Solving for Momentum in Energy Equation

[RedX]

$$\delta(\frac{p_f^2}{2m}-E_i^o-\hbar\omega)=\frac{m}{p_f} \delta(p_f-[2m(E_i^o+\hbar\omega)]^{\frac{1}{2}})$$

Shouldn't the right hand side be multiplued by 2?

 

[dextercioby]

i am sorry but according to me the fraction before the delta should disappear.

marlon

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


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

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

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

Read,Marlon...Read...

Daniel.

 

[marlon]

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

regards
marlon

 

[dextercioby]

Okay:
$$f(x)\rightarrow f(p_{f})=\frac{p_{f}^{2}}{2m}-E_{0}^{i}-\hbar\omega$$ (6)

Solving the equation
$$f(p_{f})=0$$ (7)

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

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

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

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

$$\delta (\frac{p_{f}^{2}}{2m}-E_{0}^{i}-m_{i})=\frac{m}{\sqrt{2m(E_{0}^{i}+\hbar\omega)}} \{\delta[p_{f}-\sqrt{2m(E_{0}^{i}+\hbar\omega)}]+\delta[p_{f}+\sqrt{2m(E_{0}^{i}+\hbar\omega)}]\}$$ (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 theory of distributions go hand in hand...

Daniel.

 

[marlon]

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.

marlon

i deleted my erroneous post

 

[da_willem]

thanks for the polite correction.

marlon

It's always nice to encounter the two of you in a post. All this harmony and warmth...

 

[marlon]

It's always nice to encounter the two of you in a post. All this harmony and warmth...

yes, we really are the best of friends...

marlon