4.1.1 Solution

Question: A simple form that a six-dimensional wave function can take is a product of two three-dimensional ones, as in $\psi({\skew0\vec r}_1,{\skew0\vec r}_2)=\psi_1({\skew0\vec r}_1)\psi_2({\skew0\vec r}_2)$. Show that if $\psi_1$ and $\psi_2$ are normalized, then so is $\psi$.

Answer: This is a direct consequence of the fact that integrals can be factored if their integrands can be and the limits of integration are independent of the other variable:

$$\int_{{\rm all }{\skew0\vec r}_1}\int_{{\rm all }{\skew0\... ...{\skew0\vec r}_2)\Big\vert^2 { \rm d}^3 {\skew0\vec r}_2 = 1$$