| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 1 month |
| seen | Jul 5 '12 at 1:33 | |
| stats | profile views | 7 |
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May 7 |
awarded | Supporter |
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May 7 |
accepted | Eigenvalue of $L_z$ |
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May 7 |
comment |
Eigenvalue of $L_z$ Ah, you're absolutely right. This is my fault: I was looking for justification in the pages before, not the pages after. The very next page describes exactly what you said. Thanks. |
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May 6 |
comment |
Eigenvalue of $L_z$ If it helps, you could just ignore my last sentence. In that case, I'm just wondering what allows us to claim that $\hbar \ell$ is an eigenvalue of $L_z$. In the text, this is stated and not justified (as far as I can tell). |
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May 6 |
asked | Eigenvalue of $L_z$ |
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Apr 23 |
awarded | Scholar |
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Apr 23 |
accepted | Electric dipole transitions/expectation value of position |
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Apr 23 |
comment |
Electric dipole transitions/expectation value of position You're not missing anything, it really is that simple! However, I couldn't flesh out the details until late last night. I ended up converting to Cartesian coordinates, in which case $\Psi^\ast \Psi$ is even in each of $x,y$ and $z$ since $r \mapsto \sqrt{x^2+y^2+z^2}$. Thus, $\Psi^\ast x \Psi$ is odd in $x$, and similarly for $y$ and $z$. So the integral of each is zero, which means the integral of the original thing is zero. |
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Apr 23 |
awarded | Student |
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Apr 22 |
comment |
Electric dipole transitions/expectation value of position Can someone fix the box after the $r$? I used \vec{} but as I usually do, but it apparently did not render here. |
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Apr 22 |
asked | Electric dipole transitions/expectation value of position |