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Missing h-bar in showing SHO in terms of ANHIL and CREA operators is correct

  1. Oct 10, 2014 #1
    this is the given:

    xYaMmwr.jpg

    the problem is the middle term, if the h-bar w outside the set brackets is cancelled with the h-bar w of the m/2hw, then there will be a h-bar w that is left introduced from the middle term, i.e.

    [tex]i\frac{w}{m}XP- i\frac{w}{m}PX = i\frac{w}{m}[X,P]= i\frac{w}{m}i\hbar[/tex]

    but there is no h-bar in the end result, as though h-bar w has been cancelled twice??

    thanks in advance for any help
     
  2. jcsd
  3. Oct 10, 2014 #2
    Ah so your on harmonic oscillators in susskind?! The hamiltonian is rewriten in form of a^2+b^2 with complex and imaginary numbers, yet you must add the h(bar) so that the hamiltonian stays the same.
     
  4. Oct 10, 2014 #3
    Not yet, this one was given in class, hmm not sure what you are saying, if the h-bar w outside the set bracket cancels with the one inside then how can it also be used to cancel the h-bar w that would be given from the [X,P] commutation relation?

    thanks for the reply!
     
  5. Oct 10, 2014 #4
    What I am trying to say is that you start with the last equation in your post and then use the fact that (a-bi)(a+bi)+a^+2abi-2bi-b^2=a^2+b^2 which is of the form of 1/2mw^2x^2+p^2/2m. So you don't start woth the anhil and crea form you start with this equation and reform it. The [X,P] has nothing ,that I know of, to do woth the theme at hand.
     
    Last edited: Oct 10, 2014
  6. Oct 10, 2014 #5
    ahhh i think i see what you are getting at, i'll have to run through it a few times to be sure, thanks again for your help, its appreciated!
     
  7. Oct 10, 2014 #6
    Wait here is the equation:
    1.H=(P^2+wX^2)*1/2
    We define
    2.a^2+b^2=H=(P^2+wX^2)*1/2=(a+bi)(a-bi)
    3.H=1/2(p+iwX)(p-iwX) (!note this is incomplete!)
    4.1/2(p+iwX)(p-iwX)= 1/2*(P^2+iwXP-iwPX - i^2w^2X^2)
    =1/2(P^2+iw(XP-PX)+w^2X^2
    =1/2(P^2+w^2X^2)+1/2iw(XP-PX)
    (XP-PX)=[X,P]=-ih(bar)

    5.1/2(P^2+w^2X^2)+1/2iwih(bar)
    =1/2(P^2+w^2X^2)-1/2wh(bar)

    The factored expression we started with is smaller than the actuall hamiltonian so we must add wh(bar)/2. so the equation becomes

    1/2(p+iwX)(p-iwX)+wh(bar)/2

    and then we define the creation and anhilation operators to be: (p+iwX) and (p-iwX) with some modification.

    Yet there are different versions of how to derive the anhilation operator and creation operator, for example one can define q and p and create the operators with the aid of q and p

    hope this helps...
     
  8. Oct 12, 2014 #7
    wow, thanks for your efforts! yes i see now i was working this the wrong way around, and i see now why the h-bar w must be added. Thanks once again, its appreciated!
     
  9. Oct 12, 2014 #8
    just a last question, how can you do this? it kind of comes back to my original question working the other way in that you have to remove the wh(bar)/2 term for it to be correct, why is this allowed? especially as you are trying to show one is equal to the other. thanks
     
  10. Oct 12, 2014 #9
    Ok, I will replie properly later but just for now: I don't know if you have done this already in class but the product of the operators crea and anhil is writen as N=crea*anhil so your equation would be simpler. I know this in't an answer but I shall come back later to your question and I hope I can help! Scientists went mad defining operators!:)
     
  11. Oct 14, 2014 #10

    thanks any more feedback on this would be more than welcome
     
  12. Oct 14, 2014 #11
    I believe I understand what your problem is!? The crea and anhil operators are defined as I said(either with p and q or as factors of the hamiltonian...), yet historically the crea and anhil operators are defined differnetly which changes the complete or at least part of the hamiltonian equation

    where historically they are defined with an extra factor, beeing

    +-i/2wh(bar).

    I believe one can leave out the factor, to make things simpler. Anyway I hope that helped a bit...
     
  13. Oct 14, 2014 #12
    Last edited: Oct 14, 2014
  14. Oct 16, 2014 #13
    Thats very helpful ! Think I have it now! Many thanks for all your help!
     
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