Since the wave function is an exponential equation, is there no point with zero probability density of finding a given particle? Does that justify quantum tunneling?

  • $\begingroup$ Be aware that $e^{ikx}$ is not normalizable (in the ordinary sense of the term) and so does not represent a physical state. As Griffiths writes in his QM textbook (page 60 of 2nd edition): In the case of the free particle, then, the separable solutions do no represent physically realizable states. A free particle cannot exist in a stationary state... $\endgroup$ – Alfred Centauri Jun 4 '18 at 0:26
  • $\begingroup$ No observable ever reaches any certain value. Zero is just one example. $\endgroup$ – safesphere Jun 4 '18 at 6:01
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    $\begingroup$ Related: physics.stackexchange.com/q/302269/2451 and links therein. $\endgroup$ – Qmechanic Jun 4 '18 at 6:16
  • $\begingroup$ have a look at this hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html . Note the change in the wave function ( not the energy) when tunneling $\endgroup$ – anna v Jun 4 '18 at 7:12

I think you are confusing things a bit. What you have there is the one-dimensional wavefunction for a free particle (i.e. plane wave), stripped of its time dependence. Indeed, if you try to compute its probability distribution via $P(x)=|\psi (x)|^2=\psi (x) \psi^*(x)$ then you will get 1, meaning it is position independent - it is as likely to be found in a certain place as in any other place, as you’d expect from a free particle. In jargon, a particle represented by such a wave function is completely delocalised.

To have quantum tunnelling, by definition, your particle has to “tunnel” through some potential barrier. This means the potential cannot be zero everywhere, i.e. the wavefunction cannot be that of a plane wave. Hence why your question does not make sense; a particle represented by that wavefunction has, by definition, nothing (no barrier or step) to “tunnel” through.

Where I think your confusion arises is that the free-particle wavefunction can be valid for a certain range of positions. For example, you might say that the wavefunction is $e^{ikx}$ within, say, $|x|<d/2$, and $0$ otherwise (read it as a piecewise function). This, for example, is an infinite potential well centred at $x=0$ and of diameter $d$. In this case, the potential is zero inside the well, so you can use the free particle wavefunction here, provided you apply the boundary conditions. However, there is no quantum tunnelling in infinite potential wells (indeed, one of the boundary conditions is that the wavefunction be zero at the boundaries). You start seeing quantum tunnelling, in the form of a decaying exponential in classically forbidden regions, when you start evaluating finite square wells, or potential steps/barriers.

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  • $\begingroup$ Thank you for your answer! I asked this question because I don't know how to intuitively understand why the probability of finding an electron inside or in the other side of a barrier has to be greater than zero. $\endgroup$ – Diogo Afonso Leitão Jun 3 '18 at 23:40
  • $\begingroup$ @DiogoAfonsoLeitão If you ever bought a car on credit, you know how an electron can get out of a hole. You may need to borrow money or energy to get over the hump, but you always have to pay it back. The main difference is that you also pay interest for the car while the electron enjoys an interest-free loan. $\endgroup$ – safesphere Jun 4 '18 at 5:55

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