# 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..