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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 $\left< \hat{r} | \psi \right> = \frac{1}{r} \psi$?

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  • $\begingroup$ 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

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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.

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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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