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The Calogero-Sutherland operator on the space of homogeneous symmetric polynomials in $n$ variables is defined by $$ \frac{\alpha}{2}\sum_{i=1}^n x_i^2\frac{\partial^2}{\partial x_i^2} + \frac{1}{2}\sum_{1\leqslant i < j \leqslant n} \left(\frac{x_i+x_j}{x_i-x_j}\right)\left(x_i\frac{\partial}{\partial x_i}-x_j\frac{\partial}{\partial x_j}\right). $$ It is claimed in the above link that the Jack polynomial $J^{(\alpha)}_\lambda$ of an integer partition $\lambda$ is an eigenpolynomial of this operator, with eigenvalue $$ \sum_{i=1}^n \left(\frac{\alpha}{2}\lambda_i^2 + \frac{n+1-2i}{2} \right). $$ This formula is not in agreement with the one given at page 10 of this paper, which claims that the eigenvalue is $$ \sum_{i=1}^n \left(\frac{\alpha}{2}\lambda_i^2 + \frac{n+1-2i}{2}\lambda_i \right). $$ I implemented the Jack polynomials and I'm pretty sure my implementation is correct (I started to implement them a long time ago and the implementation has been intensively tested). However, I find that the Jack polynomials are indeed eigenpolynomials of this operator but none of the two above formulas are in agreement with the eigenvalues I get. For example, for $n=3$ and $\lambda=(2,1)$, I find the eigenvalue $\alpha + 2$.

So what is the correct expression of the eigenvalue?

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There may an issue here with different definitions of the CS operator. The second expression for the eigenvalues in the OP is for a slightly different operator: \begin{align} &H=\frac{\alpha}{2}\sum_{i=1}^n \left(x_i\frac{\partial}{\partial x_i}\right)^2 + \frac{1}{2}\sum_{1\leqslant i < j \leqslant n} \left(\frac{x_i+x_j}{x_i-x_j}\right)\left(x_i\frac{\partial}{\partial x_i}-x_j\frac{\partial}{\partial x_j}\right), %&E_n=\sum_{j=1}^n\bigl(\lambda_j^2+\beta(n+1-2j)\lambda_j\bigr). \end{align} see Exact operator solution of the Calogero-Sutherland model. So this differs from the operator in the OP, which has $x_i^2\frac{\partial^2}{\partial x_i^2}$ in the first sum.

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    $\begingroup$ Right, it works. And the correct formula is the second one. Thanks. $\endgroup$ Commented Apr 30 at 22:45

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