Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivations for Schrodinger's equations for potential step

  1. Dec 24, 2014 #1
    I have been studying potential steps and barriers as well as reflection and transmission coefficients and how to derive them. Most of it makes sense to me except for one thing:

    As we know, the normal Schrodinger equation is:

    (-ħ2/2m) (∂2Ψ/∂x2) + v(x)Ψ = EΨ

    For a step potential however, my book and web resource both say that for the boundary conditions:

    v(x)= 0 for x < 0 and V0 for x ≥ 0

    the Schrodinger equations are:

    (∂2Ψ1/∂x2) + K12Ψ1(x) = 0

    and
    (∂2Ψ2/∂x2) + K22Ψ2(x) = 0

    where K1 = squrt(2mE) / ħ and K2 = squrt(2m(E- V0)) / ħ

    (the plus sign in the second one changes into a minus when the particle doesn't have enough energy to overcome the step).

    Where/How exactly did Schrodinger get these step potential equations from the original one? The step potential equations don't even seem to have the (-ħ2/2m) term in front. Can someone please explain to me why step potentials seem to have different Schrodinger equations than the normal one?
     
  2. jcsd
  3. Dec 24, 2014 #2

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    This is just basic algebra, giving ##K_1## as an example, and you should work out ##K_2## for yourself:

    $$\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi=E\psi$$ $$\frac{\partial^2}{\partial x^2}\psi=-\frac{2m}{\hbar^2}E\psi$$ $$\frac{\partial^2}{\partial x^2}\psi+\frac{2mE}{\hbar^2}\psi=0$$ $$\frac{\partial^2}{\partial x^2}\psi+K_1^2\psi=0$$
     
  4. Dec 24, 2014 #3
    I see. Thank you. It is just simple algebra. This does bring up the question however of why that was necessary in the first place. Why didn't he just leave the equation as it originally was (since these other versions are just algebraically manipulated versions of the original)? After all, the original version is just as simple to solve as the other versions. I have a hypothesis as to why this manipulation was necessary (it has to do with being able to normalize the solutions). I am not 100% sure however that this is the reason.
     
  5. Dec 24, 2014 #4

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    No, you can solve it without doing any algebraic manipulations. And this has nothing to do with normalization. But if you do these manipulations, it just makes things a little simpler when you go ahead and solve it.
     
  6. Dec 24, 2014 #5

    jtbell

    User Avatar

    Staff: Mentor

    When you finally start writing out the solutions and manipulating them algebraically, would you rather wrestle with things like ##e^{ik_1 x}## and ##e^{ik_2 x}## or with things like ##e^{i\sqrt{2mE}x/\hbar}## and ##e^{i\sqrt{2m(E-V_0)}x/\hbar}## ? :olduhh:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Derivations for Schrodinger's equations for potential step
Loading...