Overlap function of three modes

Question

In a physics paper today, I encountered a computation of the overlap of three modes. Here I'm asking what it means to compare three modes (or three vectors or three wavefunctions) in this way.

Very often we talk about the overlap of two modes. Here a "mode" is a scalar-valued function over some space ($R^1$, $R^2$, or $R^3$); in physics, a common example is a quantum mechanical wavefunction.

The overlap of two modes is just a dot product. For example, suppose we have two mode functions $u$ and $v$. Then their overlap is:

$$\left|\langle u | v \rangle\right|^2 \equiv \frac {\left(\int u(x) v(x) dx\right)^2} {\left(\int u(x)^2 dx\right)\left(\int v(x)^2 dx\right)}$$

Here we can interpret the integrals in the denominator simply as normalizations, to turn $u$ and $v$ into unit vectors.

The paper I was reading today instead involved the overlap of three modes, defining something like a dot product between three modes $u$, $v$, and $w$:

$$\Lambda \equiv \frac {V\ \left(\int u\ v\ w\ dA\right)^2} {\left(\int u^2~dA\right)\left(\int v^2~dA\right)\left(\int w^2~dV\right)}$$

where $dV$ indicates integration over volume, $V=\int dV$ is the total volume, and $dA$ indicates integration over area (part of the boundary of the integration volume). (This is (almost) equation (9) in the paper.) Here $\Lambda$ is a dimensionless "overlap" factor, with $0\leq\Lambda\leq1$.

The reason for the mixture of area and volume integrals is that $u$ and $v$ are only defined on (part of) the surface of $V$, while $w$ is defined throughout the whole volume. So, it makes sense that the normalization of $w$ is obtained by integrating over the whole volume, while the normalizations of $u$ and $v$ are obtained by integration over area. And dimensional analysis justifies the factor of $V$ in the numerator.

My question: what is this thing? It looks like a dot product, but it involves three functions/vectors/modes, so I'm not sure how to think about it. In particular, I'm not sure how to convince myself that this normalization is the correct one. In particular, why wouldn't we consider the pointwise product $u(x)v(x)$ to be a single "vector" and normalize it by dividing by $\int(uv)^2\ dA$ rather than normalizing $u$ and $v$ individually?

Answer 1

According to the paper:

$u_z$ is the component of $u$, normal to the mirror’s surface.

So the integration is actually $$\Lambda \equiv \frac {V\ \left(\iint_S u\ v (\boldsymbol{w}\cdot \boldsymbol{n}_S) dA\right)^2} {\left(\iint_S u^2~dA\right)\left(\iint_S v^2~dA\right)\left(\iiint_V \boldsymbol{w}\cdot \boldsymbol{w}~dV\right)},$$ where $\boldsymbol{n}_S$ is the unit vector normal to the surface $S$.

Qualitatively speaking, $\Lambda$ measures the "overlap" of $u$ and $v$ on the surface of interest $S$, with the $\boldsymbol{w}$'s flux density $\boldsymbol{w}\cdot \boldsymbol{n}_S$ as a weight. Notice the paper says $\boldsymbol{w}$ is the displacement, to me, the interpretation is:

Wherever there is a displacement vertical to the mirror surface, we start to consider the overlap of $u$ and $v$ at this very place (then using the inner product to add all overlaps up).

The volume integral term $\iiint_V \boldsymbol{w}\cdot \boldsymbol{w}~dV$ in the denominator looks more like a normalization of $\boldsymbol{w}$.