(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

problem 1.4 of Cohen Tannoudji:

The exercise is divided into two parts:

THE FIRST ONE (part a.) asks to find the Fourier transform of a wave function when the potential is: -αδ(x). In terms of p, E, alpha and ψ(0)

and

ask to “show that only one value of E, a negative one, is possible” Furthermore, they claim that “Only the bound state of the particle, and not the ones in which it propagates, is found by this method; why?”

2. Relevant equations

the Schrodinger equation:

-ℏ^2/2m d^2/〖dx〗^2 ψ(x)-αδ(x)ψ(x)=E ψ(x)

3. The attempt at a solution

Hi, I was solving the problem 1.4 of Cohen Tannoudji:

The exercise is divided into two parts:

THE FIRST ONE (part a.) asks to find the Fourier transform of a wave function when the potential is: -αδ(x). In terms of p, E, alpha and ψ(0)

I did the next procedure:

I wrote down the Schrodinger equation:

-ℏ^2/2m d^2/〖dx〗^2 ψ(x)-αδ(x)ψ(x)=E ψ(x) …………………………………………………………………… (i)

Then, I applied a Fourier Transform over the equation (i) What I got was:

-ℏ^2/2m 〖(ik)〗^2 ψ ̅(k)-αψ(0)=E ψ ̅(k)………………………………………………………………………….(ii)

Finally, I express ψ ̅(k) in terms of the given parameters, using the fact that p=ℏk

ψ ̅(k)=(α ψ(0))/(p^2/2m-E)…………………………………………………………………………………………………………(iii)

Now, they ask to “show that only one value of E, a negative one, is possible” Furthermore, they claim that “Only the bound state of the particle, and not the ones in which it propagates, is found by this method; why?”

What I know is that as E=V(r)+p^2/2m and for x≠0 V(r)=0, then equation (iii) diverge. But, I don’t get why there’s

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# Chapter 1 cohen tannoudji

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