1. Jun 10, 2009

pellman

Let $$H=L^2(R^n)$$

Here H is a Hilbert space, R^n is of course the nth product of the real line. what is L-squared? context is a quantum text for mathematicians.

Later the author uses $$H=L^2([0,1])$$ in another example.

2. Jun 10, 2009

nicksauce

3. Jun 10, 2009

pellman

That makes perfect sense. Thank you!

4. Jun 11, 2009

daviddoria

$$L^2([0,1])$$ means functions whose square integral from 0 to 1 converges, but what is $$L^2(R^n)$$? Functions whose square integral from $$-\infty$$ to $$\infty$$ converges? Or any sub interval of $$R^n$$?

5. Jun 11, 2009

maze

It is the integral over all of Rn. For example, if n=3, then $\int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y,z) dx dy dz$, which is also written in more compact form as $\int_{\textbf{R}^3} f(\textbf{x}) d\textbf{x}$