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[x,p] = xp - px = ih

  1. Dec 30, 2014 #1
    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. jcsd
  3. Dec 30, 2014 #2

    Doc Al

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    Really? Show how you got that.
  4. Dec 30, 2014 #3
    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.
  5. Dec 30, 2014 #4

    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.
  6. Dec 30, 2014 #5

    Doc Al

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    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.
  7. Dec 30, 2014 #6
    Bit aside, how arises I in p and x?
  8. Dec 30, 2014 #7


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    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
  9. Dec 30, 2014 #8


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    Have you heard about a Poisson bracket?
  10. Dec 30, 2014 #9

    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,
  11. Dec 30, 2014 #10


    Staff: Mentor

  12. Jan 4, 2015 #11
    Derivative is Ok, but what for I ? Moreover, how imaginary valued are connected with physical?
  13. Jan 4, 2015 #12


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    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.
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