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Suppose you do a Stern-Gerlach experiment with atoms in a $j=1$ state. There would be three separate beams ($m_j = -1, 0, 1$) coming out of the apperatus. But what would be the relative distribution of atoms in these beams?

On first thought I would say that the distribution should be $\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$. But considering the "classical" distribution of the projections of the angular momentum onto one axis I could imagine that there could be more atoms in the $m_j = 0$ state. This would still be compatible with symmetry considerations.

To state the question differently: how does the density matrix of an unpolarized beam of $j=1$ atoms look like?

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Typo(?) in the question (v1): $j=0$ should be $j=1$. –  Qmechanic Jul 12 '11 at 10:42
    
Yes, you're right of course. Thanks. –  shark.dp Jul 12 '11 at 11:42
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up vote 3 down vote accepted

The z-projection of the angular momentum of a completely unpolarized system is distributed uniformly both classically and quantum mechanically.

In quantum mechanics, the density matrix of a three state system is given by: $\rho = \frac{1}{3}(I_3+\overrightarrow{n}.\overrightarrow{\lambda})$, where $I_3$ is the three dimensional unit matrix. $\overrightarrow{n}$ is the polarization vector and $\overrightarrow{\lambda}$ is a vector of Gell-mann matrices.

In a fully polarized system $\overrightarrow{n}.\overrightarrow{n}= 1$, while in the case of a completely unpolarized system: $\overrightarrow{n} = 0$. In this case:

$\rho = \frac{1}{3}(I_3)$. Therefore, the three states are equally distributed.

In classical mechanics, the phase space of a spin system is the unit sphere. The z-component of a unit angular momentum system is given by:

$ J_z = cos(\theta)$.

Where $\theta$ is the inclination angle. A completely unpolarized classical angular momentum should be distributed uniformly with respect to the surface area of the sphere which is the classical phase space. The area of a spherical zone of thickness $h$ is $S = 2 \pi R h$, where $R=1$ is the sphere radius (This result was already known to Archimedes). Therefore the z-coordinate is uniformly distributed on the sphere and the z-component of the angular momentum expressed as:

$ J_z = \frac{z}{R}$

is linear in $z$, therefore also uniformly distributed.

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Thanks for the detailed answer. I fully agree on the quantum mechanical part. But I don't understand the reasoning for the classical counterpart: shouldn't the R in the formular for S be the "in-plane" radius (distance from the z-axis, not from the origin)? $S = 2\pi \times \sin(\theta) h$? –  shark.dp Jul 12 '11 at 12:04
    
Please see the derivation in: mathworld.wolfram.com/Zone.html –  David Bar Moshe Jul 12 '11 at 12:10
    
Okay, this came really surprising for me. I didn't expect this. Thanks for the hint and the URL. I would upvote you're answer, if I were allowed to ;-) –  shark.dp Jul 12 '11 at 12:18
    
Trying to compute this on my own, I found a much simpler derivation for this surface: With $z = R \cos(\theta)$ one can do the integration in spherical coordinates: $A = R \int_0^{2\pi} \text{d} \phi \int_{\arccos(a/R)}^{\arccos(b/R)} \text{d} \theta \sin(\theta)$ $=$ $2 \pi R \int_a^b \text{d} \left(\cos(\theta)\right) = 2 \pi R (b-a)$. Thanks again! –  shark.dp Jul 12 '11 at 12:33
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No, he is correct. $\theta$ isn't uniformly distributed with respect to the surface elements of the sphere, but $J_z$ which is proportional to $cos(\theta)$ is uniformly distributed. You can verify that according to the random variable distribution transformation law: $p_{J_z}= \frac{p_{\theta}}{|\frac{d(J_z)}{d\theta}|}$, where $p$ denotes the distribution density –  David Bar Moshe Jul 12 '11 at 15:19
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