# Why don't all observables commute in QM?

1. Jul 7, 2012

### San K

In classical physics, all observables commute and the commutator would be zero.

However this is not true in Quantum Mechanics, observables like position and momentum (time and frequency/energy) don't commute. Why?

Is it because the (probability) wave functions/forms of position and momentum can never be both "squeezed" to high degree of accuracy at the same time?

Is it because the particles are point particles and we are dealing with single dimensions? in QM and somehow they are telling us something fundamental when you get down to single dimensions.

What does it mean not to commute? in QM/maths etc.

What is the difference between variables/matrices that commute and those that don't?

Do(es) time and space commute?

Last edited: Jul 7, 2012
2. Jul 7, 2012

### tom.stoer

The observables in QM are constructed vai a method called quantization. e.g. the classical observables x and p are quantized as x and -i∂/∂x. You can easily check that they dont commute, just calculate

[x,-i∂/∂x]ψ(x) = x(-i∂/∂x)ψ(x) + (i∂/∂x)(xψ(x)) = -ixψ'(x) + ixψ'(x) + (i∂/∂x x) ψ(x) = iψ(x)

There are other observables like angular momentum which are constructed in a similar way.

The question why this method works, and "why this method?" and not anything else is difficult to answer. There's a hint when looking at matter- or de-Broglie waves. Look at a wave exp(ikx) and act on this wave with the operator i∂/∂x; the resulting ik exp(ikx) indicates that k and therefore i∂/∂x have something to do with momentum. So this is an indication that the latter one can be interpreted as momentum operator.

But honestly speaking I don't think that anybody can answer the question why quantum mechanics constructed via canonical quantization works.

Last edited: Jul 7, 2012
3. Jul 7, 2012

### San K

thanks Tom.

Could it mean that observable that commute are fundamentally the same...and convertible (as we go deeper into our understanding of reality)?.....like various forms of mass-energy...

4. Jul 7, 2012

### tom.stoer

No, they need not be the same. For example the two operators x and y in two-dim. space do commute. And x and x² commute as well.

5. Jul 7, 2012

### lugita15

Here is a recent post of mine that you may find interesting:

6. Jul 7, 2012

### Dickfore

In classical mechanics, observables are not (linear and Hermitian) operators on a Hilbert space, but regular functions on the phase space (spanned by the 2s generalized coordinates q, and corresponding canonical momenta p). The role of a commutator is taken by a Poisson bracket:
$$\left\lbrace f, g \right\rbrace = \sum_{j = 1}^{s}{\left( \frac{\partial f}{\partial q_j} \, \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \, \frac{\partial g}{\partial q_j} \right)}$$
The transition from classical to quantum mechanics is formally done by:
$$\left\lbrace f, g \right\rbrace_{\mathrm{C.M}} \rightarrow \frac{1}{i \, \hbar} \, \left[ \hat{f}, \hat{g} \right]_{\mathrm{Q.M.}}$$

7. Jul 7, 2012

This.