QM only possible in flat space?

[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 ?

 

[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.

 

[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 Δ_M 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_S² is nothing else but the angular part of the 3-dim Δ_R³, therefore the eigenfunctions of Δ_S² are the well-known Y_lm(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).

 

[tom.stoer]

The Laplacian in curved space is defined in terms of the metric as

$$\nabla = g^{ij} \partial_i \partial_j.$$

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

 

[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?

 

[tom.stoer]

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

What do you mean by non-linear evolution?

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

 

[TrickyDicky]

what do you mean by non-linear evolution?

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

 

[tom.stoer]

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

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

 

[TrickyDicky]

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

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?

 

[TrickyDicky]

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

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.

 

[tom.stoer]

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?

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_R³ you get the Δ_S² as Laplace-Beltrami operator; it takes into account the curvature due to the non-trivial metric g (if you reduce R³ to S² g_S² is nothing else but the induced metric).

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

i∂₀Y(Ω) = -Δ_S²Y(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 Y_lm(Ω), i.e. you get the usual spherical harmonics Y_lm(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_S². 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 grad_M and Δ_M. 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 grad_M and Δ_M 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_S². 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 Y_lm(Ω)

ψ(Ω) = Ʃ_lm ψ_lm Y_lm(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

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.

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_S² remain valid.

 

[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.

 

[tom.stoer]

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.

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.

 

[TrickyDicky]

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.

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).

 

[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.

 

[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

 

[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)

 

[TrickyDicky]

Of course you can construct position eigenfunctions (generalized delta-distributions) on manifolds.

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.

 

[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

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

Instead you have to use

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

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 ;-)

 

[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?

 

[tom.stoer]

Wouldn't the non-linearity of the coordinates xⁱ and ∂_i that 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?

No, of course not.

Formally there is no difference for 2-dim space, 3-dim. space, 4-dim. space-time, ... All you have to do is to specify a (Riemannian or pseudo-Riemannian) Manifold M with metric g and calculate the Laplace-Beltrami Δ_M for (M,g). Of course they will depend on dim(M) and g, so they will look different for each (M,g), but the formalism is always the same. If you do non-rel. QM on curved space (n-dim.) with flat time you have to solve

$$i\frac{d}{dt}\psi(x,t) = -\triangle_M\,\psi(x,t)$$

b/c the "geometry of t" is trivial factorization works as usual, i.e.

$$\psi(x,t) = \phi(x)\,e^{-iEt}$$

and you arrive at

$$(-\triangle_M - E)\,\phi_E(x) = 0$$

 

[TrickyDicky]

The Laplace-Beltrami operator examples in the non-flat examples of the wkipedia page refer to the n-1 space, with the manifold in n dimensions being flat (either euclidean in the S^n-1 case or minkowskian in the H^n-1 case.

 

[TrickyDicky]

If you do non-rel. QM on curved space (n-dim.) with flat time

b/c the "geometry of t" is trivial

But it wouldn't be trivial("flat") anymore so it wouldn't be standard QM, that's my point.

 

[tom.stoer]

The Laplace-Beltrami operator examples in the non-flat examples of the wkipedia page refer to the n-1 space, with the manifold in n dimensions being flat.

That doesn't matter!

You don't need any embedding of an n-dim. manifold Mⁿ in an n+k Euklidean / flat space R^n+k. You can define every Mⁿ w/o such an embedding. An ant (or any other effectively two-dim. animal) can do exactly the same maths on S² w/o knowing R³ as you can! There is no difference between a "2-dim. S²" and an "embedded S²". They are identical geometrically.

Or let's do it the other way round: for arbitrary (!) curved Mⁿ it's always possible to find an Euklidean embedding space R^n+k. So in a sense there are no curved manifolds which cannot be embedded in suitable higher-dim. Euklidean / flat manifolds.

That means that for every curved Mⁿ and QM defined on it you could start with QM on R^n+k and constrain the motion of the particle to an appropriate n-dim. submanifold (it's not always possible to write this down algebraically, but that doesn't matter).

So the difference you are constructing - the difference between curved manifolds which can be embedded in flat space and manifolds for which such an embedding is not possible - does not exist.

In that sense all manifolds are equally valid as examples.

 

[tom.stoer]

But it wouldn't be trivial("flat") anymore so it wouldn't be standard QM, that's my point.

It is standard QM; perhaps not in cartesian coordinates, perhaps not as you may know it, but it's absolutely standard - as the S² i.e. L² example should convince you.

The only difference is the complicated kinetic energy, nothing else. But even this should be familiar to you - namely when replacing p² by (p-eA)² when putting a charged particle in a magnetic field. Already in this simple case the kinetic term changes non-trivially and the particle observes something like a "non-trivial geometry".

 

[TrickyDicky]

Ok, can you show me or point me to any nonrelativistic example in S³ or H³?

 

[tom.stoer]

http://arxiv.org/abs/quant-ph/9702006
http://arxiv.org/abs/quant-ph/9511020
http://arxiv.org/abs/hep-th/9604172

As far as I remember Kleinert wrote a couple of papers about canonical and path integral quantization on Sⁿ and SO(n). So S³ is covered by the general case Sⁿ.

In addition you may find some applications in his book https://www.amazon.com/dp/9812381074/?tag=pfamazon01-20

 

[TrickyDicky]

http://arxiv.org/abs/quant-ph/9702006
http://arxiv.org/abs/quant-ph/9511020
http://arxiv.org/abs/hep-th/9604172

As far as I remember Kleinert wrote a couple of papers about canonical and path integral quantization on Sⁿ and SO(n). So S³ is covered by the general case Sⁿ.

Thanks.
Since the Laplace-Beltrami operator can be generalized to any n, do you have any idea why the wikipedia article would explain it in terms of n-1 instead of just n?, it seems an unnecessary complication.

 

[tom.stoer]

It depends where you are coming from.

If you are familiar with H ~ Δⁿ in n-dim. flat space and with its angular / radial decomposition then you see immediately that you get QM on S^n-1 by dropping the r-terms.

If on the other hand you are interested in the general case where such an embedding may not be at hand, not be known, or is too complicated, ..., then it's a worthless detour.

I think it's interesting that if you
a) constrain the r - d.o.f. on Rⁿ to S^n-1
you end up with the correct H and Δ on S^n-1 i.e. that this way is equivalent to
b) start directly on S^n-1 w/o considering a flat embedding Rⁿ with cartesian coordinates;
this equivalence is by no means trivial.

 

[TrickyDicky]

It depends where you are coming from.

If you are familiar with H ~ Δⁿ in n-dim. flat space and with its angular / radial decomposition then you see immediately that you get QM on S^n-1 by dropping the r-terms.

If on the other hand you are interested in the general case where such an embedding may not be at hand, not be known, or is too complicated, ..., then it's a worthless detour.

I think it's interesting that if you
a) constrain the r - d.o.f. on Rⁿ to S^n-1
you end up with the correct H and Δ on S^n-1 i.e. that this way is equivalent to
b) start directly on S^n-1 w/o considering a flat embedding Rⁿ with cartesian coordinates;
this equivalence is by no means trivial.

Aha, I can see what you are saying in an abstract mathematical way. But going back to physical reality I can see that starting from a E^3 manifold the Δ S^2 operator works fine and L^2 works fine etc. And to apply an Δ S^3 operator in an abstract manifold seems ok too, I just don't think it can be done in our physical 3-space unless we are considering the 4-spacetime setting.

 

[TrickyDicky]

On the other hand the Laplace operator , of which the Laplace-Beltrami op. is just a generalization, is by definition the div o grad of a function in Euclidean space .
A non-linear hamiltonian wouldn't be a linear operator so no laplacian would be used, even if the Laplace-Beltrami operator could be used to model an angular momentum in S^3 for a putative 4-space manifold.

 

[tom.stoer]

Mathematically it doesn't matter whether we talk about space opr space-time. The difference comes when writing down a SE, but this is secondary. The first step is to define the canonical position and momenta and to quantize them. This can be done in two ways

1) start on a larger Euklidean E space
derive x and p
derive H
introduce constraints C
reduce the Euklidean space E to the lower-dim. manifold (M,g) as submanifold
modify x, p, and H via Dirac's constraint formalism
derive the physical H_phys on M
replace Poisson brackets {,} by Dirac brackets {,}_D
quantize x,p,H' and replacing the Dirac brackets {,}_D by commutators [,]

2) start directly on (M,g) with physical coordinates Ω
use a special covariant gradient operator ∂_M on (M,g) as momentum
use the Laplace-Beltrami operator Δ_M on (M,g) as kinetic energy operator

In simple cases like S² (2) is rather obvious but in more complicated cases one has to apply (1) I don't whether there is a general proof that (1) and (2) are always equivalent

 

[tom.stoer]

On the other hand the Laplace operator , of which the Laplace-Beltrami op. is just a generalization, is by definition the div o grad of a function in Euclidean space .

Correct

A non-linear hamiltonian wouldn't be a linear operator so no laplacian would be used, ...

I still have no idea what this 'non-linear Hamiltonian' should be. Can you give me an example what you have in mind?

 

[TrickyDicky]

Mathematically it doesn't matter whether we talk about space opr space-time. The difference comes when writing down a SE, but this is secondary. The first step is to define the canonical position and momenta and to quantize them. This can be done in two ways

1) start on a larger Euklidean E space
derive x and p
derive H
introduce constraints C
reduce the Euklidean space E to the lower-dim. manifold (M,g) as submanifold
modify x, p, and H via Dirac's constraint formalism
derive the physical H_phys on M
replace Poisson brackets {,} by Dirac brackets {,}_D
quantize x,p,H' and replacing the Dirac brackets {,}_D by commutators [,]

2) start directly on (M,g) with physical coordinates Ω
use a special covariant gradient operator ∂_M on (M,g) as momentum
use the Laplace-Beltrami operator Δ_M on (M,g) as kinetic energy operator

In simple cases like S² (2) is rather obvious but in more complicated cases one has to apply (1) I don't whether there is a general proof that (1) and (2) are always equivalent

I agree with this.
I also see that you have a different notion of what is "QM working in curved space" than mine because for you the fact that we can apply QM to a wave function constrained to a spherical surface is an example of it, while for me that doesn't count as an example of what I mean as NRQM in curved space, which would be restricted to curved 3-space that I think is not possible in the nonrelativistic case because strictly speaking Δψ can't be used since the wave function wouldn't be in Euclidean space, however in QM ψ is defined as an abstract vector space and these are defined in R^n.

 

[TrickyDicky]

I still have no idea what this 'non-linear Hamiltonian' should be. Can you give me an example what you have in mind?

I don't know either, all i know is it wouldn't follow the QM postulates. Anyway in the real relativistic 4D spacetime is not necessary so it is not very important.