1
$\begingroup$

Consider the $n=2$ states of the hydrogen atom, which we label by $|n\,l\,m_l\rangle$. I want to calculate whether or not there should be a sign difference in these specific matrix elements of $x$:

$$\langle211|x|200\rangle \;=\; \pm \langle21-1|x|200\rangle\;?$$

Explicit Calculation

Of course, one can show that the integral itself is the same for both signs:

$$\begin{align}\langle21\pm1|x|200\rangle &\;=\;\int\psi_{21\pm1}^\ast(\textbf{r})\;x\;\psi_{200}(\textbf{r})\;d^3\textbf{r} \\ &\;=\;\iint\left(\text{functions of }r\text{, }\theta\right)\;drd\theta\int_0^{2\pi}e^{\mp i\phi}\cos\phi \;d\phi\;.\end{align}$$

where $(r,\theta,\phi)$ are the usual spherical coordinates, and the $r$ and $\theta$ integrals are the same for both signs. One can easily verify that the $\phi$ integral gives the same result for both signs, hence $\langle211|x|200\rangle = \langle21-1|x|200\rangle$.

By Wigner-Eckart Theorem

Using $\langle j_1\, m_1 ;j_2\, m_2|J\,M\rangle$ to denote the Clebsch-Gordan coefficients, the Wigner-Eckart Theorem states that:

$$\langle j\, m|\, T_q^{(k)}|j'\,m'\rangle\;=\;\langle j'\,m';k\,q|j\, m\rangle\langle j||T^{(k)}||j'\rangle\,.$$

This is where I'm not certain, and I would like to check if I've made a mistake. The $x$-component of the position vector can be written in terms of its spherical components by:

$$x \;=\; \frac{1}{\sqrt{2}}r_{-1}-\frac{1}{\sqrt{2}}r_{+1}$$

which allows us to evaluate:

$$\begin{align}\langle211|x|200\rangle &\;=\;\frac{1}{\sqrt{2}}\langle 11|r_{-1}|00\rangle \;-\;\frac{1}{\sqrt{2}}\langle 11|r_{+1}|00\rangle\\ &\;=\;\frac{1}{\sqrt{2}}\langle 00;1-1|11\rangle\langle 1||\textbf{r}||0\rangle - \frac{1}{\sqrt{2}}\langle 00;11|11\rangle\langle 1||\textbf{r}||0\rangle\\ &\;=\;-\frac{1}{\sqrt{2}}\langle 1||\textbf{r}||0\rangle\; ;\\ \\ \langle21-1|x|200\rangle &\;=\; \frac{1}{\sqrt{2}}\langle 1-1|r_{-1}|00\rangle \;-\;\frac{1}{\sqrt{2}}\langle 1-1|r_{+1}|00\rangle\\ &\;=\;\frac{1}{\sqrt{2}}\langle 00;1-1|1-1\rangle\langle 1||\textbf{r}||0\rangle - \frac{1}{\sqrt{2}}\langle 00;11|1-1\rangle\langle 1||\textbf{r}||0\rangle\\ &\;=\;\frac{1}{\sqrt{2}}\langle 1||\textbf{r}||0\rangle\; ;\end{align}$$

i.e. wrongly concluding that $\langle211|x|200\rangle = -\langle21-1|x|200\rangle$. Where did I go wrong? I'm sure I've made a careless mistake somewhere (or have fundamentally misunderstood something) but I've been staring at this for so long that I'm going crazy...

$\endgroup$

1 Answer 1

1
$\begingroup$

Your mistake is the definition of the $Y_{1,\pm 1}$ spherical harmonic.

It should be $Y_{1,\pm 1} = \mp \sqrt\frac{3}{8\pi} \sin \theta e^{\pm i \phi}$ with the $\mp$ out the front changing the sign in your "Explicit calculation".

p.s. Good luck in AQP.

$\endgroup$
1
  • 1
    $\begingroup$ Hahahaha that obvioius, huh? Thank you very much, knew I did something silly! $\endgroup$
    – DWayne
    May 28, 2023 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.