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Finding an Operator .

  1. Oct 7, 2014 #1
    Hello--I am practicing for the upcoming quiz (as part of quantum physics), but have no idea how to solve the sample problem the teacher gave (w/ solution)...

    Write out the operator A2 for A = d/dx + x
    (Hint: INCLUDE f(x) before carrying out the operations)

    so...I tried:

    A {df(x)/dx + f(x)} = d/dx{df(x)/dx + f(x)} + df(x)/dx + f(x)
    = d2f(x)/dx2 + df(x)/dx + df(x)/dx + f(x)
    = d2f(x)/dx2 + 2df(x)/dx + f(x)

    ...when the solution is A[SUP]2[/SUP] = d2/dx2 + 2xd/dx + 1 + x2

    ...where did I make mistake?????? Could someone please point out? Thanks!
     
  2. jcsd
  3. Oct 8, 2014 #2

    wle

    User Avatar

    Here, right at the beginning:
    You've expanded ##A f## as ##\frac{\mathrm{d}}{\mathrm{d}x} f + f##. It should be ##A f = \frac{\mathrm{d}}{\mathrm{d}x} f + x f##.
     
  4. Oct 8, 2014 #3
    Thanks! But, I still have trouble getting all the solution values:

    A f(x) = df(x)/dx + xf(x)
    = d (df(x)/dx + xf(x))/dx + x df(x)/dx + x2f(x)
    = d2f(x)/dx2 +x df(x)/dx + x df(x)/dx + x2f(x)
    = d2f(x)/dx2 +2x df(x)/dx + x2f(x).......I could NOT get (+1) part from the solution: A2 = d2/dx2 + 2xd/dx + 1 + x2

    ...How do I get (+1) value? Did I leave out some values while I tried to solve for A2?
     
  5. Oct 8, 2014 #4

    DrClaude

    User Avatar

    Staff: Mentor

    You didn't apply the first ##\frac{d}{dx}## correctly between the last two lines.
     
  6. Oct 8, 2014 #5
    Wait, thank you for correcting my first mistake, but I still don't see how I got the last two lines wrong.....
     
  7. Oct 8, 2014 #6
    Thank you for correcting my first mistake, forgetting to add A2 f(x), but I still cannot see how I got last two lines wrong

    .....could you pls clarify more? Thanks.
     
  8. Oct 8, 2014 #7

    DrClaude

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    Staff: Mentor

    You should get two terms out of
    $$
    \frac{d}{dx} \left( x f(x) \right)
    $$
     
  9. Oct 8, 2014 #8
    So, I should have gotten d(xf(x))dx, NOT x df(x)/dx?

    d(xf(x))dx = xdf(x)/dx + df(x)/dx?

    I redid from the beginning, but I still cannot get (+1) part:

    A2f(x) = d2f(x)/dx2 + df(x)/dx + 2xdf(x)/dx + x2f(x)

    I got df(x)/dx INSTEAD of 1......or rather f(x) in the case of A2f(x)....
     
    Last edited: Oct 8, 2014
  10. Oct 8, 2014 #9

    DrClaude

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    Staff: Mentor

    I'm not sure what you mean here. You have written yourself the first term of ##A^2 f(x)## as d (df(x)/dx + xf(x))/dx, meaning
    $$
    \frac{d}{dx} \left( \frac{d f(x)}{dx} + x f(x) \right)
    $$
    which expands to
    $$
    \frac{d}{dx} \left( \frac{d f(x)}{dx} \right) + \frac{d}{dx} \left( x f(x) \right)
    $$
    It is the second term there that you do not calculate properly.
    What is ##
    \frac{d}{dx} \left( f(x) g(x) \right)
    ## ?
     
  11. Oct 8, 2014 #10
    What is ##
    \frac{d}{dx} \left( f(x) g(x) \right)
    ## ?[/QUOTE]

    d(f(x)g(x)/dx = g(x)df(x)/dx + f(x) dg(x)/dx?
     
  12. Oct 8, 2014 #11

    DrClaude

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    Staff: Mentor

    Correct, now apply it to your problem by setting ##g(x) = x##.
     
  13. Oct 8, 2014 #12
    GOTCHA! Finally got the result:

    A2f(x) = d2f(x)/dx2 + f(x) + 2xdf(x)/dx + x2f(x)

    Also, one more question...if A = 2x d/dx...just in case...is it:

    Af(x) = 2x df(x)/dx
    OR
    =2xf(x) df(x)/dx?
     
  14. Oct 8, 2014 #13

    DrClaude

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    Staff: Mentor

    You can see that A as a combination of operators, ##\hat{A} = \hat{B} \hat{C}##, with ##\hat{B} \equiv 2x##, meaning "multiply by ##2x##," and ##\hat{C} \equiv \frac{d}{dx}##, meaning "differentiate with respect to ##x##." These operators are to be applied in sequence, starting from the right-most and moving to the left, meaning here "differentiate with respect to ##x##, then multiply the result by ##2x##." Therefore your first line above is correct.
     
  15. Oct 8, 2014 #14
    Thank you very much!!!!!!!!!!!!!
     
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