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Finite Square Well

  1. Jul 2, 2014 #1
    Hello everyone,

    I am reading about the Finite Square Well in Griffiths Quantum Mechanics Text. Right now, I am reading about the case in which the particle can be in bound states, implying that it has an energy E < 0. After some derivations, the author comes across the equation

    [itex]\tan z = \sqrt{ \left(\frac{z_0}{z} \right)^2 - 1}[/itex]

    where [itex]z = la[/itex] and [itex]z_0 = \frac{a}{\hbar} \sqrt{2mV_0}[/itex]; additionally, [itex]l= \frac{\sqrt{2m(E+V_0)}}{\hbar}[/itex].

    The author makes the claim that, "if z_0 is very large, the intersections occur just slightly below [itex]z_n = n \frac{\pi}{2}[/itex], with n odd; it follows that

    [itex]E_n + V_0 \approx \frac{n^2 \pi^2 \hbar^2}{2m(2a)^2}[/itex]" (1)

    I don't quite see this. How does z_0 being large imply (1) is a true statement? I have tried to work out the details myself, but it appears that there are a lot of missing details which the author should have included; although, this is just my opinion.
     
  2. jcsd
  3. Jul 2, 2014 #2

    WannabeNewton

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    It seems very straightforward to me. ##\frac{n\pi \hbar}{2a} \approx \sqrt{2m(E + V_0)}## so ##E + V_0 \approx \frac{n^2 \pi^2 \hbar^2}{2m(2a)^2}##. ##z_0## being very large implies that the solutions ##z## to ##\tan z = \sqrt{(z_0/z)^2 - 1}## are ##z = \frac{n\pi}{2}##. Just make a plot of the two curves on mathematica for large ##z_0## and it should be obvious to you why the intersections are as they are for large ##z_0##.
     
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