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I Schrödinger's Equation Infinite Potential Well

  1. Sep 25, 2016 #1
    Given the equation ##\frac{d^2 \psi (x)}{{dt}^2}+\frac{2m}{{\hbar}^2}(E-V(x))=0## the general solution is:
    $$\psi (x)=A_1 e^{ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}} +A_2 e^{-ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}}$$
    If we have an infinite potential well: ## V(x)=\begin{cases} \infty \quad x\ge b\\ 0 \quad a < x < b \\ \infty \quad x \le a \end{cases}## then would that mean I take the limit of x to a and b and both should equal zero? So:

    $$\lim_{x\rightarrow a}\left(A_1 e^{ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}} +A_2 e^{-ix \sqrt{\frac{2m}{{\hbar}^2}(E-V(x))}}\right)=0+\lim_{x\rightarrow a}\left( A_2 e^{x \sqrt{\frac{2m}{{\hbar}^2}(V(x)-E)}}\right)$$ where ## \lim_{x\rightarrow a}\left( A_2 e^{x \sqrt{\frac{2m}{{\hbar}^2}(V(x)-E)}}\right)## diverges. But how could that be zero? Would this mean that ##A_2## must be Zero?
     
    Last edited: Sep 25, 2016
  2. jcsd
  3. Sep 25, 2016 #2

    Nugatory

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    The solution doesn't have to be the same (that is, the values of ##A_1## and ##A_2## don't have to be the same) in all three regions. Indeed, the potential is different in the different regions so you're looking at different differential equations in the different regions, so you'd expect the solutions to be different. Call the three solutions ##\psi_L(x)## (##x\le{a}##), ##\psi_0(x)## (##a\le{x}\le{b}##) and ##\psi_R(x)## (##x\ge{b}##) and solve for them separately within their own domains. You should be able to convince yourself that ##\psi_L(x)=\psi_R(x)=0## pretty quickly, leaving only ##\psi_0(x)## to work out.

    Then apply the boundary condition ##\psi_L(a)=\psi_0(a)## and ##\psi_R(b)=\psi_0(b)##, and you'll be home. You can take ##a=0## without loss of generality and the problem becomes a bit simpler.
     
  4. Sep 25, 2016 #3
    Ah, I See now. Last question then; why is ##\psi (x) =0## when ##V(x)=\infty##?
     
  5. Sep 26, 2016 #4
    For a particle's wavefunction to have a nonzero value at a region where the potential is infinite would suggest that there is a nonzero probability amplitude associated with finding that particle in the region. How can a particle with finite energy exist in a region of space with infinite potential?
     
  6. Sep 26, 2016 #5

    Nugatory

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    In my intro QM class many years ago, the professor used the method of emphatic assertion to demonstrate the truth of this proposition. :-p

    You can get a somewhat more rigorous-looking argument by solving Schrodinger's equation for a finite positive potential, applying the boundary condition that ##\psi(\pm\infty)=0##, and then considering the limit as ##V## approaches infinity. That boundary condition can be justified as requiring that the wave function be a sensible probability distribution, meaning that its squared integral across all space is equal to unity.

    However, as a practical matter it is reasonable to look for solutions assuming that an infinite repulsive potential in a region means zero probability of the particle being found in that region, which is equivalent to saying that ##\psi=0## in that region.
     
  7. Sep 27, 2016 #6

    vanhees71

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    Don't forget that in general there are both bound states, where ##\psi(x) \rightarrow 0## for ##x \rightarrow \pm \infty## and "scattering states", where the wave function becomes sinusoidal for ##x \rightarrow \infty## and/or ##x \rightarrow -\infty##. A very good treatment of the one-dimensional time-independent Schrödinger equation can be found in Messiah's QM textbook (vol. 1, if I remember right).
     
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