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Commutation of differential operators

  1. Aug 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate the commutator [itex]\left[\frac{d}{dx},x\right][/itex]
    2. Relevant equations
    [itex]\left[A,B\right]=AB-BA[/itex]
    3. The attempt at a solution
    [itex]\left[\frac{d}{dx},x\right]=1-x\frac{d}{dx}[/itex]

    I don't know how to figure out [itex]x\frac{d}{dx}[/itex].
     
  2. jcsd
  3. Aug 24, 2012 #2
    Try using a test function here i.e. calculate [d/dx,x]f(x) and when you are done make sure you get some expression (...)f(x) again then what is between the brackets is your commutator.
     
  4. Aug 24, 2012 #3
    I actually just figured this out.
    Can somebody please just check this one I did for mistakes?

    [A,B]=[(d/dx + x),(d/dx - x)]
    (AB-BA)f=(d/dx + x)(d/dx - x) f -(d/dx - x)(d/dx + x) f
    (d/dx + x)(df/dx - xf)- (d/dx - x)(df/dx + xf)
    (df^2/dx^2 - f - x df/dx)+(x df/dx - x^2f)-(df^2/dx^2 + f +x df/dx)+(x df/dx + x^2f)
    =-2f

    --> [(d/dx + x),(d/dx - x)]=-2
     
  5. Aug 24, 2012 #4

    HallsofIvy

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    If f is any function then (d/dx)(x f)=x df/dx+ f while x(d/dx)f= xdf/dx. The difference is f so the commutator, [d/dx, x] is the identity operator.
     
  6. Aug 24, 2012 #5

    vela

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    That's fine.

    You could have also used a few properties of commutators to save yourself some work, namely [A+B,C]=[A,C]+[B,C] and [A,B]=-[B,A]. Using those, you should be able to show that [(d/dx + x), (d/dx - x)] = 2[x, d/dx], and you've probably already worked out that last one and found it to be equal to -1.
     
  7. Aug 25, 2012 #6
    Very helpful. Thanks!
     
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