Proving the Commutator w/Taylor Expansion: Rick's Problem

[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

 

[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.

 

[kcirick]

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

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.

 

[Hurkyl]

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

 

[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.

 

[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?

 

[dextercioby]

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

Daniel.

 

[kcirick]

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

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!

 

[StatMechGuy]

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

Daniel.

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.

 

[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".