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How do I evaluate [x, SinPx] commutator

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate [x, SinPx] given [x, Px]=ih

    2. Relevant equations

    Px = h/I( d/dx)

    3. The attempt at a solution
    Let f (x) be a function of x.[ x, Sin Px ] f( x) ⇒ [ x sin{( h / i )d / dx} f ( x ) -Sin { (h/i) d / dx } x f(x)] . Does anybody concur .
     
  2. jcsd
  3. Apr 14, 2013 #2

    DrClaude

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    Staff: Mentor

    What you wrote is correct, but it does not answer the problem.
     
  4. Apr 14, 2013 #3
    What is the answer?
     
  5. Apr 15, 2013 #4

    DrClaude

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    Staff: Mentor

    I won't tell you that directly. You should start by looking at the Taylor series for the sine function.
     
  6. Apr 15, 2013 #5
    after a strenous ordeal Of 48 hr and heated debate with my pals and two sleepless nights, here is what I have come up with.we know in position space momentum oprtr is given -ih d/dx.what would be the opposite.what if I had acess to a position operator in momentum space.lets say x = ih d/dPx.then what happens.
    This is what happens :[x , sinPx]f (Px)= [ ih (d / dPx) , sin Px ]f (Px) = ih [ d /dx ( fSin Px )-sin Px d/dx( f )] =
    ih [ cos Px f(Px) + sinPx f' (Px)-sin Pxf' (Px) ] = ih cos Px f(Px) . Thrfr [ x, sin Px ] = ih Cos Px
    does Anybody agree ??
     
    Last edited: Apr 15, 2013
  7. Apr 15, 2013 #6

    DrClaude

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    There are a few typos along the way (##d/dx## instead of ##d/dP_x##), but it seems ok. You should also try using the Taylor series for the sine. You should find that
    $$
    \sin \left(\frac{\hbar}{i} \frac{d}{dx} \right) x = x \sin \left(\frac{\hbar}{i} \frac{d}{dx} \right) + \frac{\hbar}{i} \cos \left(\frac{\hbar}{i} \frac{d}{dx} \right)
    $$
    which leads you to the same answer.
     
  8. Apr 15, 2013 #7
    I know.I did it that way earlier infact. But this method is more compact and satisfying as it doesnt involves any approximation.You can google out POSITION OPERATOR IN MOMENTUM SPACE.there are very good articles out there upon this.as i am a beginner today I am very happy when I came to know about this concept.
     
  9. Apr 15, 2013 #8

    DrClaude

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    Staff: Mentor

    The method I used is not approximate. You never truncate the Taylor series, you let it go to infinity, although you do not need to calculate all the terms, as the pattern is easily identifiable. It is more general, as the momentum space approach only worked because the differential operator wasn't in a function anymore. What if you had to calculate ##[ \cos(x), \sin(P_x)]##? You will also encounter things such as the exponential of operators, and the only correct way to express them is through their Taylor series.
     
  10. Apr 15, 2013 #9
    Thanks for the insight.
     
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