Commutation of position operators in three dimensions

[hokhani]

TL;DR

How to prove commutation of x, y and z operators

I don't know how to prove that the operators $\hat{x}$, $\hat{y}$ and $\hat{z}$ commute?

 

[PeterDonis]

I don't know how to prove that the operators $\hat{x}$, $\hat{y}$ and $\hat{z}$ commute?

Compute the commutator of any two of them on an arbitrary wave function. Does it vanish?

 

[Demystifier]

TL;DR Summary: How to prove commutation of x, y and z operators

I don't know how to prove that the operators $\hat{x}$, $\hat{y}$ and $\hat{z}$ commute?

It's an axiom of QM, it can't be proved. There are theories (non-commutative geometry) in which they don't commute.

 

[hokhani]

Compute the commutator of any two of them on an arbitrary wave function. Does it vanish?

Without implicit assumption of commutativity we can not obtain your result.

 

[hokhani]

It's an axiom of QM, it can't be proved. There are theories (non-commutative geometry) in which they don't commute.

Can we justify it intuitively by considering the fact that moving in one direction doesn't affect the other components? Say, we can move in x direction, keeping constant the other coordinations so the components are independent.

 

[Demystifier]

Can we justify it intuitively by considering the fact that moving in one direction doesn't affect the other components? Say, we can move in x direction, keeping constant the other coordinations so the components are independent.

Classical intuition may not be a good guide towards quantum mechanics.

You can "derive" it from the principle that QM is obtained from classical mechanics by replacing Poisson brackets with commutators. But then you may ask how to derive that principle.

 

[PeterDonis]

Without implicit assumption of commutativity we can not obtain your result.

Sure we can. Take $\hat{x}$ and $\hat{y}$ as an example. We pick the position representation (commutators are representation independent, so we can compute them in whatever representation is easiest), where we know the actions of these operators are

$$\hat{x} \psi = x \psi$$

$$\hat{y} \psi = y \psi$$

Then the commutator is

$$
[ \hat{x}, \hat{y} ] \psi = \left( xy - y x \right) \psi = 0
$$

This is the case because $x$ and $y$, the coordinates, are real numbers, and multiplication of real numbers commutes.

 

[PeterDonis]

It's an axiom of QM

The axiom is not that the given operators commute; that can be proved from our knowledge of the action of those operators in the position representation, as I did in my previous post.

The axiom, if you want to call it that, is that the coordinates in the position representation are real numbers.

There are theories (non-commutative geometry) in which they don't commute.

Do you have a reference?

 

[Demystifier]

Do you have a reference?

See e.g. the review by Szabo and references therein:
https://arxiv.org/abs/hep-th/0109162

 

[Demystifier]

Sure we can. Take $\hat{x}$ and $\hat{y}$ as an example. We pick the position representation (commutators are representation independent, so we can compute them in whatever representation is easiest), where we know the actions of these operators are

$$\hat{x} \psi = x \psi$$

$$\hat{y} \psi = y \psi$$

Here $\psi=\psi(x,y)=\langle x,y|\psi\rangle$. Thus you implicitly assume that the position representation exists, i.e. that $\hat{x}$ and $\hat{y}$ have simultaneous eigenstates $|x,y\rangle$. This is practically the same as assuming that $\hat{x}$ and $\hat{y}$ commute.

 

[anuttarasammyak]

I don't know how to prove that the operators x^, y^ and z^ commute?

Not a mathematical proof but we observe that measurement of x does not affect measurement of y, etc.

PS
In SR, very precice measurement of x of a prticle would require more than 2mc^2 energy input to the system and cause particle anti-particle pairs creation. Thus measurement of not only x but also y and z would become meaningless thereafter.

 

[PeroK]

@hokhani I think it's most important to understand the linear algebra that is fundamental to QM.

This proof is really a test of whether you understand how things are defined. The proof, as shown already in this thread, comes naturally from the definitions of "commute" and the $\hat x, \hat y$ operators.

 

[PeterDonis]

you implicitly assume that the position representation exists, i.e. that $\hat{x}$ and $\hat{y}$ have simultaneous eigenstates $|x,y\rangle$.

Isn't that part of the definition of the operators themselves?

 

[cianfa72]

I believe in post #7, using the position representation (i.e. the uncountable set of Dirac pulses centered at $(x,y)$ as Hilbert basis in position/momentum state space) the following $$\hat{x} \psi := x \psi$$ $$
\hat{y} \psi := y \psi$$
are basically the definitions of the position operators $\hat{x}$ and $\hat{y}$ respectively. Using those definitions one can prove, as you did, that they actually commute.

 

[Demystifier]

Isn't that part of the definition of the operators themselves?

It's a kind of a chicken or egg question. Yes, you can start from that definition and then derive that the operators commute. Or, you can first postulate that they commute and then derive that the operators can be represented in the position picture. Historically, Heisenberg's paper appeared before the works of Schrödinger and Dirac, so the postulate of commutativity came first. From a modern mathematical physics point of view, I think it's quite common to think of commutativity as an axiom, rather than something derived from the coordinate picture.