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While trying to find the expectation value of the radial distance $r$ of an electron in hydrogen atom in ground state the expression is:

$$\begin{aligned}\langle r\rangle &=\langle n \ell m|r| n \ell m\rangle=\langle 100|r| 100\rangle \\ &=\int r\left|\psi_{n \ell m}(r, \theta, \phi)\right|^{2} d V \end{aligned}$$

Since Hilbert space operators act on kets, What operator is $r$ in the expression: $$\begin{aligned}\langle r\rangle &=\langle n \ell m|r| n \ell m\rangle=\langle 100|r| 100\rangle~? \end{aligned}$$

Is it a component of the position operator $\mathbf x$ that is related to the radial distance?

Does it act on kets as:

$$\hat{r}|r \theta \phi\rangle=r|r \theta \phi\rangle~?$$

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You have gone awry at the end of your question, utilizing ambiguous/meaningless notation.

Here, it is best to use the caret $\hat {\mathbf v}$ to signify a unit vector, as in Sakurai-Napolitano (3.6.22,23), $$ \langle {\mathbf x}|nlm\rangle = R_{nl}(r) Y^m_l(\theta,\phi)= R_{nl}(r) \langle \hat {\mathbf x}|l,m\rangle , $$ and not an operator, as in the punchline of your question: use another symbol for operators, instead.

Consequently, by inspection, $$ \langle r\rangle =\int r^2 dr ~~ r (R_{nl}(r))^2 , $$ since the $d\Omega$ integral factored out to 1.

The linear power of r in this expression is just just a c-number, the radial value (length) of the c-number vector ${\mathbf x}= r \hat {\mathbf x}$, since you are working in the coordinate representation.

If you must think of an operator with eigenvalue r, it is $\large \mathbb r$ s.t. $$ {\large \mathbb r} ~ |{\mathbf x}\rangle= r |{\mathbf x}\rangle , $$ and you could apply the machinery of Dirac's ket notation in the above, which, however, is superfluous, and, evidently, confusing here.

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  • $\begingroup$ Thank you professor. I'm trying to work in both the wavefunction and dirac notation so that my basics get correct hence my wish to see the dirac notation. Also, Is it that, In general, functions of operators are defined such as below : $$f(\hat{x},\hat{y},\hat{z})|x,y,z \rangle=f(x,y,z) |x,y,z \rangle$$ ? Here $f$ is a real valued function of variable $x, y, z$ $\endgroup$
    – Kashmiri
    May 11, 2022 at 6:18
  • $\begingroup$ In your answer above $|\hat{ \mathbf x}\rangle$ means $|\theta, \phi\rangle$? $\endgroup$
    – Kashmiri
    May 11, 2022 at 6:22
  • $\begingroup$ Yes, of course. $\endgroup$ May 11, 2022 at 6:57
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    $\begingroup$ But don’t use carets for operators here! This is the source of your confusion! I'd write your relation here as $f({\mathbb x}, {\mathbb y}, {\mathbb z})|x,y,z\rangle= f( x, y,z)|x,y,z\rangle$. $\endgroup$ May 11, 2022 at 13:24
  • $\begingroup$ I cannot see why you say "This is the source of your confusion". I think I'm clear on operators and vectors. In my question can you please pin point the statement where I am confusing the notation so that I can learn from my mistake? Thank you professor :) $\endgroup$
    – Kashmiri
    May 12, 2022 at 4:34

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