I would like to express the wave functions for three identical particles, each with orbital angular momentum $L=1$ and spin angular momentum $S=1/2$, in terms of single-particle wave functions. In other words, I would like to obtain the Clebsch-Gordan coefficients for this problem.

The problem is discussed in Sakurai's Quantum Mechanics textbook around p. 375 and in Greiner and Muller's "Quantum Mechanics: Symmetries, 2nd Edition" on p. 300. I know I have to find the spin wave functions and the orbital angular momentum wave functions separately, and then combine them to get fully antisymmetric wave functions. I have the spin wave functions (four symmetric, 2 mixed symmetric under exchange of particles 1 and 2, and 2 mixed antisymmetric under exchange of 1 and 2), but I haven't been able to get a small enough number of angular momentum wave functions to get just 20 fully-antisymmetric total wave functions.

In Sakurai's book, p 376, eq'n (6.5.20), we see that the 20 states can be decomposed into 2 states with total angular momentum $j=1/2$, 4*3=12 with $j=3/2$, and 6 with $j=5/2$. Could anyone fill in how Sakurai got 6 for $j=5/2$, 12 for $j=3/2$, and 2 for $j=1/2$?

Most importantly, referring to my comments on Y Macdisi's answer below, could anyone answer the following question: Is the orbital angular momentum state with single particle wave functions $\alpha=1$, $\beta=0$, and $\gamma=0$ related in some way to the wave function with $\alpha=-1$, $\beta=0$, and $\gamma=0$, or $\alpha=1$, $\beta=-1$, and $\gamma=-1$, and so on? I would love it if I could just keep the first set of values for $\alpha$, $\beta$, and $\gamma$, but I see no good reason to do that.

If anyone happens to know where this problem is discussed more fully, I would appreciate a reference. Or, if anyone knows how to do this, I would appreciate help in knowing what the $j=3/2$ and $j=5/2$ states could be in terms of single-particle orbital ang. momentum states and spin states. I think I actually have the normalization factors already, I just need to know what the single-particle states are.

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    $\begingroup$ Just for the record, I don't have Sakurai in front of me, but I think you are reading the problem incorrectly - 2 identical fermions can't be in the same state, let alone 3. You probably mean 3 different fermions in the $S_z = +\frac{1}{2}$ state. Like an up, down and strange quark, or something to that effect. $\endgroup$ – DJBunk Jan 21 '13 at 20:05
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    $\begingroup$ Sorry, where did I say two or more of these fermions were in the same state? I am well aware of the Pauli exclusion principle, and, looking over my question, I can't see where I wrote anything to suggest that. Thank you for responding, whatever the case. $\endgroup$ – Impossibear Jan 21 '13 at 22:10
  • $\begingroup$ In Greiner and Muller, I'm referring to problem 9.1 on p. 300. This problem gives you the basis functions of the permutation group $S_3$ in terms of single-particle wavefunctions $\alpha$, $\beta$, $\gamma$. $\endgroup$ – Impossibear Jan 21 '13 at 22:53
  • $\begingroup$ I just wasn't sure what you meant by 'identical particles' except for same state? $\endgroup$ – DJBunk Jan 21 '13 at 23:10
  • $\begingroup$ By identical, I meant indistinguishable, I guess. Sorry for the confusion, but identical is the terminology used in every reference I've found on the problem so far. $\endgroup$ – Impossibear Jan 21 '13 at 23:46

The L=1 rep is a 3 dimensional irep of $SU(2)$; the S=1/2 rep is a 2 dimensional irep. Combining angular momentum (L=1) and spin (S=1/2) gives the tensor product rep $3 \otimes 2$ which is 6 dimensional. You are looking for the third exterior product of this : $A^3(3 \otimes 2)$ which is 20 dimensional and decomposes into $6 \oplus 4 \oplus 4 \oplus 4 \oplus 2$ irreps. These correspond to j=5/2,3/2,3/2,3/2,1/2 .

  • $\begingroup$ Thank you, Y Macdisi. I should be more specific, reading your response: Would the $j=5/2$ have to be $L_{tot}=2$, $S_{tot}=1/2$? If so, I can't see how one gets just 6 states, from the work done in Greiner and Muller's book. Would the $j=3/2$ be $L_{tot}=1$, $S_{tot}=1/2$ (for 10 of the 12), or $L_{tot}=0$, $S_{tot}=3/2$ (for 2 of the 12)? If so, I can't see how to get exactly 10 states from my $L_{tot}=1$ wavefunctions, in terms of single particle momentum states $|\pm1\rangle$ or $|0\rangle$. There just seem to be too many possible choices of single-particle angular momentum states. $\endgroup$ – Impossibear Jan 22 '13 at 15:28
  • $\begingroup$ By the way, from Greiner and Muller's book, it seems there are two mixed symmetric and two mixed antisymmetric wave functions, expressed in terms of single particle wavefunctions $\alpha$, $\beta$, and $\gamma$, where each of $\alpha$, $\beta$, and $\gamma$ can take on the values $|+1\rangle$, $|0\rangle$, or $|-1\rangle$. For instance, the two orthonormal wavefunctions which are symmetric under exchange of $\alpha$ and $\beta$ are the following: $1/2 \psi_1 = 1/2\left[\alpha\beta\gamma + \beta\alpha\gamma-\gamma\beta\alpha-\gamma\alpha\beta\right]$ $\endgroup$ – Impossibear Jan 22 '13 at 15:33
  • $\begingroup$ and $1/\sqrt{3}\psi_2' = 1/\sqrt{3}\left[1/2\left(\alpha\beta\gamma+\beta\alpha\gamma+\gamma\beta\alpha+ \gamma \alpha\beta\right)-\alpha\gamma\beta-\beta\gamma\alpha\right]$ $\endgroup$ – Impossibear Jan 22 '13 at 15:34
  • $\begingroup$ There are also two more orthonormal wave functions, which are mixed antisymmetric under exchange of $\alpha$ and $\beta$. These four mixed (anti)symmetric wave functions must have either $L_{tot}=2$ or $L_{tot}=1$, it seems. They must be multiplied by the appropriate mixed (anti)symmetric spin wave functions to eventually form fully antisymmetric total wave functions corresponding to all of the $j=5/2$ states, and most of the $j=3/2$ states. $\endgroup$ – Impossibear Jan 22 '13 at 15:38
  • $\begingroup$ Another way to think of it, is that, for three spin $1/2$ particles, the two mixed symmetric spin wave functions are the following: $|1/2,-1/2\rangle = -1/\sqrt{6}\left(2\Downarrow\Downarrow\Uparrow-\Uparrow\Downarrow\Downarrow-\Downarrow \Uparrow\Downarrow \right)$ $|1/2,1/2\rangle = 1/\sqrt{6}\left(2\Uparrow\Uparrow\Downarrow-\Uparrow\Downarrow\Uparrow-\Downarrow \Uparrow\Uparrow \right)$ $\endgroup$ – Impossibear Jan 22 '13 at 15:49

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