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Dirac delta

  1. Jan 27, 2005 #1
    [tex]\delta(\frac{p_f^2}{2m}-E_i^o-\hbar\omega)=\frac{m}{p_f} \delta(p_f-[2m(E_i^o+\hbar\omega)]^{\frac{1}{2}})[/tex]

    Shouldn't the right hand side be multiplued by 2?
  2. jcsd
  3. Jan 27, 2005 #2


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    :surprised :surprised

    Mr.QM+QFT guru,you f***ed up big time... :tongue:

    Apply the THEORY:
    [tex] f(x):R\rightarrow R [/tex](1)
    only with simple zero-s (the algebraic multiplicies of the roots need to be 1).
    Let's denote the solutions of the equation
    [tex] f(x)=0 [/tex](2)
    by [itex] (x_{\Delta})_{\Delta={1,...,N}} [/itex] (3)
    and let's assume that:
    [tex] \frac{df(x)}{dx}|_{x=x_{\Delta}} \neq 0 [/tex] (4)

    Then in the theory of distributions there can be shown that:

    [tex] \delta f(x)=\sum_{\Delta =1}^{N} \frac{\delta (x-x_{\Delta})}{|\frac{df(x)}{dx}|_{x=x_{\Delta}}|} [/tex] (5)

    Read,Marlon...Read... :rolleyes:

    Last edited: Jan 27, 2005
  4. Jan 27, 2005 #3
    is there anyone else that can solve this problem in a clear and more mature manner. It's been a while since i worked with distributions in this way and it seems quite interesting to me. Can someone tell me what i did wrong ?

    thanks in advance

  5. Jan 27, 2005 #4


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    [tex] f(x)\rightarrow f(p_{f})=\frac{p_{f}^{2}}{2m}-E_{0}^{i}-\hbar\omega [/tex] (6)

    Solving the equation
    [tex] f(p_{f})=0 [/tex] (7)

    ,yields the 2 solutions (which fortunately have the degree of multiplicity exactly 1)

    [tex] p_{f}^{1,2}=\pm \sqrt{2m(E_{0}^{i}+\hbar\omega)} [/tex] (8)

    Computing the derivative of the function on the solutions (8) of the equation (7),we get,after considering the modulus/absolute value:
    [tex] \frac{df(p_{f})}{dp_{f}}=\frac{\sqrt{2m(E_{0}^{i}+\hbar\omega)}}{m} [/tex] (9)

    Combining (8),(9) and the general formula (5) (v.prior post),we get:

    [tex] \delta (\frac{p_{f}^{2}}{2m}-E_{0}^{i}-m_{i})=\frac{m}{\sqrt{2m(E_{0}^{i}+\hbar\omega)}} \{\delta[p_{f}-\sqrt{2m(E_{0}^{i}+\hbar\omega)}]+\delta[p_{f}+\sqrt{2m(E_{0}^{i}+\hbar\omega)}]\} [/tex] (10)

    which is totally different than what the OP had posted...

    IIRC,when learning QFT,i always said to myself:theorem of residues and the thoery of distributions go hand in hand...

  6. Jan 27, 2005 #5
    Indeed, i just looked up the rule at hand. I get the same solution and i see where i went wrong in my first post. thanks for the polite correction.


    i deleted my erroneous post
  7. Jan 27, 2005 #6
    It's always nice to encounter the two of you in a post. All this harmony and warmth...
  8. Jan 27, 2005 #7
    yes, we really are the best of friends... :wink:

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