3
$\begingroup$

I was told in class that a wave function should have the following properties:

  1. Finite and single-valued
  2. Continuous
  3. Differentiable
  4. Square integrable

But if we consider the wave function in an infinite square well, the wave function isn't differentiable at the boundaries since $\Psi (x)$ is:

$$\Psi (x) =\begin{cases} \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L}), & 0<x<L \\ 0, & \text{elsewhere} \end{cases}$$

This violates one of the properties of wave functions. So how is this an acceptable wave function?

$\endgroup$
2
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/19667/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 2, 2019 at 16:02
  • $\begingroup$ Psi function need not be continuous or differentiable at all points. These are conditions that are often valid and useful assumptions when looking for solutions of the Schroedinger equation with regular potential function, but in case of infinite discontinuity of potential function, they do not hold. $\endgroup$ Commented Nov 5, 2023 at 15:26

3 Answers 3

3
$\begingroup$

If you consider the differentiability of the wavefunction at the boundary from inside an infinite square well, you find: $\frac{d\Psi(x)}{dx}$ = $\sqrt{\frac{2}{L}}\frac{\pi}{L}$ as $x\rightarrow0^{+}$, and $\frac{d\Psi(x)}{dx}$ = $-\sqrt{\frac{2}{L}}\frac{\pi}{L}$ as $x\rightarrow L^{-}$. Outside the well, you obtain: $\frac{d\Psi(x)}{dx} = 0$ at $x=0$ and $x=L$. The derivatives do not match, hence the function is not differentiable at either boundary of the well.

This is easily seen by graphing the hybrid wavefunction (for say the ground state, $n=1$) and noting the sharp points.

enter image description here

The infinite square well is one of the first problems taught in undergraduate QM. It is an attempt at taking a free particle and trapping or localizing it over a finite domain, in order to obtain a normalizable wavefunction and demonstrate novel effects like energy quantization.

The problem is, you physically cannot produce a free particle over some finite domain, and then on either side arbitrarily "clamp it down" by imposing an infinite potential to instantaneously drive the wavefunction to zero. This is why you end up violating basic properties of the wavefunction, as you have mentioned.

You could, for instance, use a Fourier transform to build up a wave packet from a sum of plane waves of varying amplitude and momentum, approximating a sinusoidal solution over $0<x<L$, however, the probability of finding the particle outside of this domain would not sharply change to zero, as suggested by the infinite square well problem.

Finally, other issues arise as a result of this problem. Namely, the relationship $E=\frac{p^2}{2m}$ does not hold, where $p=\hbar k$, $k$ the wavenumber associated with the momentum, and you end up obtaining a continuous not discrete distribution of momenta, even though the energy is quantized. Carefully reading the Wikipedia article https://en.wikipedia.org/wiki/Particle_in_a_box helps elucidate these issues.

$\endgroup$
1
$\begingroup$

Like with a plane wave, the wave function you are giving is only a "limiting case". In reality you should start with finitely high walls and impose all four requirements that you list. Then, as you take the limit of higher and higher walls, you will obtain your wave function. However, think of it only as a limiting case. In reality, walls will always have finite height.

$\endgroup$
1
$\begingroup$

As it is written in the first answer, "the problem is, you physically cannot produce a free particle over some finite domain, and then on either side arbitrarily clamp it down by imposing an infinite potential to instantaneously drive the wavefunction to zero". Thus the infinite potential wall is not physical realizable, and so to study the situation in a theoretical way we drop one of the requirements of the wave function, in order to get an approximation of the wave function when the well is not infinite but really high.

Imposing the derivative to be continuous, would mean that we have a wave function $ \psi(x)$ such that $\psi'(0)=\psi'(L)=0$ combined with the fact that $\psi(0)=\psi(L)=0$. It is easy to see that $\psi(x)=0 $ is a solution satisfying both the boundary conditions and the Schrodinger equation. Thus for existence and uniqueness theorem for a linear, 2nd order ODE we would only have the null eigefunction $\psi(x) = \psi = 0$ satisfying the Schrodinger Equation.

This is a related question, expanding on the existence and uniqueness theorem for a linear 2nd order ODE:Wavefunction and its derivative vanish at same point

$\endgroup$
4
  • $\begingroup$ This only pushes the question further down the line: Why shouldn't the conclusion then be that there are no non-trivial solution? $\endgroup$
    – Trebor
    Commented Nov 4, 2023 at 4:45
  • $\begingroup$ @Trebor as it is written in the first answer, " the problem is, you physically cannot produce a free particle over some finite domain, and then on either side arbitrarily clamp it down by imposing an infinite potential to instantaneously drive the wavefunction to zero". So the infinite potential wall is not physical realizable, and so to study the situation in a theoretical way we drop one of the requirements of the wave function, in order to get an approximation of the wave function when the well is not infinite but really high. $\endgroup$
    – Luigi
    Commented Nov 5, 2023 at 8:09
  • $\begingroup$ Exactly, so not including this part in the answer would make it incomplete. If you edit this comment into the answer I'll retract my downvote. $\endgroup$
    – Trebor
    Commented Nov 5, 2023 at 8:49
  • $\begingroup$ @Trebor modified as suggestd. Thaks a lot for improving it! $\endgroup$
    – Luigi
    Commented Nov 5, 2023 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.