I have a problem that was given as

$$\left< r | \psi \right> = \dots$$

I was midway through doing the problem when I realized we might have problems because we are working with a radius and not $x$, and I was taking integrals from $-\infty$ to $\infty$, and negative radii don't make sense. This got me to thinking whether the $\left< \psi(r) | \psi(r) \right>$ normalization is even the same $\int_{-\infty}^\infty \psi^* \psi dr$ as before. I believe this must not be the case because:

  1. Negative radii don't make sense

  2. If we assume that the normalization equation is as above starting with some volume, then converting to radius would require change of $dV$ to $dr$.

So I'm assuming then, that for a $\psi(r)$ to be normalized:

$$\left< \psi(r) | \psi(r) \right> = \int_0^\infty \psi^*(r) \psi(r) r^2 dr = 1$$

So then what about the expected value equation? For some operator $\hat{O}$:

$$\left< \hat{O} \right> = \frac{\left< \psi | \hat{O} | \psi \right>}{\left< \psi | \psi \right>}$$

If we have a wave equation $\psi(r)$, is the denominator like the above (including $r^2$)? What about the numerator terms? Does the $\left< \psi \right|$ include $r$?

Therefore, does that mean that:

$$\left< r | \psi \right> = f \implies \psi = rf $$

$$\left< \psi | \psi \right> = \int_0^\infty f^* f r^2 dr$$

This doesn't really make sense to me if we consider $r$ as an operator. If $\hat{r}$ is an operator, does that mean that $\left< \hat{r} | \psi \right> = \frac{1}{r} \psi$?

  • $\begingroup$ is it possible that $r$ is just the notation used for the position coordinate? Or are you sure it represents a radius? $\endgroup$ Nov 2, 2021 at 9:55
  • $\begingroup$ The question was given as $\left< \mathbf{r} | \psi \right> = f(r)$ where $f(r)$ was a function of radius (e.g. $e^{- \alpha r^2}$). But inside the bra it was a $\mathbf{r}$ as in a position vector. $\endgroup$ Nov 2, 2021 at 10:08

2 Answers 2


So you don't have $\langle{r}|\psi\rangle$, you have $\langle{\boldsymbol{r}}|\psi\rangle$, so you integrate over all possible values of $\boldsymbol{r}$: $-\infty<x,y,z<\infty$, and if then you need to switch from the cartesian system of coordinates to the spherical system of coordinates, you do it the standard way.


The problem is not with your interpretation of $\langle r\vert \psi\rangle=\psi(r)$ but with converting your measure, which is not $dx$ but $r^2 dr$ in spherical (once you’ve integrated angles). 2-d cylindrical it would be $r dr$.

The average value would then be $\int dr \, r^2 \,\hat O(r) \vert\psi(r)\vert^2$.


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