What are plane waves in Bethe ansatz

Question

I study Bethe ansatz, although my background is mathematics not physics. Can somebody explain to me what is plane waves? I have seen in many papers this expression that "The idea of the Bethe ansatz is that eigenfunctions are a linear combination of plane waves", or superposition of plane waves.

My understanding is for each particle a complex number which is called plane wave has been defined! then an eigenfunction is a combination of these complex numbers over all possible permutations. But why is such a complex number for each particle defined, and why is it called a plane wave?

Answer 1

I suppose that you consider the Lieb-Liniger model which describes a one-dimensional gas of bosons with a contact interaction, i.e. particles feel each other only when they are at the same point. In quantum mechanics, we are interested in the stationary states which are solutions of the Schrödinger equation. In the absence of any interaction between particles, the Schrödinger equation reads $$-{\hbar^2\over 2m}\sum_i{\partial^2\over\partial x_i^2}\psi(x_1,x_2,\ldots,x_n)=E\psi(x_1,x_2,\ldots,x_n)$$ and can be easily solved exactly. The solutions are $$\psi(x_1,x_2,\ldots,x_n)=Ae^{\imath\sum_i k_ix_i}$$ Each term $e^{i\imath k_ix_i}$ is called a plane-wave (very common in various other fields of physics : sound, electromagnetic waves, $\ldots$). The quantities $k_1,k_2,\ldots k_N$ are called wavevectors (even though they are not vectors since a 1D problem is considered) and they are real numbers because the square modulus of the wavefunction can be interpreted as a probability and should therefore satisfy the normalization condition $$\int |\psi(x_1,x_2,\ldots,x_n)|^2\prod_i dx_i=1$$ The total energy is given by $$E={\hbar^2\over 2m}\sum_i k_i^2$$ With interaction, Bethe considered the following ansatz $$\psi(x_1,x_2,\ldots,x_n)=\sum_{\cal P}a_{\cal P}e^{\imath \sum_i k_{{\cal P}_i}x_i}$$ where $\cal P$ is a permutation of $\{1,2,\ldots N\}$, i.e. a linear superposition of plane waves. Physically, the idea is that particles are free when they are at differents points so their wavefunction should be a plane wave. The superposition arises because of the interactions. It turns out that this ansatz is an exact solution of the Schrödinger for some particular values of $a_{\cal P}$.

Answer 2

This depends a bit on the context, but in essence you have a position variable $x$ (which is typically real but it can be discrete), and any function of the form $$ f(x) = e^{ikx}, $$ where $k$ is a real number, is called a plane wave.

This isn't just "some complex number": it's a complex-valued function of position. Moreover, each particle has some wavefunction $f:S\to\mathbb C$, which is a complex-valued function of your position space $S$ (which could be $\mathbb R$, or could be discrete), but not all such functions are plane waves: you could have $f(x) = 1/(x+i)$, say, or $f(x) = e^{ix^2}$, neither of which are plane waves.

The term plane wave comes from the fact that if you give them a time dependence, and you look at it in three dimensions, $$f(x,y,z,t) = e^{ikx}e^{-i\omega t} = e^{i(kx-\omega t)},$$ then $f$ oscillates harmonically with position, and its wavefronts (the surfaces of constant phase) form planes which move with time. For further details, consult a solid optics or advanced-EM textbook.