I want to verify explicitly that for $N$ particles in two dimensions the function $f(x)=g(x)h(x)$ where

$$g(x)=\prod_{i\neq j} \vert x_i-x_j \vert^{2\beta/N}$$


$$h(x)=e^{-\beta \sum_{i=1}^N \vert x_i \vert^2}$$

is an eigenfunction to the Hamiltonian $H$ with eigenvalue zero, i.e. satisfies $Hf=0$.

The physical motivation for this question is that $f$ is the Gibbs measure of a classical electron gas and this one can be revcovered as the ground state of a certain Hamiltonian.

The Hamiltonian reads

$$ H = \left(-\Delta + \beta^2 \vert x \vert^2 - 4\beta N \right)+ \frac{\beta^2}{N^2}\sum_{i=1}^N \left( \sum_{j \neq i} \frac{x_i-x_j}{\vert x_i-x_j \vert^2} \right)^2-\beta^2(N-1).$$

What I managed was to show that

$$g(x)\left( -\Delta +\beta^2 \vert x \vert^2 - 4\beta N \right)h(x)=0.$$

But this was only the "easy part" of the game. Does anybody see how to show in full generality $$Hf=0?$$

Please let me know if you have any questions.

  • $\begingroup$ What goes wrong if you simply write down $Hf$? What issue are you having exactly? $\endgroup$ – John Donne Jan 18 '20 at 8:51
  • $\begingroup$ Is $x$ meant to be the vector containing the $x_i$? $\endgroup$ – Arthur Morris Jan 18 '20 at 14:49
  • 1
    $\begingroup$ @ArthurMorris correct, $x=(x_1,..,x_N)$ where $x_i$ is again a vector in two dimensions describing the position of the $i$ th particle. $\endgroup$ – Xin Wang Jan 18 '20 at 15:22
  • $\begingroup$ Is this from some reference? $\endgroup$ – Qmechanic Jan 19 '20 at 21:50

Here's what I could do. Not complete, but my workings were too long for a comment.

Firstly, I'm not sure how you managed to get that $$ (-\Delta +\beta^2\mathbf{x}^2-4\beta N)h(\mathbf{x}) = 0. $$ My working out was \begin{align} \Delta h(\mathbf{x})&= \sum_k \Delta_kh(\mathbf{x}) \\ &= \sum_k\nabla_k\cdot\nabla_k e^{-\beta\sum_k \mathbf{x}_k^2} \\ &= \sum_k\nabla_k\cdot\left((-2\beta\mathbf{x}_k) e^{-\beta\sum_k \mathbf{x}_k^2} \right)\\ & = -2\beta\sum_k\left[ e^{-\beta\sum_k \mathbf{x}_k^2}(\underbrace{\nabla_k\cdot\mathbf{x}_k}_{2}) +\mathbf{x}_k\cdot(-2\beta\mathbf{x}_k) e^{-\beta\sum_k \mathbf{x}_k^2} \right]\\ &=(-4\beta N+4\beta^2 \mathbf{x}^2)h(\mathbf{x}) \end{align} where I have used that $\nabla_k\cdot\mathbf{x}_k = 2$ because the system is $2$ dimensional. If I've gone wrong here then please do let me know!

As for that horrible looking sum, I have it mostly figured out I think. Firstly, we need that \begin{align} \frac{\partial}{\partial y_i} |\mathbf{y}-\mathbf{a}|^n &= \frac{\partial}{\partial y_i}\left(\sum_j (y_j-a_j)^2\right)^{\frac{n}{2}}\\ &=\frac{n}{2}\left(\sum_j (y_j-a_j)^2\right)^{\frac{n}{2}-1}\cdot2(y_i-a_i)\\ &=n|\mathbf{y}-\mathbf{a}|^{n-2} \cdot(y_i-a_i) \end{align} so that $$ \nabla|\mathbf{y}-\mathbf{a}|^n = n|\mathbf{y}-\mathbf{a}|^{n}\cdot\frac{\mathbf{y}-\mathbf{a}}{|\mathbf{y}-\mathbf{a}|^2}. $$ Now separate the product $g(\mathbf{x})$ into two parts: those containing $\mathbf{x}_k$, and those not. It is then possible to differentiate it, giving \begin{align} \nabla_k g(\mathbf{x}) &= \frac{\partial}{\partial \mathbf{x}_k}\prod_{i \neq j} |\mathbf{x}_i-\mathbf{x}_j|^{\frac{2\beta}{N}}\\ &=\prod_{i \neq j, (i, j \neq k)} |\mathbf{x}_i-\mathbf{x}_j|^{\frac{2\beta}{N}}\nabla_k\prod_{l \neq k} |\mathbf{x}_k-\mathbf{x}_l|^{\frac{4\beta}{N}}\\ &= \underbrace{\prod_{i \neq j, (i, j \neq k)} |\mathbf{x}_i-\mathbf{x}_j|^{\frac{2\beta}{N}}\cdot\prod_{l \neq k} |\mathbf{x}_k-\mathbf{x}_l|^{\frac{4\beta}{N}}}_{g(\mathbf{x})}\sum_{m\neq k} \left[\frac{4\beta}{N} \frac{\mathbf{x}_k-\mathbf{x}_m}{|\mathbf{x}_k-\mathbf{x}_m|^2}\right]\\ & = \frac{4\beta}{N}g(\mathbf{x}) \sum_{m\neq k} \left[\frac{\mathbf{x}_k-\mathbf{x}_m}{|\mathbf{x}_k-\mathbf{x}_m|^2}\right] \end{align} Differentiating again: \begin{align} \Delta g = \sum_k \Delta_k g &= \sum_k \nabla_k\cdot\nabla_k g \\ &= \sum_k \nabla_k\cdot\frac{4\beta}{N} \sum_{m\neq k} \left[\frac{\mathbf{x}_k-\mathbf{x}_m}{|\mathbf{x}_k-\mathbf{x}_m|^2}\right]g(\mathbf{x})\\ &=\frac{16\beta^2}{N^2} \sum_k \left(\sum_{m\neq k} \left[\frac{\mathbf{x}_k-\mathbf{x}_m}{|\mathbf{x}_k-\mathbf{x}_m|^2}\right]\right)^2g(\mathbf{x})\\ &\quad+\frac{4\beta}{N}g(\mathbf{x})\sum_k\underbrace{\left[\frac{1}{|\mathbf{x}_k-\mathbf{x}_m|^2} \underbrace{\nabla_k\cdot(\mathbf{x}_k-\mathbf{x}_m)}_{2}+(\mathbf{x}_k-\mathbf{x}_m)\cdot(-2)\frac{\mathbf{x}_k-\mathbf{x}_m}{|\mathbf{x}_k-\mathbf{x}_m|^4}\right]}_{0}\\ &=\frac{16\beta^2}{N^2} \sum_k \left(\sum_{m\neq k} \left[\frac{\mathbf{x}_k-\mathbf{x}_m}{|\mathbf{x}_k-\mathbf{x}_m|^2}\right]\right)^2g(\mathbf{x}) \end{align} The last line makes me think you have made a mistake with the exponent in $g$ - are you sure it is $2\beta/N$ and not $\beta/2N$? That would fix the factor of $16$.

From here I get a bit stuck. If the Hamiltonian were instead $$ \mathcal{H} = (-\Delta +4\beta\mathbf{x}^2-4\beta N)+\frac{\beta^2}{N^2} \sum_k \left(\sum_{m\neq k} \left[\frac{\mathbf{x}_k-\mathbf{x}_m}{|\mathbf{x}_k-\mathbf{x}_m|^2}\right]\right)^2 $$ and the exponent of $g$ were $\beta/2N$ then the problem would be solved, but the extra term on the end confuses me. There's also no guarantee that the above calculation is correct! Have a look and let me know if you have any trouble understanding my working, or if you spot a mistake.


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