# Normalization of time-independent Schroedinger equation in Spherical Coordinates

I have a short question about the time-independent Schroedinger equation in Spherical coordinates:

$$\psi(r, \theta, \phi) = R(r)Y(\theta, \phi)$$ then the normalization condition becomes $$\int |\psi|^2 r^2 \sin \theta dr d \theta d \phi = \int |R|^2 r^2 dr \int |Y|^2 \sin \theta d \theta d \phi = 1$$

In literature I am using it states that we can then normalize $R$ and $Y$ separately: $$\int^{\infty}_{0}|R|^2r^2 dr = 1~~~\text{and}~~~\int^{2 \pi}_{0}\int^{\pi}_{0}|Y|^2 \sin \theta d \theta d \phi =1$$

I just want to make sure I understand the reason we have the freedom to do the separate normalization of $R$ and $Y$. The way I understand it is that given a state $| \psi \rangle$, this is no different to the state $c| \psi \rangle$, where $c$ is a constant. Since we interpret $|\psi(x)|^2dx$ as the probability density we will find the particle between $x$ and $x + dx$ we require that that the integral is $1$. So in a sense it is scaling the wave function until it meets the requirement that it is physically realizable. In the case above we are doing the same and for conveience we will scale the integrals involving $R$ and $Y$ separately and hence $\psi$ is normalized. Am I missing the actual reason or does this contain the main points?

## 2 Answers

I've not done this in a long while, but it seems to me the physics would dictate it. I think you are right.

Each part represents a total probability that the particle is at some coordinate value. The first part for r, and the second part for the angular coordinates. Each has to be 1, physically, unless something strange is happening. If one was prob at a certain r, sure the other one at certain angles could vary, but not for the total probability

If one was less than 1, the other would have to be more than 1, which would make no physical sense.

Your normalization condition, $$\int_0^\infty\vert{}R\vert^2r^2\mathrm{d}r\int_0^\pi\int_0^{2\pi}\vert{}Y\vert^2\sin\theta\mathrm{d}\theta\mathrm{d}\phi = 1,$$ can be easily translated to $$C_R C_Y = 1$$ where $$C_R \equiv \int_0^\infty\vert{}R\vert^2r^2\mathrm{d}r, \quad C_Y \equiv \int_0^\pi\int_0^{2\pi}\vert{}Y\vert^2\sin\theta\mathrm{d}\theta\mathrm{d}\phi.$$

Now, $$C_R$$ and $$C_Y$$ must be different from zero, otherwise, $$R$$ and/or $$Y$$ would also be identically zero. Since they are non-zero the solution is simply $$C_Y = 1/C_R$$ for any real strictly positive real number $$C_R$$.

Since the wave-function is given by $$\psi(r,\theta,\phi) = R(r)Y(\theta,\phi)$$ we can always rewrite it as $$\psi(r,\theta,\phi) = \bar{R}(r)\bar{Y}(\theta,\phi),$$ where $$\bar{R}(r) \equiv \frac{R(r)}{\sqrt{C_R}}, \quad \bar{Y}(\theta,\phi) \equiv \sqrt{C_R}Y(\theta,\phi).$$ This means that the individual normalization is actually irrelevant, and therefore, you can choose $$C_R = 1$$ and $$C_Y = 1$$ for simplicity.

If we insist in not using the individually normalized form, calculating a marginal distribution in $$r$$ we obtain $$P(\theta,\phi) = \int_0^\infty\vert\psi\vert^2r^2\mathrm{d}r = C_R \vert{}Y(\theta,\phi)\vert^2 = \vert{}\bar{Y}(\theta,\phi)\vert^2.$$ Thus, the advantage of normalizing each function is that in that case they will coincide with the part of the wave function related to the respective marginal probability distribution.