# Momentum operator and comutators

1. Aug 9, 2007

### Ene Dene

I'm having trouble understanding the folowing:
(I'll write h for h/(2Pi))
Momentum in x direction is represented by operator
$$p=-ih\frac{d}{dx}$$.
So comutator
$$[x,p]=xp-px=x(-ih)\frac{d}{dx}-(-ih)\frac{dx}{dx}=0+ih*1=ih.$$
Now here comes the part that I don't understand. I'll calculate [x,p^2]:
$$[x,p^2]=xp^2-p^2x=xpp-ppx$$,
but from:
$$[x,p]=xp-px=ih$$
I have:
$$[x,p^2]=xpp-ppx=(ih+px)p-p(-ih+xp)=2ih+pxp-pxp=2ih$$
As it should be. But why can't I say:
$$p^2=pp$$,
$$(-ih)\frac{d}{dx}*(-ih)\frac{d}{dx}=-h^2\frac{d^2}{dx^2}$$
And write:
$$[x,p^2]=xp^2-p^2x=x*(-h^2\frac{d^2}{dx^2})+h^2\frac{d^2}{dx^2}(x)=0+0=0$$

2. Aug 9, 2007

### Dick

I don't know how you are reducing the last line to zero. [x,p^2] in this representation is an operator. It needs to operate on something. Work out [x,p^2](f(x)) for an arbitrary function f(x).

3. Aug 9, 2007

### malawi_glenn

try to apply the comutator on a arbitrary wavefunction$$\psi$$. Remember that $p^2$ is not the p-operator squared, it is performed twice on a wavefunction:

$$p^2 \psi=-ih\frac{d}{dx} \left(-ih\frac{d}{dx} \psi \right)$$

Edit: Dick was 5 seconds before me =)

4. Aug 9, 2007

### Ene Dene

Yes, of course, I understand now! Thank you!

5. Aug 10, 2007

### olgranpappy

the third equal sign is the above equation is incorrect (you didn't distribute the p correctly). The equation should read:

$$[x,p^2]=xpp-ppx=(ih+px)p-p(-ih+xp)=2ihp+pxp-pxp=2ihp$$

In general
$$[x,f(p)]=i\hbar f'(p)$$
and the above is a special case with f(p)=p^2.

Last edited: Aug 10, 2007
6. Aug 10, 2007

### CompuChip

You better also had done the first one that way, as not to get confused by what the differential operator acts on.

Let f denote a function of x, then
$$[x,p]f = xpf-pxf=x(-ih)\frac{df}{dx}-(-ih)\frac{dxf}{dx}= -i h x f' + i h (f + x f') = i h f$$
so dropping the function we see $$[x, p] = i h$$.

And this method will give you the correct result for any commutator (just remember, the differentiation operator works as far as possible to the right, so $\frac{d}{dx} x f$ really is $\frac{d (x f)}{dx}$ and not $\left( \frac{dx}{dx} \right) \cdot f$).

Last edited: Aug 10, 2007
7. Jan 28, 2011

### Geometrick

How come there is no h^2 term after we do it this way? Using the definition, I don't see why I only get an h and not h^2.

Last edited: Jan 28, 2011