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Finding potential given some conditions

  1. Oct 16, 2013 #1

    CAF123

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    Gold Member

    1. The problem statement, all variables and given/known data
    Denote the ground state and the first excited state of a 1D quantum system by ##\psi_{0}## and ##\psi_{1}##. If it is given that $$\psi_{1}(x) = x \psi_{0}(x)\,\,\,\text{and}\,\,V(0)=0$$ find the potential V(x).

    2. Relevant equations
    TISE

    3. The attempt at a solution
    If I sub ##\psi_1## into the TISE I get ##-\frac{\hbar^2}{2m} \psi_1'' + V(x)\psi_1 = E_1 \psi_1##(1). I can replace ##\psi_1## with ##x\psi_o## to give after differentiation, ##-\frac{\hbar^2}{2m} \left(2\psi_o' + x\psi_o''\right) + V(x)x\psi_o = E_1x\psi_o##. (2)

    But I also know that ##-\frac{\hbar^2}{2m} \psi_0'' + V(x)\psi_0 = E_2\psi_0## Rearrange (2) yields $$-\frac{\hbar^2}{m}\psi_o' + x\left(-\frac{\hbar^2}{2m}\psi_o'' + V(x)\psi_o)\right) = E_1x\psi_o.$$The coefficient of x is precisely ##E_2\psi_o## so I end up with a diff eq for ##\psi_o##: $$\psi_o' = -\frac{m(E_2 - E_1)}{\hbar^2}x\psi_0 \Rightarrow \psi_o = K \exp\left(-\frac{m(E_1-E_2)}{\hbar^2} \frac{x^2}{2}\right)$$, K a constant.

    So it then easy to get ##\psi_1## by multiplying ##\psi_o## by x. Subbing ##\psi_1## back into (1) and differentiating twice I get that $$V(x) = \frac{m}{2}\frac{(E_1-E_2)^2}{\hbar^2}x^2$$, neglecting the constant terms. Does this seem okay?
     
    Last edited: Oct 16, 2013
  2. jcsd
  3. Oct 16, 2013 #2

    vanhees71

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    Looks good. It's the harmonic oscillator with the frequency
    [tex]\omega^2=\frac{(E_1-E_2)^2}{\hbar^2}[/tex]
    as it should be.
     
  4. Oct 16, 2013 #3

    CAF123

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    Hi vanhees71, I was wondering how the answer I got here differs from the potential in my other thread '1D Potential Step QM Problem'. Is it just the case that in that thread V was the harmonic oscillator to a restricted interval, whereas in this case it holds for all x? Hence the potentials in both threads are different.
     
  5. Oct 16, 2013 #4

    BruceW

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    eh? In the 1D potential step problem, I'm guessing the potential was simply a step potential. But here you have a completely different potential. different potential gives a different wavefunction.

    edit: ah whoops, I didn't read your post carefully enough. OK so in your other thread, the potential was like a harmonic oscillator potential over some interval and some larger constant outside that interval? Well, still that potential is different to this potential, as you say, which is why you get different answers. In other words, different system, different answer.
     
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