# Quantiation rule

1. Jan 29, 2007

### xboy

Are the quantization rules for momentum and energy axioms of quantum mechanics? Do we have such quantization rules because the expressions for energy,momentum and angular momentum they furnish are conserved?Or are there any other reason?

2. Jan 29, 2007

### dextercioby

Nope, the Born-Jordan commutation relatons are axiomatized, not the quantization prescrptions for classical observables.

3. Jan 29, 2007

### xboy

so what's the deal here?why do we quantize the classical variables in the prescribed way and no other?

4. Jan 29, 2007

### dextercioby

Because they're essentially unique, by the Stone-Von Neumann theorem. The only trouble we get is when ordering comes into place for noncommuting operators.

5. Jan 29, 2007

### xboy

ok,so why do we have the operator for momentum that we have?why that particular operator and no other?

6. Jan 29, 2007

### StatMechGuy

Dexter, I like your answers, but they are so mathematical it gets confusing at times. Anyway...

Okay, so let's just pick the standard position space representation of momentum and position operators, and a wave function. Now suppose that instead of $$p = -i \hbar \partial_x$$ we pick $$p = - i \hbar \partial_x + f(x)$$. This most certainly satisfies the canonical commutation relations. But we also like to have momentum be the generator of space translations, so that
$$\psi(x + a) = \exp[i p a/\hbar] \psi(x)$$
Okay, now pretend that our momentum operator is the new one, and we want to look at the momentum operator acting on the translated wave function. Since all we did was shift our coordinates, we expect the action of the momentum on the wave function to be the same, yes?

Well,
$$-i \hbar \partial_x \psi(x + a) = -i \hbar \partial_x \exp[i/\hbar (-i \hbar \partial_x + f(x) ) a] \psi(x)$$
Do you see that this adds terms that go like $$f'(x)$$ to the momentum operator? This is why you can't add an arbitrary function of x to the momentum operator. If you add a constant to it, all you do is change the wave function by a phase that you can't detect anyway, so this gets the job done.

Last edited: Jan 29, 2007
7. Jan 30, 2007

### xboy

No,my question is why do we call it 'momentum'?what has this got to do with momentum as defined in classical mechanics?

8. Jan 30, 2007

### dextercioby

Then why do we call

$$\hat{x}=i\hbar\frac{\partial}{\partial p}$$

"coordinate", what does it have to do with the "x" in classical mechanics ?

9. Jan 30, 2007

### masudr

In classical mechanics, we have $\{x,p\}=1$, and since we have the similar relation $[x,p]=i\hbar$ in quantum mechanics, we are inspired to identify position and momentum with those operators.

Furthermore, momentum is the conserved quantity given space translation symmetry in both formalisms, and so again we are inspired to define them as we have done.

10. Jan 30, 2007

### Gokul43201

Staff Emeritus
Did you read the post above this carefully? Classically, momentum is the generator of infinitesimal translations. You want to retain that in the quantum case.

11. Jan 30, 2007

### reilly

There are a lot of clues in the contact transformation approach to classical mechanics -- see Goldstein, for example. But, I like to think of the formulation of operator-based QM as inspired guess work. Note that it took many brilliant physicists quite a few years to get this stuff straight. Indeed it is not obvious.

I've been told that I'm in the hard-nosed, empirically based group of physicists -- watch out, we may rule the world someday. That is, we use the operators for p, x, J, etc, because they work, and are at the very essence of QM. My personal bias is that axioms are great in math, but not appropriate for physics. I prefer "well tested assumptions" as the catch all phrase. Strictly a matter of taste.
Regards,
Reilly Atkinson