Commutation relation of the position and momentum operators

[Newbie says Hi]

Homework Statement

I've just initiated a self-study on quantum mechanics and am in need of a little help.

The position and momentum operators do not commute. According to my book which attemps to demonstrate this property,
(1) $$\hat{p} \hat{x} \psi = \hat{p} x \psi = -i \hbar \frac{\partial}{\partial x}(x \psi)$$

OK, so far I was following. I'm not understaing how they get to this next equation from the previous one...
(2) $$= -i \hbar \left( \psi + x \frac{\partial \psi}{\partial x}\right)$$

To be more precise, where did the second psi symbol come from and where did the + symbol come from in equation 2?

Homework Equations

$$\hat{x} = x$$
$$\hat{p} = -i \hbar \frac{\partial}{\partial x}$$

The Attempt at a Solution

When I attempt to multiply the momentum and position operators, I get:
(3) $$\hat{p} \hat{x} \psi = \hat{p} x \psi = \left( -i \hbar \frac{\partial}{\partial x} \right) x \psi$$

But how do I go from equation 3 and arrive at equation 2? Or can I?

 

[Meir Achuz]

psi also depends on x. This leads to the second derivative.
You may need a live teacher to learn QM.

 

[Newbie says Hi]

psi also depends on x. This leads to the second derivative.
You may need a live teacher to learn QM.

OK, I understand the first part, but I am not following why it "leads to the second derivative." Could you please explain in more detail? Or preferably, could you show me the next few steps after equation (1)?

 

[dextercioby]

What book are you using ? Throw it away. One postulates the commutation relations, they cannot be proved.

 

[Newbie says Hi]

What book are you using ? Throw it away. One postulates the commutation relations, they cannot be proved.

The author of the book is demonstrating that $$\hat{p} \hat{x} \psi does NOT equal \hat{x} \hat{p} \psi$$ .

Sorry if I miscommunicated this point in the first post.

 

[Newbie says Hi]

I understand the solution to why $$\hat{x} \hat{p} \psi$$ equals what it equals. My question just has to do with how the author gets the solution that he does for the $$\hat{p} \hat{x} \psi$$

 

[cristo]

Are you asking how you get from the left to the right hand side of this? $$\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi = -i \hbar \left( \psi + x \frac{\partial \psi}{\partial x}\right)$$.

If so, then note that $x\psi$ is a product, each of which is dependent on x, and so we must use the product rule. So, starting from the left $$\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi =-i\hbar\frac{\partial}{\partial x}(x\psi)=-i\hbar(x\frac{\partial\psi}{\partial x}+\psi)$$

 

[nrqed]

Homework Statement

I've just initiated a self-study on quantum mechanics and am in need of a little help.

The position and momentum operators do not commute. According to my book which attemps to demonstrate this property,
(1) $$\hat{p} \hat{x} \psi = \hat{p} x \psi = -i \hbar \frac{\partial}{\partial x}(x \psi)$$

OK, so far I was following. I'm not understaing how they get to this next equation from the previous one...
(2) $$= -i \hbar \left( \psi + x \frac{\partial \psi}{\partial x}\right)$$

To be more precise, where did the second psi symbol come from and where did the + symbol come from in equation 2?

It's simply the product rule. If $f (x)$ is a function of x, then $$\frac{d}{dx} ( x f(x) ) = 1 \times f(x) + x \times \frac{df(x)}{dx}$$
, right?

What book are you using? I would highly recommend Shankar and, as a supplement, Griffiths.

 

[Newbie says Hi]

Great. Thanks a LOT cristo and nrqed!

I love this site! :)

What book are you using? I would highly recommend Shankar and, as a supplement, Griffiths.

I am currently using two books: "Quantum Mechanics for Undergraduates" by Norbury as well as Griffiths (both books use the same template and complement each other strongly). I also have Shankar, however, after going through the first chapter, I found it was too math heavy... perhaps it is aimed at graduates?