Bethe ansatz wavefunction vs plane waves

Question

I am reading Negele & Orland's "Quantum many-particle systems". In problem 1.9 you show that the (Bethe ansatz) wave function

$$ \psi(\{x \}) = \exp \left( - \alpha \sum_{i < j}^N |x_i - x_j| \right) $$

solves the Schrödinger equation for $H= - \sum_{k=1}^N \frac{\partial^2}{\partial x_k^2}$ (free Hamiltonian, in units where $\hbar^2/2m=1$). In particular the above state has negative energy, so it is a bound state.

Question: usually when you diagonalize a free Hamiltionian what you get is plane waves. Why is it that $\psi$ also diagonalizes $H$ although it is not a plane wave (in general)? Intuitively I would say that you can decompose $\psi$ in a sum of plane waves with different wavenumbers (with different directions and magnitudes), and therefore their sum would neither be a momentum eigenstate nor a Hamiltonian eigenstate. But on the other hand direct computation shows that it is indeed a Hamiltonian eigenstate. Why?

Answer 1

There is no theorem telling you that there is just a single way of describing a space of functions. Think about the eigenvalue equation $H\psi = \lambda\psi$ just as a differential equation. Whether you solve it in momentum space or in coordinate space is not important. In the approach above they have decided to stay with coordinate space. And as is known from differential equations, the space of solutions to the homogeneous problem can be described by many different basis. Also notice how the boundary conditions play a role. The Ansatz above goes to zero whenever one of the coordinates grows too much.

To be clearer take first the case of just one particle and try to Fourier transform back the usual plane wave... See what you get. Compare with the Ansatz at least pictorially.

For the case of two particles you could track down possible divergences, write the Ansatz as $$\psi(\{x\}) = \exp\left[-\alpha(\theta(x_1-x_2)(x_1-x_2)+\theta(x_2-x_1)(x_2-x_1)) \right]$$ where $\theta(x)$ is the Heaviside function (in order to represent the absolute value). Taking a first derivative leads us to: $$\begin{align} \frac{d\psi}{dx_1} &= -\alpha\bigg(\delta(x_1-x_2)(x_1-x_2)+\theta(x_1-x_2) -(x_1\leftrightarrow x_2)\bigg)\psi(\{x\})\end{align}$$ Let us take the second derivative $$\begin{align} \frac{d^2\psi}{dx_1^2} &= -\alpha\bigg(\delta'(x_1-x_2)(x_1-x_2)+\delta(x_1-x_2)+\delta(x_1-x_2)\\ &\qquad +\delta'(x_2-x_1)(x_2-x_1) +\delta(x_2-x_1) + \delta(x_2-x_1)\bigg) \psi(\{x\})\\ &\qquad + \alpha^2 \bigg(\delta(x_1-x_2)+\theta(x_1-x_2) -(x_1\leftrightarrow x_2)\bigg)^2\psi(\{x\})\\[7pt] &= -4\alpha\delta(x_1-x_2)\psi(\{x\})\\ &\qquad + \alpha^2 \bigg(\delta(x_1-x_2) (x_1-x_2) +\theta(x_1-x_2) -(x_1\leftrightarrow x_2)\bigg)^2\psi(\{x\}) \end{align} $$ where $'$ on the delta is to be understood in the distributional sense and I have used the fact that $\delta$ is an even distribution. We can now imagine particle number two is fixed at $x_2=0$ by some external reason. We would expect the problem to reduce to the single particle case at least for $x_1\neq 0$, $$H\psi = -\frac{d^2\psi}{dx_1^2}\bigg|_{x_2=0} = \left\{ \begin{array}{ll} -\alpha^2\psi(\{x\}) & \qquad\text{for } x_1 >= 0 \\ -\alpha^2\psi(\{x\}) & \qquad \text{for }x_1 < 0 \end{array}\right. $$ which checks out.