# No general quantization recipe?

1. Oct 18, 2012

### wdlang

there is in general no exact recipe for quantization

what does this mean?

i think there is only one (correct) quantum mechanics

2. Oct 18, 2012

### Jano L.

I understand it in this way: there is no known way to derive the principles and results of quantum-mechanical theory from known classical theory. Rather, quantum or wave mechanics is a supplement to classical mechanics, containing some new ideas. These are hard to obtain simply by some mathematical transformation, "quantizing", classical theory.

3. Oct 18, 2012

### tom.stoer

There are several ambiguities when quantizing a classical theory

The simplest one is operator ordering: classically

$$p^2 = p\,f(x)\,p\,f^{-1}(x)$$

but when introducing operators this is no longer true b/c they do no longer commute.

There may be symmetry principles to overcome these difficulties. Constructing p² for curved manifolds M with metric g one uses the Laplace-Beltrami operator

$$\vec{p}^2 \to -\Delta_g$$

But this is only a guiding principle and other choices are possible. In other cases there may be no such principles at all.

Last edited: Oct 18, 2012
4. Oct 19, 2012

### Staff: Mentor

Its got to do with operator ordering. In classical physics q and p commute but in QM it doesn't. So you write down the classical Hamiltonian and reasonably assume the corresponding quantum Hamiltonian is the same - but how do you determine the ordering in the quantum Hamiltonian if the classical one contains qp? In practice it usually doesn't occur but in principle it can - and does on occasion.

Over and above that however is a number of different Hamiltonian's can lead to the same Hamiltonian in the classical limit - see:
http://arxiv.org/pdf/gr-qc/9904042v1.pdf

And issues like that do occur more often in practice eg the Aharonov-Bohm eﬀect - in particular an issue is does the momentum operator correspond to p=mv or to p as the generalized momentum. It turns out to be the generalized momentum but there is nothing in the usual rules that determines that. This is one reason the approach by Ballentine in QM - A Modern Development is better - he derives the form of the Hamiltonian when a field is present and there is no ambiguity.

Thanks
Bill

Last edited: Oct 19, 2012
5. Oct 20, 2012

### tom.stoer

There are other ambiguities as well. Suppose you want to quantize a free particle on a circle, i.e. you replace x with an angle θ in [0,2π]. Obviously you start with plane waves ψ respecting periodicity in θ:

$$u_n(\theta) = e^{i n \theta}$$

with

$$u_n(\theta+2\pi) = u_n(\theta)$$

But there are unitarily inequivalent representations labelled by a parameter δ with

$$u_{n,\delta}(\theta) = e^{i (n + \delta) \theta}$$

and "twisted" boundary conditions

$$u_{n,\delta}(\theta+2\pi) = e^{i (n + \delta) (\theta + 2\pi)} = e^{i\delta\theta}\,u_{n,\delta}(\theta)$$

Obvously these twisted boundary conditions cannot be ruled out due to a classical analogy b/c there is none. In addition they cannot be ruled our quantum mechanically b/c all sectors labelled by δ are equivalent and there is no a priory reason to select δ=0 - except for "aesthetic prejudices".

Please note that this δ changes the spectrum of the momentun = angular momentum operator

$$-i\partial_\theta \, u_{n,\delta}(\theta) = (n + \delta)\,u_{n,\delta}(\theta)$$

and is therefore not irrelevant physically.

Instead of introducing wave functions with twisted boundary conditions we can use wave functions with δ=0 but a "shifted" momentum operator

$$p_\theta \to -i\partial_\theta + \delta$$

which has the same effect when acting on wave functions with δ=0, i.e. a shift in the momentum. Please note that this new momentum operator has the same commutation relations b/c the constant δ does not affect them.

I think this is another example where quantum mechanics cannot be derived from classical mechanics w/o ambiguities. Therefore we do not arrive at "one quantum theory for a particle on a circle" but at a "familiy of inequivalent quantum theories labelled by δ".