Why Does the Delta Potential Only Allow a Single Negative Energy State?

[mielgosez]

Homework Statement

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?”

Homework Equations

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

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

 

[vela]

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)

You have to take care with the constants. The Fourier transform of the potential term is
-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \alpha\delta(x)\psi(x)e^{-ikx}\,dkNote also that your method will give you \bar{\varphi}(k), not \bar{\varphi}(p).

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”

Use the fact that
\varphi(x) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \bar{\varphi}(p)e^{ipx/\hbar}\,dpto calculate \varphi(0).

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