The particle has mass m and energy E0 = V.

The solution for the particle in the classically forbidden step region is W(x)=Cexp(-Kx). I know how to find K = sqrt(2m(V-E))/hbar.

Logically the probability is |W(x)|^2 so the probability of it penetrating to a least d is found by integrating the |w(x)|^2 between d and infinity however this leaves a factor of C^2 that I don't know how to find or cancel.


1 Answer 1


The factor of $C$ will come the constraint of normalisation. That is, the total probability of finding your particle everywhere, must be equal to $1$. This is the mathematical statement of "your particle exists somewhere, always". So you find $C$ by setting, $$\int_D|\psi|^2dx=1,$$ where $D$ represents the domain of your problem. Note for this to work, you need a wavefuntion that satisfies $\psi \rightarrow 0$ as $x \rightarrow \pm \infty$. In this case, you can achieve that by making the problem symmetrical in $x$, and your normalisation will become: $$2\int^{\infty}_{w}|\psi_{\textrm{out}}|^2dx + \int^{w}_{-w}|\psi_{\textrm{in}}|^2dx= 1,$$ where $w$ is now the width of your finite well, and the subscripts on the wavefunctions denote the difference from those in the classically forbidden region, and the classically allowed region. You may also need to use the condition of continuity of the wavefunction, to match constants on $\psi_{\textrm{in}}$ and $\psi_{\textrm{out}}$, that is the two functions must be equal to each other at $w$.

Note finally, that this intuitively makes sense. If I have a very long 'box', then the chance of finding the particle in any one region must be much smaller, than say if I have a very small box. So it does make sense that the probability of finding the particle tunneling, should depend on the size of the box.


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