How do you convert SRTM HGT elevations from WGS84 into meters above sea level?

The axiom that probability density is symmetric: $$ |\psi(\mathbf r_1,\mathbf r_2)|^2 = |\psi(\mathbf r_2,\mathbf r_1)|^2 $$ only implies $$ \psi(\mathbf{r}_1,\mathbf{r}_2) = e^{i\delta(\mathbf{r}_2,\mathbf{r}_1)}\psi(\mathbf{r}_2,\mathbf{r}_1)~~~(*) $$

where $\delta(\mathbf{a},\mathbf{b})$ is arbitrary real function of $\mathbf{a},\mathbf{b}$. It does not imply that $\delta$ is constant throughout the whole configuration space $\mathbf{r}_1,\mathbf{r}_2 \in \mathbb{R}^6$ and so it is not possible generally to obtain

$$ e^{i2\delta} \psi $$ after application of the transposition operator to (*). What is obtained instead is $$ \psi(\mathbf{r}_2,\mathbf{r}_1) = e^{i\delta(\mathbf{r}_1,\mathbf{r}_2)}\psi(\mathbf{r}_1,\mathbf{r}_2) $$ From these two relations it follows $$ e^{i\delta(\mathbf{r}_1,\mathbf{r}_2)}.e^{i\delta(\mathbf{r}_2,\mathbf{r}_1)} = 1 $$ so $$ \delta(\mathbf{r}_1,\mathbf{r}_2) + \delta(\mathbf{r}_2,\mathbf{r}_1) = k.2\pi $$ but it does not follow that $e^{i\delta}$ has definite value for all $\mathbf r_1,\mathbf r_2$.

To obtain symmetric and antisymmetric functions, one has to make stronger assumption. For example, if we assume that all multiples of wave function represent the same state, and postulate that transposition does not change the state, then we have $$ \psi(\mathbf{r}_1,\mathbf{r}_2) = e^{i\delta}\psi(\mathbf{r}_2,\mathbf{r}_1)~~~(*) $$ with $\delta$ constant and the rest of the usual argument can be used to derive $e^{i\delta} = \pm 1$.