[x,p] = xp - px = ih

  • Thread starter leonmate
  • Start date
  • #1
84
1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
Doc Al
Mentor
44,892
1,144
I can see that the px = i * hbar
Really? Show how you got that.
 
  • #3
309
38
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
84
1
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
Doc Al
Mentor
44,892
1,144
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.
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
3
0
Bit aside, how arises I in p and x?
 
  • #7
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
Bit aside, how arises I in p and x?
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:
  • #8
dextercioby
Science Advisor
Homework Helper
Insights Author
12,981
540
Bit aside, how arises I in p and x?
Have you heard about a Poisson bracket?
 
  • #9
98
5
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
 
  • #11
3
0
Derivative is Ok, but what for I ? Moreover, how imaginary valued are connected with physical?
 
  • #12
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
Derivative is Ok, but what for I ?
What do you get when you use the product rule to evaluate the last expression in post #7?

Moreover, how imaginary valued are connected with physical?
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.
 

Related Threads on [x,p] = xp - px = ih

  • Last Post
2
Replies
27
Views
2K
  • Last Post
Replies
22
Views
7K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
640
  • Last Post
Replies
1
Views
935
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
23
Views
4K
Replies
8
Views
945
Top