# [x,p] = xp - px = ih

1. Dec 30, 2014

### leonmate

xp - px = x * (-i*hbar*d/dx) - (-i * hbar * d/dx (x)) = i * hbar

I can see that the px = i * hbar
But, why does xp = 0 ... ? I just get - i * hbar everytime I try, Wolfram Alpha agrees...

Can anyone show me why this isn't the case please?

2. Dec 30, 2014

### Staff: Mentor

Really? Show how you got that.

3. Dec 30, 2014

### bolbteppa

The annoying thing about commutators is that, because they are operators, when you do it out explicitly you have to apply them to a function to make them work out, so apply [x,p] to f and see what happens.

4. Dec 30, 2014

### leonmate

Ok,

What I've been doing is just d/dx ( const * x ) = x
That works when we apply p to x. But not x to p I suppose.

5. Dec 30, 2014

### Staff: Mentor

These are operator relationships. To calculate what you call "px", you need to evaluate:
$$\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi$$
Note that you apply x first, then the derivative. So you'll need to use the product rule.

6. Dec 30, 2014

### mac_alleb

Bit aside, how arises I in p and x?

7. Dec 30, 2014

### Fredrik

Staff Emeritus
It shows up when you apply the product rule for derivatives:
$$(\hat p\hat x f)(x)=(\hat p(\hat x f))(x) = -i(\hat x f)'(x) =-i\frac{d}{dx}(\hat x f(x)) =-i\frac{d}{dx}(xf(x))=\cdots.$$

Last edited: Dec 30, 2014
8. Dec 30, 2014

### dextercioby

Have you heard about a Poisson bracket?

9. Dec 30, 2014

### USeptim

Hi,

Roughtly speaking, the derivative operator "p" acts over everything to the right and the wave function it's always implicitly the last term, so you really have:

pxΨ= i * hbar ∂x(xΨ) = i * hbar * [x * ∂x(Ψ) + Ψ * ∂x(x) ]

Where Ψ is the wave function in the evaluated point, implicitly, Ψ(x).

x(x) = 1 so the conmutation is [x,p]Ψ = (xp - px)Ψ = i*h_bar Ψ.

Best regards,
Sergio

10. Dec 30, 2014

### Staff: Mentor

11. Jan 4, 2015

### mac_alleb

Derivative is Ok, but what for I ? Moreover, how imaginary valued are connected with physical?

12. Jan 4, 2015

### Fredrik

Staff Emeritus
What do you get when you use the product rule to evaluate the last expression in post #7?

There are many such connections. For example, $|\psi(x)|^2\Delta x$ is approximately equal to the probability that a particle detector near $x$ that's covering a region of size $\Delta x$ will detect a particle.