# Can we have discontinuous wavefunctions in the Infinite Square well?

The energy eigenstates of the infinite square well problem look like the Fourier basis of L2 on the interval of the well. So then we should be able to for example make square waves that are an infinite linear combination of those energy eigenstates. But since linear superpositions of solutions are solutions, does this mean that the square function is an allowable wavefunction that a particle can be in in this problem? The issue I'm having is the fact that the time-independent Schrödinger equation (TISE) involves the derivative of the wavefunction, but the square function has none, so how can it be a solution to the equation?

• It probably makes sense if you extend the meaning of the Schrödinger equation in terms of distributions. Sep 24, 2012 at 11:42

In the $L^2$ calculus that is relevant for the wave functions from the Hilbert space, it is not really true that the "derivative of a discontinuous function doesn't exist".

For example, if a wave function $\psi(x)$ obeys $$\psi(x)=a,\quad x\lt 0$$ and $$\psi(x)=b,\quad x\gt 0$$ then the value of $\psi(0)$ is physically irrelevant as long as it is finite – it doesn't affect the equivalence class of the $L^2$ functions – and the derivative of the wave function is $$\psi'(x) = (b-a)\delta(x).$$ The derivative of a step function is a multiple of the Dirac delta-function, a function that vanishes everywhere except for $x=0$ where it's "infinite enough" so that the integral equals one. It's really a distribution, not a function, but it has very natural Fourier transform (namely a constant function) and one may deal with such elements of the Hilbert space in all other bases of the Hilbert space.

A reason why discontinuous functions with a step don't appear in real-world physical problems is that the expectation value of the kinetic energy is infinite. It's really because the expectation value of the kinetic energy is proportional to the integral of $\psi'(x)$ and the integral of $\delta(x)^2$ diverges because $\delta(x)^2$ is "much more infinite" at zero than $\delta(x)$ itself.

As long as we know that the kinetic energy is smaller than a certain finite bound, we also know that the wave function won't have this sort of a discontinuity. But that doesn't mean that we should never calculate with such discontinuous wave functions – and even wave functions equal to distributions such as delta-functions and their derivatives. In fact, physicists work with them all the time because such wave functions are very natural mathematically, even though none of them may really appear as a "strictly realistic" state vector that satisfies all the normalizability and finite-energy criteria. Realistic wave functions such as wave packets may often be expressed as an integral of such unsmooth wave functions. The integration "smears them out" and eliminates the singular features.

• ...and one may deal with such elements of the Hilbert space.... It is false! a Dirac delta is not an element of $L^2$. Generalized functions defining functionals on $L^2$ can live in a different space. This is the case of the Dirac delta. May 20, 2021 at 8:56
• It's a normal sloppy terminology used by physicists. If you need to know, the generalized Hilbert space that allows Dirac-like distributions is called the rigged Hilbert space. May 21, 2021 at 9:04
• I know quite well QM with rigged Hilbert spaces. However, even without adding that level of math machinery, I do not think it is a good idea to write that $\delta(x)$ is an element of $L^2$. Moreover, the whole story of delta functions could be avoided from the beginning by just stating explicitly the reasons an $L^2$ function in the domain of the Hamiltonian cannot have discontinuities. May 21, 2021 at 10:52
• The delta-function surely isn't square-integrable and I am making no such claim. However, it is an excellent powerful tool as a generalized basis vector to deal with the Hilbert spaces of square-integrable functions. - I don't want to avoid delta-functions. I think that they should be taught and any teaching of this math-relevant-for-physics that tries to avoid them is myopic. May 22, 2021 at 11:09
• A substantial part of my answer WAS about the fact that delta squared is not integrable, so the delta function is not an element of L2. But the statement that I made is that the delta function is an important tool in the calculus of the L2 functions and I insist it is a true and important statement. May 25, 2021 at 4:29

I) Well, in general there is a notion of weak solutions, i.e. one may e.g. rewrite the time independent Schrödinger equation from differential form

$$\tag{1} -\frac{\hbar^2}{2m} \psi^{\prime\prime}(x) ~=~ (E-V(x))\psi(x)$$

into an integral equation

$$\tag{2} \frac{\hbar^2}{2m} \psi(x) ~=~ \int^x \!\!dx^{\prime} \int^{x^{\prime}} \!\!dx^{\prime\prime}~ (V(x^{\prime\prime})-E)\psi(x^{\prime\prime})$$

where the wave-function $\psi(x)$ do no longer have to be twice differentiable for the eq.(2) to make sense (as opposed to eq.(1)).

(OP is asking a more general question involving the time dependent Schrödinger equation, see section V, but let us for simplicity first consider the time independent Schrödinger equation.)

In the following, we will always assume that the wave-function $\psi\in{\cal L}^2(\mathbb{R})$ is square integrable (=normalizable).

Note however, that if one assumes that $V,\psi\in{\cal L}^2_{\rm loc}(\mathbb{R})$ are both locally square integrable, then one may deduce (via a sort of bootstrap argument, cf. this Phys.SE answer) that a solution $\psi$ is in fact differentiable with continuous derivative, $\psi\in C^1(\mathbb{R})$.

II) Next note that the infinite square well potential

$$\tag{3} V(x)~=~\left\{\begin{array}{ccc} 0 &\text{for}& |x| < a \cr\cr \infty &\text{for}& |x| \geq a \end{array} \right.$$

is an idealization of the finite square well potential$^1$

$$\tag{4} V(x)~=~V_0 ~\theta(|x|-a),$$

where $V_0$ is a very large positive constant, much bigger than the energy of the particle that we would like to study.

III) On one hand, the finite square well potential (4) is indeed locally square integrable, so the solutions $\psi$ are $C^1(\mathbb{R})$. One may show that a normalizable solution $\psi$ only exists for certain discrete energy levels $E_n$ (which depend on the parameter $V_0$). One may moreover show in the limit $V_0\to \infty$, that these energy eigen-functions $\psi_n$ i) remain continuous functions, and ii) that they vanish outside the well $|x|\geq a$.

IV) On the other hand, the infinite square well potential (3) is not locally square integrable, but one may show that the restriction of the solutions $\psi$ to the well $|x|< a$ must be $C^1(]-a,a[)$, i.e. that $\psi$ could at most have a discontinuity at the potential walls $x=\pm a$. Outside the well $|x|>a$, the integral eq. (2) is not well-defined due to the infinite potential. On physical grounds based on experience from the finite case (4), we now declare that the energy eigen-functions $\psi_n$ should be i) continuous and ii) that they should vanish outside the well $|x|\geq a$. In particular, it is physically reasonable to impose Dirichlet boundary condition $\psi_n(x\!=\!\pm a)=0$ at the potential walls.

V) Now let us return to OP's question. Yes, the Hilbert space is $H=L^2([-a,a])$ with basis $\psi_n$ given by the energy eigen-functions $\psi_n$ from Section IV. Square integrable infinite linear combinations

$$\tag{5} \psi~=~\sum_{n=1}^{\infty} c_n \psi_n ~\in~ H$$

do not need to be continuous.

Example: The odd and discontinuous wave-function

$$\tag{6} \psi(x)~=~\left\{\begin{array}{ccc} A~{\rm sgn}(x) &\text{for}& |x| < a \cr\cr 0 &\text{for}& |x| \geq a \end{array} \right.$$

is normalizable $\psi\in H=L^2([-a,a])$. It can be written as an infinite series

$$\tag{7} \psi(x)~=~\sum_{n=1}^{\infty} c_n \psi_n(x) ~=~\frac{4A}{\pi}\sum_{k=0}^{\infty}\frac{1}{2k+1} \sin\frac{(2k+1)\pi x}{a}$$

of energy eigen-functions, and the series converges both $x$-pointwise and in $L^2$-norm. So purely from the perspective of the Hilbert space $H=L^2([-a,a])$, the discontinuous wave-function (6) is perfectly fine.

However, one may show that the kinetic energy of the wave-function (6) is infinite, and thus not physically acceptable. Arguments along these lines naturally leads one to take a closer look of the Hamiltonian operator, which is an unbounded operator, defined on a pertinent domain, and study its self-adjoint extension. We leave that to the interested reader.

$^1$The finite potential (4) is itself also an idealization, but we will ignore that here.

In an infinite 1D well the energy of the $n$th mode is proportional to $n^2$. However if you're summing modes to make a square well the factor for the $n$th mode is 1/$n$. That means to make a square wave would require an infinite amount of energy, which is not surprising since it would have an infinite first derivative.