# Hilbert Space Cross Products

1. Jan 7, 2016

### DuckAmuck

The equivalent of a dot product in Hilbert space is:
$$\langle f | g \rangle = \int f(x) g(x) dx$$
And you can find the angle between functions/vectors f and g via:
$$\theta = arccos\left( \frac{\langle f | g \rangle}{\sqrt{\langle f|f \rangle \langle g|g \rangle}} \right)$$

So is it possible to come up with a cross product between two Hilbert space vectors? I know with discrete dimensionalities, you can only have vectors result from cross products in 3D and 7D, but is it possible here?
If not, is it possible to get something akin to a bivector, like with 4D cross products?

Is it possible to get something like:
$$\langle f | \times |g \rangle = | h \rangle$$
where
$$\langle f |h \rangle = 0$$
$$\langle g |h \rangle = 0$$

Maybe the cross product is a more complicated object akin to a bivector, like something of the form:
$$\langle f | \times |g \rangle = | h \rangle | k \rangle$$
or
$$\langle f | \times |g \rangle = | h \rangle \langle k |$$
Is there a formal way to construct a cross product like this in Hilbert space?

As an aside: if
$$cos \theta = q$$
then
$$sin \theta = \sqrt{1 - q^2}$$

And for cross products, the maginitude is $$|A \times B| = |A||B|sin \theta$$

So it seems to follow from the second equation in this post, that the magnitude of a Hilbert cross product is:
$$| \langle f | \times |g \rangle | = \sqrt{ \langle f|f \rangle \langle g|g \rangle - \langle f | g \rangle^2 }$$

Last edited: Jan 7, 2016
2. Jan 7, 2016

### thephystudent

In 3D, the cross-product is the Hodge dual of the wedge product of the two vectors. If the vectors are 1D, their wedge product is 2D, and in an N-dimensional space the hodge dual of this would be (N-2)-dimensional. So at least it's not just a vector(ket), it might be something tensorial but I don't know if such a thing is defined in any way in a quantum Hilbert space.