QM only possible in flat space?

[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.

 

[tom.stoer]

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 ...

It can be used; you have to use the 3-dim. Laplace-Beltrami operator for the manifold (M,g)

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.

Not Euklidean Rⁿ, but a Hilbert space; that's a big difference.

I think we should get rid of coordinates and use coordinate-free notation.

Δ_M is a linear operator defined on tangent space of M in a point P. Δ_M can be applied to arbitrary scalar functions f(P): M → R. If we apply this to non-rel. QM on M then Δ_M becomes the Hamiltonian H (for a free particle = w/o interaction terms) and f becomes a wave function defined over M. So f must live in a Hilbert space, which means we need a measure (integral) to define the scalar product

(f,g) → ∫_M f* g

on M. If all this can be constructed then we are ready to do QM on M.

Now Δ_M is typically highly non-linear in any coordinate representation (chart) on M, but as an operator it acts linearly on functions f on M, and therefore it can potentially act as a linear operator on a Hilbert space defined as function space over M.

Honestly, I do have no idea how to go beyond linear functional analysis. Neither do I know the maths, nor am I able to guess a physical application. I know that there are research programs (non-linear QM) trying to replace the linear SE with a non-linear one to describre the (non-unitary) "collapse of the wave function", but afaik all those proposals have failed. I would call this "beyond mainstream" whereas QM on curved manifolds is standard

 

[TrickyDicky]

An interesting discussion can be found in the second answer in this page:
http://physics.stackexchange.com/questions/5773/quantum-mechanics-on-a-manifold

In the intrinsic approach it is mentioned the ambiguity of the derivative operator in curved space, so I guess this is a not completely settled thing.

 

[tom.stoer]

You are right. The Laplace-Beltrami Operator is dictated by a classical symmetry consideration but of course it is by no means clear whether this is "correct" in some fundamental sense. That's why it's interesting to derive the equations via constraint quantization and to check whether the results of the two approaches coincide.

Keep in mind: already in flat Euklidean space and cartesian coordinates it is by no means clear why the ansatz

Hψ = -∂²ψ

is the correct one; why not

Hψ = -(x^a ∂ x^b ∂ x^c ψ)

with a+b+c=0 ?

The answer is a) symmetry and b) b/c it works.

Quantization on curved manifolds is not really more ambiguous that quantization on flat space. But the ambiguities become more obvious and pressing - and we do not have many real systems at hand to check experimentally ;-)

 

[TrickyDicky]

I was trying to find any physical example in NRQM that applies the Laplace-Beltrami op. for general manifolds in a situation that is not a stationary solution(not harmonic) but I'm not having any success. Any hints?

 

[tom.stoer]

no hints, I am sorry

but it should be clear the time-dep. SE still holds

i\frac{d}{dt}\,\psi = \left(-\frac{1}{2m}\triangle_\mathcal{M} + V\right)\,\psi

what are you interested in? calculation of propagators?

 

[TrickyDicky]

no hints, I am sorry

but it should be clear the time-dep. SE still holds


i\frac{d}{dt}\,\psi = \left(-\frac{1}{2m}\triangle_\mathcal{M} + V\right)\,\psi

Certainly, but I find hard to find a situation that can be solved within the nonrelativistic setting that is not stationary and still has the symmetry necessary to use the \triangle_\mathcal{M}, it would seem to me the introduction of a time asymmetry prevents it or at least I have only seen examples of its use related with harmonic functions.
Even in the QFT example you gave me about the Hawking radiation the problem is about the static Black hole.

what are you interested in? calculation of propagators?

Yes, but I'm certainly in a -previous to being able to do any calculation-phase at this moment.

 

[tom.stoer]

You could add a second particle moving on the same manifold M (i.e. S²) plus a two-particle interaction via a scalar potential respecting the symmetry of M.

i\frac{d}{dt}\,\psi_1\otimes\psi_2 = \left(-\frac{1}{2m}\triangle_1 -\frac{1}{2m}\triangle_2 + V_{12}\right)\,\psi_1\otimes\psi_2

 

[TrickyDicky]

You could add a second particle moving on the same manifold M (i.e. S²) plus a two-particle interaction via a scalar potential respecting the symmetry of M.

i\frac{d}{dt}\,\psi_1\otimes\psi_2 = \left(-\frac{1}{2m}\triangle_1 -\frac{1}{2m}\triangle_2 + V_{12}\right)\,\psi_1\otimes\psi_2

If we add a second particle it wouldn't be the same manifold, we'd double the dimensions.

 

[tom.stoer]

If we add a second particle it wouldn't be the same manifold, we'd double the dimensions.

yes, strictly speaking you are right; nevertheless I would say that Tom and TrickyDicky live in the same R³ space, not in R⁶

 

[TrickyDicky]

really?? Maybe, but don't tell the relativity forum guys

 

[tom.stoer]

For 2-dim systems these ideas can be useful in nanotechnology and surface science; I guess the solid-state guys could be interested as well; you could also look for "QM on (Riemann) manifolds" or "QM on (Riemann) surfaces".

For 3-dim systems (not R³) the only (trivial) example I know is the three-torus T³ which is often used to regulate the IR limit and to get a discrete spectrum. I cannot think about any application in non-rel. QM but this may be due to my own ignorance;-)

 

[TrickyDicky]

Thanks.