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Chapter 1 cohen tannoudji

  1. Oct 10, 2011 #1
    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
     
  2. jcsd
  3. Oct 12, 2011 #2

    vela

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    You have to take care with the constants. The Fourier transform of the potential term is
    [tex]-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \alpha\delta(x)\psi(x)e^{-ikx}\,dk[/tex]Note also that your method will give you [itex]\bar{\varphi}(k)[/itex], not [itex]\bar{\varphi}(p)[/itex].
    Use the fact that
    [tex]\varphi(x) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty \bar{\varphi}(p)e^{ipx/\hbar}\,dp[/tex]to calculate [itex]\varphi(0)[/itex].
     
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