TITLE: Why is the Vandermonde determinant harmonic?
QUESTION [43 upvotes]: It can be checked that the Vandermonde determinant defined as 
$$V(\alpha_1, \cdots, \alpha_n) = \prod_{1 \le i < j \le n}(\alpha_i-\alpha_j) $$
is a harmonic function, that is $\Delta V = 0$ where $\Delta$ is the Laplace operator. Is there a deeper or more intuitive reason why this fact should hold? The straightforward proof of just computing the derivatives and checking doesn't provide any insights.

REPLY [2 votes]: The other answers already answered the question. I just wanted to point out that this type of question is also part of the Dyson Brownian motion topic. You can see that from the answers already stated, for example Carlo Beenakker. If you look up Dyson Brownian motion you can get to this old blog post of Tao
https://terrytao.wordpress.com/2010/01/18/254a-notes-3b-brownian-motion-and-dyson-brownian-motion/
Around equation (9) he starts to address the type of algebra you mention. In your notation, his equation (11) says
$$
\frac{\partial}{\partial \alpha_i} V\, =\, \sum_{\substack{j\\j\neq i}} \frac{V}{\alpha_i-\alpha_j}\, .
$$
Applying that again and the derivative of $1/x$ you get
$$
\frac{\partial^2}{\partial \alpha_i^2} V\, =\, \sum_{\substack{j,k\\|\{i,j,k\}|=3}} \frac{V}{(\alpha_i-\alpha_j)(\alpha_i-\alpha_k)}\, .
$$
Then see his Exercise 25 and the simple algebraic fact that facilitates it. Of course this might not be the deeper reason you want. And then you have all the other deeper connections mentioned by the other folks who already gave you answers.