# Commutator Problem

1. Oct 17, 2006

### kcirick

Question:
If g(p) can be Taylor expanded in polynomials, then prove that:

$$\left[x, g\left(p\right)\right] = i\hbar \frac{dg}{dp}$$

To start, I multiply the wave function $\Psi$ and expand the commutator:

$$\left( xg\left(p\right)-g\left(p\right)x \right)\Psi$$

then expand g(p) using Taylor:

$$x\left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)\Psi + \left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)x\Psi$$

Then I'm stuck... I'm nowhere near getting $i\hbar$ or $\frac{dg}{dp}$

Any help would be appreciated. Thank you!
-Rick

2. Oct 17, 2006

### StatusX

Prove that's true for g(p)=p^n. Then write out the taylor series of g and apply the commutator to each term.

3. Oct 17, 2006

### kcirick

That was my previous question... I already proved that

$$\left[x, p^{n}\right]=ihnp^{n-1}$$

But I don't see how that is helpful or how I'm going to use that to do my original question.

4. Oct 17, 2006

### Hurkyl

Staff Emeritus
g(a) + g'(a) (p - a) is almost never the Taylor expansion of g(p).

5. Oct 17, 2006

### StatusX

[a,b+c]=[a,b]+[a,c]. Use this to show the formula is true for any polynomial. Then take the limit to show it's true of the taylor series (ie, $g(p)=a_0 + a_1 p + a_2 p^2 + ...$). This part might be a little tricky, and I don't know how rigorous you need to be, but to show two operators are equal, you need to show they agree on every function.

6. Oct 18, 2006

### kcirick

Solution

So here's what I have:

Taylor expansion of g(p) can be expressed as the following:
$$g\left(p\right) \approx a_{o}+a_{1}p+a_{2}p^{2}+\cdots+a_{n}p^{n}$$
So applying the commutator:
$$\left[x, g\left(p\right)] = [x, a_{o}+a_{1}p+a_{2}p^{2}+\cdots+a_{n}p^{n} = \left[x, a_{o}\right]+\left[x, a_{1}p\right]+\left[x, a_{2}p^{2}\right]+\cdots+\left[x, a_{n}p^{n}\right]$$
$$= 0 + i\hbar a_{1} \frac{dp}{dx} + i\hbar a_{2} \frac{dp^{2}}{dx} + \cdots + i\hbar a_{n} \frac{dp^{n}}{dx}$$
since $\left[x, p^{n}\right] = i\hbar\frac{dp^{n}}{dx}$.

So:
$$\left[x, g\left(p\right)\right] = i\hbar\frac{d}{dx}\left(a{o}+a_{1}p+a_{2}p^{2}+\cdots+a_{n}p^{n}\right)=i\hbar\frac{dg\left(p\right)}{dx}$$

I think I did that okay. What do you think?

7. Oct 18, 2006

### dextercioby

Evaluate the Poisson bracket between the 2 functions and then quantize the result via Dirac's prescription.

Daniel.

8. Oct 18, 2006

### kcirick

I thought I had it, but it turned out that I was getting the wrong answer. Here's what I did:

$$\left[x, p^{n}\right]\Psi = \left(xp^{n} - p^{n}x\right)\Psi = x\left(-i\hbar\frac{d}{dx}\right)^{n}\Psi - \left(-i\hbar\right)^{n} \left(\frac{d^{n}x}{dx^{n}}\Psi + x\frac{d^{n}}{dx^{n}}\Psi\right)$$
$$= -\left(-i\hbar\right)^{n}\frac{d^{n-1}}{dx^{n-1}}\Psi = i\hbar\left(-i\hbar\frac{d}{dx}\right)^{n-1}\Psi$$

But then I get $i\hbar p^{n-1}$ as an answer, and I should be getting $i\hbar np^{n-1}$ . Where did I go wrong?

Thanks!

9. Oct 18, 2006

### StatMechGuy

This works too, but if you want to do it purely in the framework of QM, I'd recommend trying to do this in the momentum representation, where $$x = - i \hbar \partial_x$$ and have your trial function sitting up front.

Commutators are invariant under unitary changes of coordinates.

10. Oct 18, 2006

### StatusX

The idea is to start from the fundamental commutation relation $[x,p]=-i\hbar$ and not worry about what p is. Then there is a rule that [A,BC] = [A,B]C + B[A,C], so you can use induction to get $[x,p^n]=-i\hbar n p^{n-1}$. Finally, apply this to the taylor series, and you'll end up with what they want you to prove. But this shows that $x=-i\hbar\partial/\partial p$, and symmetrically, $p=-i\hbar\partial/\partial x$, so from that simple commutation relation you've found what p "is".

Last edited: Oct 18, 2006