Understanding the Commutator [x, p^2]?

[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

 

[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).

 

[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 =)

 

[Ene Dene]

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

 

[olgranpappy]

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

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

<br /> [x,p^2]=xpp-ppx=(ih+px)p-p(-ih+xp)=2ihp+pxp-pxp=2ihp<br />

In general
<br /> [x,f(p)]=i\hbar f&#039;(p)<br />
and the above is a special case with f(p)=p^2.

 

[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&#039; + i h (f + x f&#039;) = 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).

 

[Geometrick]

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 =)

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.