Heisenberg uncertainty principle in R^n

[kittensies]

Homework Statement

$$\phi(x)$$ is in Schwartz space, and $${\int|\phi(x)|^2dx=1$$.
I need to show that $$(\int_{R^n}|x|^2|\phi(x)|^2dx)(\int_{R^n}|\xi|^2|\phi(\xi)|^2d\xi)\geq \dfrac{n^2}{16\pi^2}$$

Homework Equations

Heisenberg uncertainty in one dimension:
$$(\int_{-\infty}^{\infty}|x|^2|\phi(x)|^2dx)(\int__{-\infty}^{\infty}|\xi|^2|\phi(\xi)|^2d\xi)\geq \dfrac{1}{16\pi^2}$$

Plancherel's theorem, probably: For any function in S(R), $$||f||=||\hat{f}||=(\int|f(x)|^2)^{-1/2}$$

Cauchy Schwartz inequality

Anything else?

The Attempt at a Solution

If I expand the integral, I get
$$(\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}(x_1+x_2+...+x_n)|\phi(x_1,x_2...)|^2dx_1 dx_2...d_x_n)$$

I guess I need to show that this is at least $$n/4\pi$$

If I expand the polynomial and decompose, I get

$$(\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}(x_1)|\phi(x_1,x_2...)|^2dx_1 dx_2...d_x_n)+(\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}(x_2)|\phi(x_1,x_2...)|^2dx_1 dx_2...d_x_n)+...+(\int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}(x_n)|\phi(x_1,x_2...)|^2dx_1 dx_2...d_x_n)$$

But each of the integrals evaluated individually is greater than $$1/16\pi \int_{-\infty}^{\infty}1dx_1... dx_{i-1} dx_{x+1}...d_x_n)$$ which blows up into $$1/16\pi \Pi_{i\neq j}x_i.$$I have a feeling that's not what's supposed to happen.

 

[dextercioby]

I can give you a hint to convert the functions in R^n from cartesian to spherical coordinates.