# Bethe ansatz wavefunction vs plane waves

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?

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.
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.
• OK so it is just like another basis for the "space of plane waves"... However, I think there are a few things which are not precise. First, the case of one particle the Ansatz I gave is degenerate: $\psi = 0$ because there are no pairs to sum over. Second, in the case of two particles, I think you missed a factor of $(x_1 - x_2)$ multiplying $\delta(x_1 - x_2)$ in $d \psi/dx_1$. That makes the deltas not cancel in $d^2 \psi / d x_1^2 |_{x_2=0}$. Commented Jun 22, 2020 at 9:59
• Moreover, I don't know what you meant by "If you care about the cases where there is possible contact of coordinates, x1=x2 you will start getting the delta terms which are indeed just the inverse Fourier transform of plane waves". If you have a contact term when $x_1=x_2$, what you have is a delta function in coordinate space $\delta(x_1-x_2)$, and not in momentum space (which would be the Fourier transform of a plane wave $e^{ikx}$ ). Commented Jun 22, 2020 at 10:02