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Quantum tunneling

  1. Nov 21, 2003 #1
    Hi everyone!
    I'm having some problem calculating the probability for a particle to penetrate a barrier (potential well). This is a math assignment in school, and we haven't learned anything about this area, so I may be fumbling in the dark completely.

    Anyway, we have the Schrödinger time-independent equation:

    [tex]\frac{h^2}{2m}\frac{d^2}{dx^2}\psi (x)+U(x)\psi (x) = E \psi (x)[/tex]

    The equation before the barrier is [tex]\psi (x) = e^{ikx} + Re^{-ikx}[/tex] (incoming and reflected wave).
    The equation in the barrier is [tex]\psi (x) = Ae^{imx} + Be^{-imx}[/tex]
    The equation after the barrier is [tex]\psi (x) = Te^{ikx}[/tex]

    If I've calculated right, the constants (k and m) should be:
    [tex]k = +- i \frac{\sqrt{2mE}}{h}, x < a or x > b[/tex]
    [tex]m = +- i \frac{\sqrt{2m(E-U_0)}}{h}[/tex]

    We know that [tex]U(x) = 0[/tex] when [tex]x < a[/tex] or [tex]x > b[/tex] (outside of the barrier) and [tex]U(x) = U_0, a < x < b [/tex] (inside the barrier).

    We have for unknown variables - A,B,R and T. We are supposed to get the probability from [tex]|T|^2[/tex]. We was instructed to derive the functions so that we got 4 function, which should yield an equation system when we attach the functions to eachother. So we set:

    [tex]e^{ika} + Re^{-ika} = Ae^{ima} + Be^{-ima}[/tex]
    [tex]Te^{ikb} = Ae^{imb} + Be^{-imb}[/tex]
    [tex]ke^{ika} - Rkae^{-ika} = mAe^{ima} - mBe^{-ima}[/tex]
    [tex]kTe^{ikb} = imAe^{imb} - mBe^{-imb}[/tex]

    How can I solve this system? Of course, I do not expect you to do this for me, I was just hoping you could help me do it.

    Oh, and english isn't my native language, so if the lingo is messed up sometimes - sorry. And I hope the tex stuff works...

    Thanks in advance,
    Nille
     
  2. jcsd
  3. Nov 21, 2003 #2
    Things are looking pretty good so far. For your function before the barrier, it is usual to include normalization on everything, including the first exponential term (unless you have been told specifically not to).
    [tex]\psi=Ie^{ikx}+Re^{-ikx}[/tex]

    Other than that, it is a matter of dregding through the equations. It is most likely duable using matrices, or the tedious way. In that case, you will want to solve for A and B in terms of T, then use those relations to solve for I in terms of T. Your transmission probability is then

    [tex]P=\frac{|T|^2}{|I|^2}[/tex]

    Also, I would choose a different letter for your energy constant inside the barrier, so you don't confuse it with the mass m.
     
  4. Nov 21, 2003 #3
    Also, I am guessing you are considering the case where 0<E<V. In that case, you might want to check your energy constants. They should probably be

    [tex]
    k=\frac{\sqrt{2mE}}{\hbar}
    [/tex]
    [tex]
    \kappa=\frac{\sqrt{2m(U_0-E)}}{\hbar}
    [/tex]

    Your solution inside the well would then be

    [tex]
    \psi=Ae^{\kappa x}+Be^{-\kappa x}
    [/tex]
     
  5. Nov 21, 2003 #4
    That was fast! Thanks!!

    Do you have any tips on how to solve the equation system? I tried today, and it seams it is hard to do it...

    Thanks!
    Nille
     
  6. Nov 29, 2003 #5
    try to correct and solve this problem

    the Schrödinger time-independent equation is written as:
    [tex]-\frac{h^2}{2m}\frac{d^2}{dx^2}\psi (x)+U(x)\psi (x) = E \psi (x)[/tex]
    Here,I would like to demonstrate the case of 0<E<U.Consider a barrier with a potential of U and width of a.For simplifying,the barrier is from zero to a in x-axis.
    The equations at the boundary of the barrier are some different from nille40's.
    x=0,[tex]\psi (x) & \psi (x)^'[/tex] are continuous:
    [tex]1+R=A+B[/tex]
    [tex]\frac{ik}{m}(1-R)=A-B[/tex]
    x=a
    [tex]Ae^{mb}+Be^{-mb}=Te^{ikb}[/tex]
    [tex]Ae^{mb}-Be^{-mb}=\frac{ik}{m}Te^{ikb}[/tex]
    Usually,R and T are concerned.To Solve T and R is not so difficult in above equations.
    May this can give you some idea.
    good luck.
     
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