# Infinite Potential Well Energy for Piece-wise Constant Wave Function

I'm trying to compute the expectation value of energy for a certain state in an infinite potential well but I'm getting contradictory answers.

The well has potential \begin{align} V(x) = \left\{ \begin{array}{lr} 0 & : 0 < x < L\\ \infty & : \text{ elsewhere} \end{array} \right. \end{align} which has eigenstates $\phi_n(x) = \sqrt{\frac{2}{L}} \sin(\frac{n \pi x}{L})$ with corresponding energies $E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$.

Now, consider the state \begin{align} \psi(x) = \left\{ \begin{array}{lr} 0 & : 0 < x < L/2\\ \sqrt{\frac{2}{L}} & : L/2 \leq x < L\\ 0 &: \text{ elsewhere} \end{array} \right. \end{align} We want to compute $\langle H \rangle$ for this state. One way to do this is simply using the definition: $\langle H \rangle = \int_{0}^L \psi^*(x) H \psi(x) dx$. The problem with this though is that $\psi(x)$ is piecewise constant, and therefore this will give you $0$. The other option is to expand $\psi$ in terms of the energy eigenbasis by computing the coefficients $c_n = \langle \phi_n | \psi \rangle$, and getting $\langle H \rangle = \sum_{n=1}^\infty | c_n |^2 E_n$. As $E_n > 0$ for every $n$, this quantity will be strictly greater than $0$ and therefore will differ from the previous answer.

What is the discrepancy? It surely has to do with the discontinuity at $x = L/2$, but I can't figure out how to deal with it.

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Note that your piecewise-constant $\psi(x)$ is not continuous, and is therefore not a physically available wavefunction. Try repeating the calculation using $C^1$ functions that converge to $\psi$ and you will get unboundedly-increasing contributions from the "glue" at 0 and L. –  Emilio Pisanty Jun 2 '13 at 16:31
More generally, representing discontinuous functions using Fourier series is a nontrivial affair and leads to the Gibbs phenomenon. –  Emilio Pisanty Jun 2 '13 at 16:35

The expectation value

$$\tag{1} \langle \psi | V| \psi \rangle~=~0$$

of the potential energy operator $V$ is indeed zero, but the expectation value

$$\tag{2} \langle \psi |K| \psi \rangle~=\frac{\hbar^2}{2m} \int_{\mathbb{R}}\!dx~ |\psi^{\prime}(x)|^2 ~=~+\infty$$

of the kinetic energy operator $K$ is actually infinite for the wave function

$$\tag{3} \psi(x)~=~ \sqrt{\frac{2}{L}}\left(\theta(x-\frac{L}{2}) -\theta(x-L)\right), \qquad x\in \mathbb{R}.$$

Here $\theta$ is the Heaviside step function.

The kinetic energy operator $K:=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ is an example of an unbounded operator, which only make sense on its domain ${\cal D}_K\subsetneq {\cal H}$ inside the Hilbert space ${\cal H}:=L^{2}(\mathbb{R})$ of square Lebesgue integrable functions. In particular, it is a non-trivial mathematical problem how to apply the differential operator $K$ to the non-differentiable wave function (3).

The infinite result (2) can be seen (at the physical level of rigor) in at least three ways (ordered with the computationally simplest calculation first):

1. Plug the Heaviside step function into eq. (2) to get an integral over the square of a pair of Dirac delta function situated at $x=\frac{L}{2}$ and $x=L$. This is strictly speaking mathematically ill-defined. Physically, it makes sense to assign the integral the value infinite, cf. this Phys.SE post.

2. Calculate the overlaps $c_n=\langle \phi_n | \psi \rangle$, and show than the sum
$$\tag{4} \langle \psi |H| \psi \rangle=\sum_{n=1}^{\infty}|c_n|^2 E_n ~=~+\infty$$ diverges. This infinite conclusion seems physically robust, since all terms in the series (4) are non-negative.

3. By regularization, as Emilio Pisanty suggests in a comment. Define a regularized wavefunction $\psi_{\varepsilon} \in C^1(\mathbb{R})$ in such a way that (i) it converges $\psi_{\varepsilon}\to \psi$ for $\varepsilon\to 0^{+}$, (ii) the expectation value $\langle \psi_{\varepsilon} |K| \psi_{\varepsilon} \rangle$ is easy to compute and finite for $\varepsilon>0$. Show that $\langle \psi_{\varepsilon} |K|\psi_{\varepsilon} \rangle\to +\infty$ diverges for $\varepsilon\to 0^{+}$.

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Thanks, my issue with approach (1) is, as you say, it doesn't make sense to me mathematically. I was thinking in terms of Lebesgue integrals, where $\psi'$ is $0$ almost everywhere and therefore its integral (or the integral of its square) is therefore $0$ (is there something wrong with this thought process?). Approach (2) is the most straight forward (in my opinion) as it can be done in a brute force way. –  user1394629 Jun 2 '13 at 18:01
This requirement that physical wave functions be continuously differentiable is confusing to me. It could just be my lack of understanding of infinite dimensional QM, but take any 1D wave function $\psi$ (make it as nice as you like) and choose $x^*$ so that $\psi(x^*) \neq 0$. Now measure if the particle is to the left or right of $x^*$. Supposing you find the particle to the right of $x^*$, what is the state after measurement? Being used to finite dim., I assume you 'project $\psi$' (set $psi(x) = 0$ for $x < x^*$) and normalize. This function won't be continuous but I can't see the problem. –  user1394629 Jun 2 '13 at 18:16
The problem is that the wave function $\psi$ in general has to belong to the domain ${\cal D}_A$ of the self-adjoint operator $A$ if one wants to measure the expectation value $\langle \psi |A| \psi \rangle$. –  Qmechanic Jun 2 '13 at 19:15