# Why can't a particle penetrate an infinite potential barrier?

I am studying basic quantum theory. My question is:

Why can't a particle penetrate an infinite potential barrier?

The reasoning that I have applied is that particles under consideration have finite energy. So, to cross an infinite potential barrier the particle requires infinite energy. But I cannot think of the mathematical relation between potential and energy so that indeed I am convinced that to cross an infinite potential barrier the particle needs infinite energy.

What is the relation between the potential and energy of quantum mechanical particles?

• You start with the tunneling probability knowing that it is exponentially small with the finite barrier height, therefore if the latter is infinite the former is zero. Once you see this you may use the infinite high potential well as a mathematical model for an impenetrable barrier. Commented Dec 19, 2018 at 15:53

Imagine a finite potential well of the form

$$V(x) = \begin{cases} 0 & |x| < L/2 \\ V_0 & {\rm otherwise}\end{cases}$$

You can solve Schrodinger's equation in the usual way, by splitting the domain in three parts, the resulting wave function will look something like this

$$\psi(x) = \begin{cases} \psi_1(x) & x < L/2 \\ \psi_2(x) & |x| \leq L/2 \\ \psi_3(x) & x > L/2\end{cases}$$

Inside the box $$\psi_2(x) \sim e^{\pm ikx}$$, but outside the box you will find

$$\psi_3(x) \sim e^{-\alpha x}$$

where

$$\alpha = \frac{\sqrt{2m(V_0 - E)}}{\hbar}$$

Now calculate the limit $$V_0\to\infty$$ (infinity potential barrier), and you will see that $$\psi_3(x)\to 0$$, same as $$\psi_1(x)$$. So in that sense the particle cannot penetrate the barrier and remains confined in the region $$|x| \leq L/2$$

• I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way. Commented Dec 19, 2018 at 16:10
• While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable). Commented Dec 19, 2018 at 18:07
• Hard to fathom why these critical comments were made. This is the correct physical answer, and it should have been accepted, instead of the other one which proves, uselessly, that either the wavefunction would be zero or else the total energy would be infinite. And next what? Would we have to proove then that the energy cannot be infinite? The leson is that, often, an argument based on a physical limit is more valuable than a mathematical "proof". Commented Apr 2, 2019 at 18:56
• @Kostas Thanks for the feedback, I certainly agree with your last statement ;) Commented Apr 2, 2019 at 19:53

The relation between the particle's wave function $$\psi(x)$$, potential $$V(x)$$ and energy is $$E = \int dx\ \psi^*(x)\left(-\frac{\hbar^2}{2m}\psi''(x) + V(x)\psi(x)\right) \quad \label((*)$$ Suppose $$V(x)$$ is bounded from below and is equal to $$+\infty$$ on some interval $$[x_1,x_2]$$. If $$\psi(x)\neq 0$$ for $$x\in[x_1,x_2]$$, then the energy $$E$$ is infinite. The term containing second derivative is always non-negative, so it can not compensate this infinity.

Update. This relation is well known in the quantum mechanics. I didn't mention that the norm of a wave function is usually taken to be $$1$$: $$\int dx\ \psi^*(x)\psi(x) = 1$$ Under this condition the Schrodinger equation $$-\frac{\hbar^2}{2m}\psi''(x) + V(x)\psi(x) = E\psi(x)$$ been multiplied by $$\psi^*(x)$$ and integrated by $$x$$ gives the relation (*).

The term $$-\frac{\hbar^2}{2m}\int dx\ \psi^*(x)\psi''(x)$$ corresponds to the kinetic energy of a particle, so it must be non-negative. Indeed, integration by parts leads to the following manifestly non-negative expression $$\frac{\hbar^2}{2m}\int dx\ \psi'^*(x)\psi'(x).$$ By the way, quantity $$\psi''(x)/\psi(x)$$ can be either positive or negative.

• Can I arrive at the same conclusion from the following equation: $-\frac{\hbar^2}{2m}\frac{1}{\psi (x)}\frac{\partial ^2\psi (x)}{\partial x^2}+V(x)=E$ Commented Dec 19, 2018 at 17:25
• Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx\ \psi^*(x)\left(-\frac{\hbar^2}{2m}\psi''(x) + V(x)\psi(x)\right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where? Commented Dec 19, 2018 at 17:34
• @Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $\psi''(x)/\psi(x)$ term in this case. I'll update my answer.
– Gec
Commented Dec 19, 2018 at 17:38
• Thirdly, since $\psi''(x)$ is non negative, the term $-\frac{\hbar^2}{2m}\psi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=\infty$ Commented Dec 19, 2018 at 17:39
• @Soumee I have updated my answer.
– Gec
Commented Dec 19, 2018 at 17:59

Gec's answer is the one I would consider rigorous, but the intuitive answer is this:

Suppose you put a particle detector in the barrier. How often do you expect to measure a particle there? Answer: never, because if you did this then the particle after measurement would be in a position eigenstate that would force you to conclude it had infinite (expectation value of) energy. And we're disallowing that.

The only states where there is no chance to ever measure the particle in the barrier are those with $$\psi=0$$ inside the barrier (or at least over a dense subset).

It depends on how you solve the problem. Imagine a parabolic equation, and you solve with finite elements method.

$$f''(x)=\dot{f}(x)$$

The basis functions were $$\phi_m(x)$$ triangles at the discretized points $$\phi_m(x_n)=\delta_{mn}$$ and the phi's are linear from $$x_{m\pm 1}$$ to $$x_m$$. $$f(x)=\sum_m a_m(t)\phi_m(x)$$ Then the weak formulation :

$$\int f''(x)\phi_n(x)dx=\int \dot{f}(x)\phi_n(x)dx\\ \sum_m a_m(t)\int\phi''_m(x)\phi_n(x)dx=\sum_m\dot{a}_m(t)\int\phi_m(x)\phi_n(x)dx$$

after integrating by parts leads to the matrix equation $$\sum_n A_{mn}a_n(t)=\sum_n B_{mn}(a_n(t+h)-a_n(t))$$

Were A, the rigidity matrix, and B are tridiagonal matrices.

Hence the function at the next time step is $$a_n(t+h)=(\sum_m (hB^{-1}A+\mathbb{1})_{nm}a_{m}(t)$$.

The inverse of B is a full matrix, hence the function at point $$x_n$$ at the next time step depends on the values at all points at the previous time step.

Hence the solution in the other side becomes non zero and with this global method there is tunneling.

However it is a mathematical artefact because the base functions are not orthogonal.

To explain physically how the particle passes through, the exit is maybe that the wavefunction has countably infinite spatial nodes with probability 0. Since a measurement is finite it can only give rational results. But in the continuum the wavefunction oscillates between the nodes but it is measured zero since Q is dense in R.