Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose we have two probability measures $P_1$ and $P_2$ on some Riemannian manifold $(\Sigma,g)$. There are many potential distance measures between $P_1$ and $P_2$:

  • The Wasserstein distance
  • For $P_1$, $P_2$ given by density functions $f,g\in L^2$, we can use the functional distance $\|f-g\|_2$ (or any other $p$-norm)
  • The obvious generalization of histogram-related distances from the machine learning literature

Instead of representing a probability distribution directly, however, we can write a probability amplitude function $\psi:\Sigma\rightarrow\mathbb{C}$. That is, we assume $\psi\in L^2(\Sigma,\mathbb{C})$ with $\int_\Sigma |\psi|^2=1$. So, $\psi$ looks like the square root of a probability density.

Is there a natural distance metric on the space of such probability amplitude functions?

I realize that $L_2$ distances work here, but they don't have much meaning in this probabilistic view of the world, since $\exists c\in\mathbb{C}$ with $|c|=1$ such that $\|\psi-c\psi\|_2\neq0$, even though $\psi$ and $c\psi$ yield the same distribution function $|\psi|^2$. I'm looking for a norm that can be computed in terms of $\psi$ directly rather than having to first resort to $|\psi|^2$ directly. Writing in terms of Laplace-Beltrami or Dirac operator eigenfunctions is fair game.

share|cite|improve this question
Yes, it can distinguish between these cases. In fact, this may help clarify the role of the complex "redundancy" in expressing probability distributions using amplitude functions. – Justin Oct 4 '12 at 16:55
Should it distinguish between $\psi$ and $c\psi$ with $|c|=1$, but $c$ not constant, like in quantum mechanics? If it should, one natural thing to look at is $\Vert \psi \otimes \psi^\ast - \phi \otimes \phi^\ast \Vert$ for some operator norm. If not, and you are really talking about distributions disguised this way, it is unclear what exactly you want... – Alexander Shamov Oct 4 '12 at 16:55
Reminds me a bit of the Hellinger distance ( – Dirk May 16 '14 at 5:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.