GR 8677927796770177 | # Login | Register

GR9277 #28
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{28}
A system is known to be in the normalized state described by the wave function
$\psi(\theta,\psi)=\frac{1}{\sqrt{30}}(5Y^3_4+Y^3_6-2Y^0_6$,
where the $Y^m_l(\theta,\phi)$ are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number m=3 is

1. 0
2. 1/15
3. 1/6
4. 1/3
5. 13/15

Quantum Mechanics$\Rightarrow$}Probability

One doesn't actually need to know much (if anything) about spherical harmonics to solve this problem. One needs only the relation $P=\sum_i |\langle Y^3_i |\psi (\theta,\phi)\rangle|^2$. Since the problem asks for states where $m=3$, and it gives the form of spherical harmonics employed as $Y^m_l$, one can eliminate the third term after the dot-product.

So, the given wave function $\psi(\theta,\psi)=\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6 - 2Y^0_6\right)$ gets dot-product'ed like $|\langle Y^3_i |\psi (\theta,\phi)\rangle|^2\left(\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6\right)\right)\left(\frac{1}{\sqrt{30}}\left( 5 Y^3_4 + Y^3_6 - 2Y^0_6\right)\right)=\frac{25+1}{30}=\frac{13}{15}$, as in choice (E).

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
mpdude8
2012-04-19 19:55:26
One can also deduce that the particle is more likely than not to have m = 5 by looking at the normalized wave function. E is the only possibility that fits.
tera
2006-08-21 04:37:41
No dirac symbols not anything. The Possibility is c1^2 + c2^2 as ina a simple mathematical problem in our case 25/30+1/30=26/30
 jmason862009-09-09 22:12:22 agreed, the probability of finding a system in a given state is always just the sum of the squares of the coefficients in the wavefunction.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$