambiguities in h²).

So you can use (2) w/o ever considering (1). You can do quantum mechanics on S² using H = -Δ. Of course S² is curved, but you wil never need that. You have perfectly linear quantum mechanical euqations (with linear I mean linear in ψ, not linear in xi). You can construct wave pakets ψ(Ω) as superpositions of Ylm(Ω)

ψ(Ω) = Ʃlm ψlm Ylm(Ω)

You can calculate matrix elements <ψ'|A|ψ> for linear operators A(Ω). You can do evberything you know from standard quantum mechanics, i.e. in that special case for spherical harmonics.

This is possible for every manifold (M,g), but of course it may become awfully complex. It does not require the embedding like S² in R³; you can immediately start from (M,g) w/o considering such an embedding. You can calculate propagators and path integrals on (M,g). In case of Sn and group manifolds you will get closed expressions which generalize the S² case.

So the curvature on M does not change the formalism of QM, it only changes the construction of the operators p, H, etc. They remain linear operators on the Hilbert space i.e. on the wave functions ψ, but they are nin-linear in position space, i.e. in xi and ∂i

That doesn't change anything. It doesn't matter whether you construct S² as embedding in R³ with an induced metric or whether you start with S² w/o any reference to an Euclidean embedding space. The results are the same.

Here the embedding is a nice example b/c you see immediately how the r-terms drop out due to the constraint; all what remains are the Ω-terms.

But if you start directly with S² w/o ever referring to R³ and any embedding nothing will change. All properties of Δ remain valid.

Last edited: May 2, 2012
12. May 2, 2012

### K^2

It's still a good observation. Even though you can have linear Hamiltonian in curved space-time, the moment you include Hamiltonian in equations for the metric, you are screwed. Einstein's Field Equation is non-linear, so if an object you are considering is the source of the curvature, Hamiltonian is non-linear and QM breaks down.

And that's kind of the biggest problem we have in physics at the moment. QM and GR just can't play together.

13. May 2, 2012

### tom.stoer

Again, the non-linearity is due to the manifold, i.e. the coordinates xi and ∂i. The quantum mechanical equations and all operators derived from xi and ∂i remain linear operators on Hilbert space!

You can look at several different approaches to quantum gravity, e.g. strings, loops, asymptotic safety; in all these cases the relevant operators or functionals are represented linear (!) on Hilbert space. In addition these approaches show that quantum mechanics and gravity can play together. All approaches are by no means complete, but they show how to quantize gravity w/o giving up the fundamental properties of quantum mechanics, i.e. Hilbert space structure and linear operators.

Last edited: May 2, 2012
14. May 2, 2012

### TrickyDicky

There might be a little misunderstanding here, your examples seem to be either eigenfunction problems that are linear by definition (spherical harmonics, orbital angular moment problems, and anyway manifolds are locally linear by definition regardless their curvature) or imply spacetime, I'm restricting my case to the time-dependent dynamic evolution of a wave packet dictated by SE and its non-linearity in position representation when the position space is non-linear, what I mean is that it would seem like such a quantum particle could no longer have a well defined position eigenstate if the position space is non-linear (nevermind here the fact that in Euclidean space the wave packet is also delocalized after some time due to linear dispersion).

Last edited: May 2, 2012
15. May 2, 2012

### TrickyDicky

Ok, I realize that in the OP I also implied spacetime, I guess I mixed the physical spacetime situation with the non-relativistic QM approach that separates a 3D-space on one hand and a time evolution parameter on the other.

16. May 2, 2012

### tom.stoer

Of course you can construct position eigenfunctions (generalized delta-distributions) on manifolds. For closed manifolds like S² you may also construct localoied, dispersion-free coherent states.

If you want to look at 4-dim. space-time with curvature the Schrödinger equation is no longer a good starting point. But you can use e.g. the generalized Dirac equation satisfying local Lorentz covariance with a generalized Dirac operator. What you need is a so called spin manifold.

http://en.wikipedia.org/wiki/Spin_manifold

17. May 2, 2012

### vanhees71

You can also do relativistic quantum-field theory in a curved background spacetime, but it's a physically delicate issue. A nice review paper about this is

B. S. DeWitt, Quantum Field Theory in Curved Spacetime, Phys. Rept. 19, 295 (1975)

18. May 2, 2012

### TrickyDicky

Ok, so you say it is perfectly possible to use the linear time-dep. Schrodinger equation i.e. in S^3 or H^3 manifolds, no need for the non-linear Schrodinger equation for example, right?
I still can't understand how you can have a linear time evolution hamiltonian in 3-dimensios if the manifold is a intrinsically curved 3-space but you are the expert here.

19. May 2, 2012

### tom.stoer

Yes, this is what I am saying.

Take the S² example which is intrinsically curved. Write down the free Hamiltonian

$$H = \triangle_{S^2}(\Omega)$$

and solve

$$(H - E) \phi_E(\Omega) = 0$$

$$\psi(\Omega, t) = e^{-iHt}\,\psi(\Omega, 0)$$

This looks terribly complicated ... but you have all ingredients at hand since your first QM course:

$$\triangle_{S^2}(\Omega) \;\Rightarrow\;\vec{L}^2$$

$$\phi_E(\Omega) \;\Rightarrow\;Y_{lm}(\Omega)$$

and everything is fine (you will never think about any non-linearity in the SE).

I agree that things become more complicated when space-time is curved b/c it's by no means clear that you always may have something like "curved space" and "flat time". That means that in those cases you will not be allowed to write

$$\Box = \partial_0^2 - \triangle_\mathcal{M}$$

$$\Box_\mathcal{M}\,f = \frac{1}{\sqrt{|\text{det}g}|} \, \partial_\mu \, (\sqrt{|\text{det}g|}\,g^{\mu \nu}\,\partial_\nu\,f)$$

where the indices run over 0..3 and you have something like a generalized Laplace-Beltrami-Operator for pseudo-Riemannian manifolds (pseudo means that the metric is diag(-+++).

It's exactly this operator that is responsible for all the crazy stuff happening in QFT on non-flat space-time like Hawking radiation. Have a look at

http://www.itp.uni-hannover.de/~giulini/papers/BlackHoleSeminar/Hawking_CMP_1975.pdf

Hawking is using a conformally invariant expression b/c he is interested in massless fields; his expression reads

$$\Box\,f \;\Rightarrow\;g^{\mu \nu}\,f_{;\;\mu \nu}$$

I have to think about the differences between these two box-symbols - but I guess Hawking is right ;-)

Last edited: May 2, 2012
20. May 2, 2012

### TrickyDicky

Tom, I think I understood that, but you keep going back to the S^2 example and the spacetime 4D examples. I'm restricting it to the non-relativistic case, with maximum number of dimensions equal to three, and intrinsic curvature in 3 dim, not in 2.
Wouldn't the non-linearity of the coordinates xi and ∂ithat are explicit in the Hamiltonian produce a non-linear hamiltonian so that Hilbert space is no longer usable and therefore we'd be out of QM in this case?