Very Basic QM problem: Commuter of position and momentum operators

[Pinu7]

I'm not exactly sure if this belongs in introductory or advanced physics help.

Homework Statement

In my book, the author was explaining the proof of the Uncertainty relation between po
position and momentum.

It simply stated that [x,p]= ih(h is reduced)
But when I tried to verify it I got -ih. I now it would give the same result, but it still won't be good for me to mess up a fundamental concept so early.

Homework Equations

$$\hat{p}$$=-ih d/dx
$$\hat{x}$$=$$\hat{x}$$
[A,B]=AB-BA

The Attempt at a Solution

[x,p]$$\left|\psi$$>=(xp-pa)$$\left|\psi$$>=xp$$\left|\psi$$>-pa$$\left|\psi$$>

It would become this:
-ihx d$$\psi$$/dx -ihx-(-ihx d$$\psi$$/dx)=
-ihx

Which is not the answer my book gave me.

 

[Dick]

Your answer isn't right. The px part means you have to differentiate x*psi. Use the product rule.

 

[Tian WJ]

Hello, Pinu7 ! I wish I could do some help.

Firstly, I'd like to tell you that, the given attempt solution is incorrect, for the sign of $i\hbar x$ is plus rather than minus. It is a slight carelessness in extending the derivative of product. Now, I will re-perform the calculating in details. And please translate the tex codes yourself.

For one-dimensional simplified case, as put forward in the question (or for the x-component of 3-dimensional analyses):
$$
\hat{x}=x
$$
$$
\hat{p}= -i\hbar \frac{d}{dx}
$$
Hence, with the quantum Poisson bracket operator:

[\hat{x}, \hat{p}] \phi = \hat{x} \hat{p} \phi - \hat{p} \hat{x} \phi

= -i\hbar x \frac{d}{dx} \phi -( -i\hbar \frac{d}{dx} (x \phi))

= -i\hbar x \frac{d}{dx} \phi -( -i\hbar \phi -i\hbar x \frac{d}{dx} \phi ) (Look Out! Your Carelessness Happens Here!)

= i\hbar \phi

Just as the textbook gives.