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Quantum tunneling for dummies

  1. Nov 11, 2012 #1

    I am a physics student but still a layman in QM. I got a task to explain Quantum tunneling and I did my researchs for 3 days (I have read almost all threads here regarding to the topic). I found many different interpretations but I believe the Schrodinger's Equation is the correct answer.

    What I want to be helped is: How can I show that the tunneling probabilities to another side is non-zero with the equation?

    And the actual "probability" is |ψ|2 or the coefficient |T|2? sorry there were so many version of explanations, I really got confused....

  2. jcsd
  3. Nov 12, 2012 #2


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    Science Advisor

    The interpretation is clear; the particle can tunnel through a potential well which is forbidden in classical mechanics.

    Have a look at http://en.wikipedia.org/wiki/Quantum_tunnelling

    There are different ways to address the problem, but I think the most transparent one is start with a problem that can be solved exactly. Consider a free particle with a non-vanishing potential V(x) = V0 for x in [-L, +L] and V(x) = 0 otherwise. Let the energy of the particle (the energy as eigenvalue of the Schrödinger equation) be E < V0, so classically the particle would be confined to x < -L and could never be observed in the region x > L. You have plane waves outside the potential step, where the relation between momentum and energy is given by p²/2m = E. For an incoming particle from x = -∞ you have
    1) a wave propagating solely in positive x-direction for x > L
    2) an incoming wave travelling in positive x-direction for x < -L; and a reflected wave travelling into negative x-direction for x < -L
    So for x > L you have a wave function exp(ikx), for negative x you have a superposition of exp(ikx) and exp(-ikx). Now you can solve the Schrödinger equation explicitly (I think this is standard and is decsribed in many textbooks and e.g.
    3) in the region with non-vanishing potential you have an exponential decay like exp(-λx) b/c the the relation p²/2m = E-V0 results in an imaginary wave number k = iλ.

    Another idea is to use the WKB approximation; one can think of an arbitrary potential consisting of an infinite number of infinitesimal steps. Then the solution to integrate over the region with non-vanishing potential [x1, x2]; the tunneling is exponentially suppressed with

    [tex]\exp\left[ -\frac{\sqrt{2m}}{\hbar}\int_{x_1}^{x_2}dx\,\sqrt{V(x)-E}\right][/tex]
  4. Nov 12, 2012 #3
    Thanks a lot! And thanks not treating me as a QM layman!

    Basically I know what are you trying to say, the questions are:

    1) how can you assume that there is "a wave propagating solely in positive x-direction for x > L"? I mean if you are the 1st person to investigate this problem, and you do not know the effect of tunneling, the wave with E<V0 should be reflected at x=-L, why there will be a wave traveling at x>L?

    2) another noob question is: what do you mean by "solve the Schrödinger equation"? To get what result? to get ψ? or ψ2 or the other coefficients? And once I "solved" the equation, how do I know this is the evidence of tunneling?

    What I always trying to do is:
    1st, to confirm know what kind of quantities (some said |ψ|2? others said T2, or these 2 are the same thing?) represent probability of finding a particle;
    2nd, compute it in an equation under boundry condition(ie. Schrodinger's?) and get a non-zero result. This will satisfy and convince me.

    I am sorry pls be patient with me, I really need to know these. Thanks again.
  5. Nov 12, 2012 #4


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    You assume for the sake of argument that there is a "source" of some kind on the left, at some far away x < -L, and no "source" on the right (x > L). You start by assuming the most general possible outcome, namely that some of the wave is reflected and some of it is transmitted. That is, you don't prejudice the outcome by assuming the classical one which is no wave on the right. If the classical outcome is indeed the correct one, your solution will show it, that is, the amplitude of the outgoing wave on the right will turn out to be zero.

    The Schrödinger equation is a differential equation: it gives the relationship between ψ(x,t) and its derivatives. (Do you know any calculus?) The solution of a differential equation is a function, in this case ψ(x,t). In this case you find that ψ ≠ 0 for x > L, which leads to a probability density |ψ|2 > 0 for x > L. This indicates tunnelling because the situation was set up so that there is no "source" for the wave on that side of the barrier. It has to come from the left, through the barrier.

    I'd better point out that ψ ≠ 0 inside the barrier, also. You get a continuous function to the left, inside and right of the barrier, as in the example here:


    In general, |ψ(x,t)|2 gives you the probability density for finding the the particle at a particular point x at time t. We usually solve the tunneling problem using simple plane waves on both sides of the barrier. For these, |ψ(x,t)|2 is a constant, namely |A|2, the square of the amplitude of the wave. The transmission probabliity is the square of the ratio of the amplitudes of the wave going out to the right and the wave coming in from the left.

    T = |A(outgoing to right) / A(incoming from left)|2

    Some books may use T for the ratio of the amplitudes, in which case the transmission probability is |T|2.

    For further details, you should look for a QM textbook. Setting up the math is fairly simple, but getting the amplitudes and the transmission probability requires a lot of grinding algebra. I think most books leave the details of the algebra as an exercise for the student. This book shows a lot of the details (you still have to fill in some steps):


    Somewhere on the web there might be lecture notes that present some of the details.
    Last edited by a moderator: May 6, 2017
  6. Nov 12, 2012 #5
    Since the potential is defined differently in three regions of x axis, you have to solve Schrodinger's time-independent equation Hψ = Eψ for three regions: A) x<-L , B) -L[itex]\leq[/itex]x[itex]\leq[/itex]L and C) x>L . Since the solution for regions A and C give a non-zero ψ (which exponentially decays as you get away grom region B), you deduce that that there is a possibility to detect the particle outside the classical allowed region B. This is the tunneling effect: it is a direct mathematical consequence of the Schrodinger’s equation. Note that as V0 goes to infinity, the solutions in regions A and C go to zero.
  7. Nov 12, 2012 #6
    Thanks for all your replies, espectially for jtell, very good explanation! I cried.
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