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Form of Solution to Schrodinger Equation

  1. Mar 17, 2014 #1
    Hey guys, I just wanna ask on how do you determine the form of wave function for Schrodinger equation of finite potential well and potential barrier.

    Why is it ψ(x) = Ae^ikx + Be^-ikx (x < -a)
    ψ(x) = Fe^ikx (x > a)
    ψ(x) = Ce^μx + De^-μx (-a < x < a)
    for k^2 = (2mE)/(hbarr), μ^2 = (2m(V - E))/(hbarr) and E < V

    The potential barrier is 0 at x < -a and x > +a and V at -a < x < a.

    I know that there are two cases (E < V and E > V). What I really don't know is how do you get the solution form of the Schrodinger equation.

    And for a potential well:

    Why is it ψ(x) = Ae^kx + Be^-kx (x < -a)
    ψ(x) = Fe^kx + Ge^-kx (x > a)
    ψ(x) = Csin(μx) + Dcos(μx) (-a < x < a)

    The potential well is 0 at x < -a and x > +a and -V at -a < x < a.
  2. jcsd
  3. Mar 17, 2014 #2


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    Staff: Mentor

    You [STRIKE]guess[/STRIKE] select a suitable ansatz (see http://en.wikipedia.org/wiki/Ansatz), then insert your [strike]guess[/STRIKE] carefully chosen ansatz back into the equation to see if it's a good solution.

    That's how many differential equations are solved.
  4. Mar 17, 2014 #3


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    Well, in this case there's no need to guess the solution but you can derive it directly from the time-independent Schrödinger equation, which obviously is the one the OP likes to solve. Setting [itex]\hbar=1[/itex] for convenience, it reads
    [tex]-\frac{1}{2m} \psi''(x)+V(x) \psi(x)=E \psi(x),[/tex]
    where [itex]E \in \mathbb{R}[/itex] is an eigenvalue of the Hamilton operator.

    To find the possible eigenvalues, you have to solve this equation and find the values for [itex]E[/itex] such that the resulting wave functions become either square-integrable (leading to the discrete part of the Hamiltonian's spectrum and the corresponding the bound-state solutions of the time-independent Schrödinger equation) or functions that are normalizable to [itex]\delta[/itex] distributions (continuous part of the Hamiltonian's spectrum and scattering solutions of the time-independent Schrödinger equation).

    Here, the potential is piece-wise constant and it's pretty easy to solve the differential equation in the three regions separately. Then you have to make the wave function and its first derivative continuous at the places where the potential makes (a) jump(s).

    These conditions give unique (generalized) eigenvalues [itex]E[/itex] and the associated (generalized) eigenfunctions!
  5. Mar 17, 2014 #4

    at start you want to separate 3 different regions because you got different V potential as you can see in the picture.So generaly you solve 3 Schrodinger equations for the 3 regions.
    The results must ofcourse unite to 1 wave function keeping in mind that a wave function(and its derivative)must be continuous.

    Anyway The first thought is how much energy you start with, as you said there are the tow cases of E < V and E > V

    for E<V

    you got in Schrodinger for the 3 regions

    1st region x<-L ) (-h^2/2m)ψ1''+V*ψ1=Eψ1 ψ1''=d^2ψ(x)/dx^2 , h=h(barr)

    2cond -L<x<L ) (-h^2/2m)ψ2''=Eψ2

    3rd x>L ) (-h^2/2m)ψ3''+V*ψ3=Eψ3

    So now we solve them

    1st) ψ1''=(2m/h^2)(V-E)ψ1 , (2m/h^2)>0 ,(V-E)>0 using k^2=(2m/h^2)(V-E) you get ψ1''=k^2ψ1 solve→ ψ1(x)= Ae^kx + Be^-kx

    for that solution if we take x → -∞ we get ψ1=∞ and you dont expect that cause you are in the potential region and its pushing you away, or you can say that you want ψ to be Square-integrable function , so B=0

    2cond) (-h^2/2m)ψ2''=(2m/h^2)(-E)ψ2 now (2m/h^2)(-E)<0 using γ^2=(2m/h^2)(E) you get
    ψ2''=-γ^2 ψ2 solve→ ψ2(x) = Ce^iγx + De^-iγx

    now to get the form you said use Eulers Eq and you ll get ψ(x) = Vsin(γx) + Mcos(γx)
    (mind V and M are now complex)

    3rd) solve same way as the first with same k^2
    and you get ψ(x) = Fe^kx + Ge^-kx (x > a) for the same reason you dont expct to get too big
    So F=0

    after that to get the whole solution you just use boundary conditions.

    Now for E>V you get different kind of problem witch is scatering one.The difference applies essentially that now we dont worry if ψ is Square-integrable function
  6. Mar 17, 2014 #5


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    Staff: Mentor

    Have you had a course in differential equations yet? The time-independent SE in each region is a linear homogeneous differential equation which can be solved using the method of "characteristic polynomials." A Google search on "linear differential equation characteristic polynomial" will give you several sites that describe this method.

    In practice, physicists usually use Nugatory's "guess and try it" method first. If that fails, we use a more formal method like characteristic polynomials, or the power-series method (which you might see when you get to the SE for the simple harmonic oscillator), etc.
  7. Mar 18, 2014 #6
    Oh, I see. I could solve it or I could guess and test it if it satisfies the Schrodinger equation for that region. Thanks Nugatory. vanhees71, SyLba, and jtbell. Really helpful answers.
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