# Very Basic QM problem: Commuter of position and momentum operators

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

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.