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Raising and lowering operators

  1. May 17, 2008 #1
    [SOLVED] raising and lowering operators

    1. The problem statement, all variables and given/known data

    [​IMG]

    3. The attempt at a solution

    I expand a+ and a-, introduce the wavefunction and then substitute the values given at the very end to give:

    [​IMG]

    This is where I'm stuck. I can't see how epsilon = 2mE/hbar comes into the equation. Are my initial calculations even correct?
     
  2. jcsd
  3. May 18, 2008 #2

    cristo

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    No, I don't think your calculations are right. You are dealing with operators, so you've missed a term when expanding out. What you should do is first apply a- to psi, then apply a+ to that (you'll see any extra term drops out by the product rule).

    In order to put the equation in terms of epsilon, I would use the Schroedinger equation that you're given in the question.
     
    Last edited: May 18, 2008
  4. May 18, 2008 #3
    i get the same thing if i apply a- to psi and then a+ to that. Am I still doing something wrong?

    [​IMG]
     
  5. May 18, 2008 #4

    malawi_glenn

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    yes, you do wrong

    [tex] A_-\psi (x) = \frac{d\psi}{dx}+ \tanh (x)\psi (x) [/tex]

    then [tex] \frac{d}{dx}(\frac{d\psi}{dx}+ \tanh (x)\psi (x)) = \text{??} [/tex]
     
  6. May 18, 2008 #5

    cristo

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    You're not applying the operator a- to psi properly:

    [tex]\hat{A}_-\psi(x)=\left(\frac{d}{dx}+\tanh x\right)\psi=\frac{d\psi}{dx}+\tanh x\cdot\psi [/tex]

    Now apply a+ to this.
     
  7. May 18, 2008 #6
    oh, for some reason I was under the impression the wavefuction is not performed on tanh(x).

    I will try again
     
  8. May 18, 2008 #7
    Ok, cleared that up

    [​IMG]

    Now I can see from the Schrodinger equation that two of my terms appear similar. I'm not clear as to where to progress from here.
     
  9. May 18, 2008 #8

    cristo

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    I think you should look at your last line again. I agree with your first two terms, but not the others. Remember that you are expanding:

    [tex]\left(-\frac{d}{dx}+\tanh x \right)\left(\frac{d\psi}{dx}+\tanh x\cdot\psi\right) [/tex].

    Do this term by term, and then simplify after; it reduces the chance of making mistakes. You should get some cancellations.
     
    Last edited: May 18, 2008
  10. May 18, 2008 #9
    Cristo, in my notes it says this..

    "In obtaining this result it is importatnt to note that we are dealing with operators, and hence the order in which we operate is critical, whence: (A + B)(C + D) = AC + BC + AD + BD
     
  11. May 18, 2008 #10

    cristo

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    Ok, so if you expand the brackets according to that rule you get

    [tex]-\frac{d}{dx}\left(\frac{d\psi}{dx}\right)+\tanh x\cdot\frac{d\psi}{dx}-\frac{d}{dx}\left(\tanh x\cdot\psi\right)+\tanh x\cdot\tanh x\cdot \psi \hskip3cm (1)[/tex]

    Can you simplify this? Note that the brackets in the third term is a product.
     
    Last edited: May 18, 2008
  12. May 18, 2008 #11
    Because of the order of the expansion, I am not 100% sure if the second and third terms can cancel. I think thats what you are hinting at, but I have heard conflicting ideas.

    Also,two hints are provided:
    1) d/dx (tanhx) = sech^2(x)
    2) tanh^2(x) - sech^2(x) = 1-2sech^2(x)

    Using these two, I substituted to give what is shown in post #7.
     
  13. May 18, 2008 #12

    cristo

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    The second and third terms will not cancel entirely! The third term is a product; what would the solution to this be:

    [tex]\frac{d}{dx}(u\cdot v)[/tex] where u and v are functions of x?
     
  14. May 18, 2008 #13
    using product rule, evaluates to

    (d/dx)(tanh(x)(ψ) + (dψ/dx)(tanh(x))

    using the equation given above, will yield
    ψsech^2(x) + (dψ/dx)(tanh(x))
     
  15. May 18, 2008 #14

    cristo

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    Good. Now plug that into Eq. (1) (post #10)
     
  16. May 18, 2008 #15
    now this is where the schrodinger equation comes in?
     
  17. May 18, 2008 #16

    cristo

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    Yes.
     
  18. May 18, 2008 #17
    Got it.

    Thanks
     
  19. May 18, 2008 #18

    cristo

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    Well done; you're welcome.
     
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