# How to get the commutation relation of q and p

1. Apr 27, 2014

### chern

We all know that quantum theory is based on the commutation relation and superposition principle. The trouble haunting me long time is that how to "get" the famous commutation relation? Could anybody give me an explanation?

2. Apr 27, 2014

### vanhees71

The commutation rules follow from symmetry considerations. The basic quantities are defined as generators of the symmetries of space and time. The components of momentum wrt. to a Cartesian basis build the generators for spatial translations. This implies the commutation relations with a position operator, provided the latter exists. This is the case for all massive particles and for massless particles with spin $\leq 1/2$.

3. Apr 27, 2014

### atyy

The commutation relation is a fundamental postulate and cannot be derived.

One way to guess is it is from the classical commutation relation (Poisson brackets) of canonically conjugate observables. Also, from the uncertainty relation. However, since quantum mechanics is more fundamental, we now usually see the classical commutation relation between canonically conjugate variables as the classical limit of the quantum commutation relation. The uncertainty relation is also a consequence of the commutation relation.

The violation of Bell's inequality is nowdays seen as summarizing quantum weirdness, and within quantum mechanics the violation requires non-commuting observables and entanglement.

4. Apr 27, 2014

### MisterX

Well it depends on what you mean by p and q, I suppose. I'm going to assume you mean to ask about the commutation for regular position and momentum.

It has to do with the basic construction of quantum mechanics along with

momentum $\Leftrightarrow \hbar \mathbf{k}$ (wavevector)
Energy $\Leftrightarrow \hbar \omega$ (frequency)

This leads us to the Schrodinger equation. We simply define the operator $\mathbf{p}$ so that it extracts the momentum from a plane wave $e^{ik x - i\omega t}$

So we want the operator $p$ to extract the k as a coefficient like so:
$p\, e^{ikx - i\omega t} = \hbar k \,e^{ikx - i\omega t}$

This can be done if we define p as
$p \equiv -i\hbar \frac{\partial}{\partial x}$

Since the derivative brings down a factor of $ik$, and$-i\hbar \, ik = \hbar k$.

This results in the commutation relations for position and momentum. Note that $\frac{\partial}{\partial x} (x f(x)) \neq x \frac{\partial}{\partial x} f(x)$.

\begin{align*} (xp - px) f(x) &= -i\hbar\left( x \frac{\partial}{\partial x} f(x) - \frac{\partial}{\partial x} (x f(x)) \right) \\ &=-i\hbar \left( x \frac{\partial}{\partial x} f(x) - \left[f(x) + x \frac{\partial}{\partial x} f(x)\right] \right) \\
& = -i\hbar\left(-f(x) \right) \\ &= i\hbar f(x)\end{align*}

So we conclude then that $\left[x,\, p\right] = i\hbar$. The product rule was used to get from the first line to the second line.

5. Apr 27, 2014

### Staff: Mentor

That's not quite true - strictly speaking it is, but many would say its determined by the symmetries of the POR ie quantum probabilities are invariant between frames. Actually its the same reason for classical mechanics as well except the symmetries are in the Lagrangian and not probabilities - but that is another story.

See chapter 3 - Ballentine - Quantum Mechanics - A Modern Development:
https://www.amazon.com/Quantum-Mechanics-A-Modern-Development/dp/9810241054

As usual Vanhees is spot on.

Thanks
Bill

Last edited by a moderator: May 6, 2017
6. Apr 28, 2014