In QM operators, why [YPz,YPx]=0, [ZPy,ZPy]=0 [X,Py]=0 [Y,Pz]=0, etc

1. Nov 14, 2007

mudyos

in the quantum mechanical operators :
why :
[YPz,YPx]=0
[ZPy,ZPy]=0
[X,Py]=0
[Y,Pz]=0
[Z,Py]=0

2. Nov 14, 2007

CompuChip

For the second:
What is the commutator of any operator with itself?

For the last three:
Can you prove that [X Px] = i h ?
Now do the same argument for [X Py].

For the first one:
Reduce it to Y^2 [Pz Px] using that [Y Pz] = [Y Px] = 0 (which you have just proven) and calculate [Pz Px] in the same way.

Last edited: Nov 14, 2007
3. Nov 14, 2007

pellman

Isn't [X Px] = i h an axiom, CompuChip?

4. Nov 14, 2007

OOO

Usually one thinks of this as something that nature implements (through de Broglie's relation). If one accepts it as a mathematical axiom a priori then the question "why?" wouldn't make much sense indeed.

5. Nov 14, 2007

mudyos

thank

6. Nov 15, 2007

CompuChip

Hmm, yes it is usually taken as an axiom isn't it.
But actually the way I first learned it was: the p-operator in x-space can be written as $$- i \hbar \partial/\partial x$$
and from this you can show that (indeed)
$$[x p_x] f = - i \hbar x \partial f/\partial x + i \hbar \partial/\partial x (x f) = \cdots = i \hbar f$$
and it was this reasoning that I wanted muydos to continue to [X Py] (which would then of course give 0).

7. Nov 16, 2007

mudyos

[Y,Px] = YPx-PxY
PxY=
$$P_x Y f(x) = i \hbar \partial /\partial x Y f(x)$$
$$P_x Y f(x) = i \hbar Y \partial f/\partial x + i \hbar \partial Y/\partial x f(x)$$
YPx=
$$Y P_x f(x) = i \hbar Y \partial f/\partial x$$

[Y,Px] = YPx-PxY
$$= i \hbar Y \partial f/\partial x - i \hbar Y \partial f/\partial x - i \hbar \partial Y/\partial x f(x)$$

[Y,Px] = $$- i \hbar \partial Y/\partial x f(x)$$
but
[Y,Px] = 0

$$- i \hbar \partial Y/\partial x f(x) = ?$$ is thes =0 . why?

8. Nov 16, 2007

mudyos

why [Y,Px] = 0

[Y,Px] = YPx-PxY
PxY=
$$P_x Y f(x) = i \hbar \partial /\partial x Y f(x)$$
$$P_x Y f(x) = i \hbar Y \partial f/\partial x + i \hbar \partial Y/\partial x f(x)$$
YPx=
$$Y P_x f(x) = i \hbar Y \partial f/\partial x$$

[Y,Px] = YPx-PxY
$$= i \hbar Y \partial f/\partial x - i \hbar Y \partial f/\partial x - i \hbar \partial Y/\partial x f(x)$$

[Y,Px] = $$- i \hbar \partial Y/\partial x f(x)$$
but
[Y,Px] = 0

$$- i \hbar \partial Y/\partial x f(x) = ?$$ is thes =0 . why?

9. Nov 16, 2007

ZapperZ

Staff Emeritus
dY/dx is zero because this "Y" is a position operator for the y coordinate. In other words, one can think of "Y" as a function of only y, and no other variable. Thus, its derivative with respect to other variable, such as x, is zero.

Zz.

10. Nov 16, 2007

mudyos

thak you

11. Nov 16, 2007

omyojj

physically..if certain obsevables commute, there exist states for which they all have particular values..(there exist eigenstate of p_x, y, |p_x>|y> having their eigenvalues p_x and y respectively..) So a mesurement of p_x, y would give only particular value p_x, y respectively..