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The $2p_0$ and $1s$ wavefunctions for hydrogen;

$ \psi_{2p_0} = \dfrac{1}{4\sqrt{2\pi}} \left(\dfrac{Z}{a_b}\right)^{\frac{3}{2}} \dfrac{Z r}{a_b} e^{\frac{-Zr}{2a_b}} \cos(\theta) $

$ \psi_{1s} = \dfrac{1}{\sqrt{\pi}} \left(\dfrac{Z}{a_b}\right)^{\frac{3}{2}} e^{\frac{-Zr}{a_b}} $

Then calculating the dipole element; splitting the dipole $-er$ into x,y,z components gives you;

$ -e\langle \psi_{2p_0}|x|\psi_{1s}\rangle =0 $

$ -e\langle \psi_{2p0}|y|\psi_{1s}\rangle = 0 $

$ -e\langle \psi_{2p0}|z|\psi_{1s}\rangle = -e2^{\frac{7}{2}} a_{b}\frac{1}{3^5} $

of which the $x$ and $y$ are zero, due to symmetry, and the z component is non zero due to being asymmetric.

Why can the dipole element not be calculated by using: $ -e\langle \psi_{2p0}|r|\psi_{1s}\rangle $ ?

As; $ -e\langle \psi_{2p0}|r|\psi_{1s}\rangle = -e\int^{-\infty}_{-\infty}\frac{1}{4\sqrt{2\pi}} (\frac{Z}{a_b})^{\frac{3}{2}} \frac{Z r}{a_b} e^{\frac{-Zr}{2a_b}} \cos(\theta)r \frac{1}{\sqrt{\pi}} (\frac{Z}{a_b})^{\frac{3}{2}} e^{\frac{-Zr}{a_b}} d^{3}r $

$= \frac{-e}{4\pi\sqrt{2}}\frac{Z}{a_b}^{4} \int^{\infty}_{-\infty} r^2 e^{-\frac{3Zr}{2a_b}} \cos(\theta) d^{3}r $

Then changing the integral to spherical polar coordinates;

$= \frac{-e}{4\pi\sqrt{2}}\frac{Z}{a_b}^{4} \int^{\infty}_{0}\int^{\pi}_{0}\int^{2\pi}_{0} r^2 e^{-\frac{3Zr}{2a_b}}\cos(\theta)r^{2}\sin(\theta) drd\theta d\phi $

$= \frac{-e}{4\pi\sqrt{2}}\frac{Z}{a_b}^{4} \int^{\infty}_{0}\int^{\pi}_{0}\int^{2\pi}_{0} r^4 e^{-\frac{3Zr}{2a_b}} \cos(\theta)\sin(\theta) dr d\theta d\phi $

Which of course, evaluates to zero because $\int^{\pi}_{0}\sin(\theta)\cos(\theta) = 0$ But we know that the 2p0 to 1s transition does happen.

So the question is, why is setting the dipole to $-er$ wrong?

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  • $\begingroup$ $r$, being the radial coordinate, has no knowledge of direction. Another way to see this is to note that the dipole matrix element has to be a vector. Calculating the $r$ matrix element gives you just one number. Perhaps more to the point: you want to find the matrix element of $\vec{r}$, not $r$. $\endgroup$ – garyp Jun 23 '16 at 12:04
  • $\begingroup$ Doh. Well that certainly makes sense. Just the difference between $\vec{r}= x\vec{i}+y\vec{j}+z\vec{k}$, and $r=\sqrt{x^2 + y^2 + z^2}$. Thank you for needed injection of sense. $\endgroup$ – user94877 Jun 23 '16 at 12:10
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Following Garyp's comment;

"r , being the radial coordinate, has no knowledge of direction. Another way to see this is to note that the dipole matrix element has to be a vector. Calculating the rr matrix element gives you just one number. Perhaps more to the point: you want to find the matrix element of r⃗ r→, not rr. – garyp"

The dipole element should be; $-e< \psi_{2p0}|\vec{r}|\psi_{1s}> = -e\int^{-\infty}_{-\infty}\frac{1}{4\sqrt{2\pi}} (\frac{Z}{a_b})^{\frac{3}{2}} \frac{Z r}{a_b} e^{\frac{-Zr}{2a_b}} cos(\theta)\vec{r} \frac{1}{\sqrt{\pi}} (\frac{Z}{a_b})^{\frac{3}{2}} e^{\frac{-Zr}{a_b}} d^{3}r$

The difference being: $\vec{r}= x\vec{i}+y\vec{j}+z\vec{k}$

compared to: $r=\sqrt{x^2 + y^2 + z^2}$

Thus the dipole element becomes,

$-e< \psi_{2p0}|\vec{r}|\psi_{1s}> = \frac{-e}{4\pi\sqrt{2}}\frac{Z}{a_b}^{4} \int^{\infty}_{-\infty} r \vec{r} e^{-\frac{3Zr}{2a_b}} cos(\theta) d^{3}r$

$ = \frac{-e}{4\pi\sqrt{2}}\frac{Z}{a_b}^{4} \int^{\infty}_{-\infty} r (x\vec{i}+y\vec{j}+z\vec{k}) e^{-\frac{3Zr}{2a_b}} cos(\theta) d^{3}r$

Which is then the same as the seperated polarisations from the original question.

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