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Barrier tunneling - Intro to QM

  1. Mar 26, 2008 #1

    Defennder

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    1. The problem statement, all variables and given/known data

    Ok, maybe I'm slow at grasping this, but why is the potential energy U of an electron while it is tunneling through a 1 dimensional potential barrier of height V0 = V0?

    2. Relevant equations

    Schrodinger equation in 1D (using semiclassical approximation)

    3. The attempt at a solution

    I'm assuming that the wave number [tex]k = \frac{\sqrt{2m(E-V_{0})}}{\hbar}[/tex] implies that U = V0 where U is the potential energy expression in the schrodinger equation.
     
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  3. Mar 26, 2008 #2

    Kurdt

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    Its not that the electron has that potential its more a consequence of the wave function having to be continuous everywhere. So for the finite square well solution you have essentially two regions, one where there is no potential and the other where it is constant. For energies less than the potential the wavefunction penetrates slightly into the classically forbidden region and decays exponentially. This depends on the strength of the potential, but does not mean the electron is at that potential.
     
  4. Mar 26, 2008 #3

    Defennder

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    But if the electron doesn't necessarily have that potential, then why does the E-V0 term appear in the term for k? The potential barrier is given graphically as a wall, so this isn't exactly like the finite square well.
     
  5. Mar 26, 2008 #4

    Kurdt

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    Its as a consequence of how the Schrödinger equation is set up in the region with the potential barrier, and then solved subject to the boundary conditions. I'm sure you've seen this so I'm not too sure how to explain it for you.
     
  6. Mar 26, 2008 #5

    Defennder

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    I'll try to put it more explicitly. This is how I believe it's done. This is with reference to region II in the picture, where I'm solving the Schrodinger equation in that region.

    [​IMG]

    [tex]-\frac{{\hbar}^2}{2m}\frac{d^2\psi}{dx^2} + U\psi = E\psi [/tex]

    [tex]-\frac{{\hbar}^2}{2m}\frac{d^2\psi}{dx^2}=(E-U)\psi[/tex]

    [tex]\frac{d^2\psi}{dx^2}=\frac{-2m(E-U)}{{\hbar}^2}\psi [/tex]

    This a 2nd order ODE with the characteristic equation [tex]\lambda=\pm\frac{\sqrt{2m(U-E)}}{\hbar}[/tex], and the general solution is given by [tex]\psi(x)=Ce^{kx}+De^{-kx}[/tex], where [tex]k=\lambda[/tex]

    This is where I don't understand at all. The textbook and my notes replace U in the above equation with V0 the height of the barrier which implies that U = V0 when the electron is tunneling through the barrier. This is also done before we solve for unknown C and D by applying boundary conditions and normalizing such that the probability of finding the electron where is 1, I don't see how the boundary conditions is relevant to this. The question is, why is U=V0 in the barrier?
     
  7. Mar 28, 2008 #6

    Astronuc

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    It would seem that the problem states that U = V0, which would be an idealized (somewhat arbitrary) model. Then Schrödinger's equation in 1-D becomes a simple and straightforward linear diff. eq. with constant coefficients. This is a square well, square barrier (i.e. with constant amplitude) problem.
     
    Last edited: Mar 28, 2008
  8. Mar 28, 2008 #7

    Defennder

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    Thanks for replying, Astronuc. This is a rather awkward way to get my thread noticed, though. I understand that U in the above derivation is a constant rather a function of x, which makes it possible such that it is easy to solve, but I don't understand why set U=V0. If it's supposed to be idealised, why does that assumption make it ideal? What are they trying to show?

    After all, that is the mininum energy the electron must have if it managed to 'climb' up the barrier. And this minimum energy is inclusive of both potential + kinetic energy. And if the electron has V0 as potential energy, and yet its total E is less than the V0 (hence the need to tunnel through rather than surmount the barrier), does this imply that its kinetic energy is negative?

    PS. Thanks to your reply, the thread is at the top again.
     
    Last edited: Mar 28, 2008
  9. Mar 28, 2008 #8

    siddharth

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    As astronuc pointed out, the potential in the region II is by definition V0. That's how the well is constructed! The electron does not "possess" that potential, rather region 2 has a potential of V0.

    Ah, no. The classical concepts of "climbing" the barrier and having fixed turning points do not hold. Instead, when E<0, you have bound states and when E>0 you have scattering states.
     
  10. Mar 28, 2008 #9

    Defennder

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    Doesn't the U term in the 1D Schrodinger equation represent the potential energy of the electron? If so, then setting U = V0 seems to be saying that the electron posseses that potential energy in that region.

    What do you mean by bound states and scattering states? I'm trying to understand this from the perspective of a course in microelectronics. I presume E is the total energy of the electron?
     
  11. Mar 28, 2008 #10

    siddharth

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    Exactly. My point is, the electron does not have "intrinsic" potential energy. The well is made in such a way that the potential energy of the electron is kept constant in that region.

    Yes, E is the energy of the electron. In classical mechanics, if particle is stuck in a well, where E < (V at the boundaries), it is restricted in that region so that the particle moves only in that region.

    In QM, since the potentials go to zero at infinity, if E < (V at the boundaries = 0), the states corresponding to this energy are called bound states. Note that in this problem, even if E < V, the electron can tunnel through the barrier
     
    Last edited: Mar 28, 2008
  12. Mar 28, 2008 #11

    Defennder

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    Ok, I see that the assumption made is that the well keeps the potential energy of the electron constant at V0, which is the height of the barrier. More importantly, why do they assume that the barriers of the well fixes the potential energy to be that of the barrier height? Why not some other fixed arbitrary (or unknown) value? But back to the original point which I raised earlier, if E<V0, and U+KE(kinetic energy)=E, and U=V0, doesn't this imply that the KE of the electron is negative? If so, what does that mean?

    What do you mean by the potentials diminishing to zero at infinity? Are you referring to the wavefunction which decays exponentially deeper into the barrier? How is this related to the potential energy of the electron, which as you said above is 'fixed' by the barrier?
     
    Last edited: Mar 28, 2008
  13. Mar 28, 2008 #12

    siddharth

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    I don't understand. If you set some other value for the barrier potential, that value would be the PE of the electron in the barrier.

    Yeah, and that's where QM and CM differ. If you could measure the en in the barrier you would get a negative KE. However, the wavefunction inside decays exponentially, and experimentally I don't think you can.
     
  14. Mar 28, 2008 #13

    Kurdt

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    Observation of a particle tunneling seems impossible as we can only detect positive kinetic energy.
     
  15. Mar 28, 2008 #14

    Defennder

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    Ok, I think what I understand is roughly as follows: the barrier sets the electrons PE to its height and that the KE of the electron when tunneling is negative, though not measurable experimentally. The reason why the PE of the electron can be set by the barrier to be V0 is because PE is largely arbitrary, that's why we can assume PE=0 for a free electron.

    Am I wrong somewhere?
     
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