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Is the following correct, when adding 3 angular momenta/spins: \begin{align} 1\otimes 1\otimes \frac{1}{2}&=\left(1\otimes 1\right)\otimes \frac{1}{2} \\ &=\left(2\oplus 1\oplus 0\right)\otimes \frac{1}{2} \\ &=\left(2\otimes \frac{1}{2}\right)\oplus \left(1\otimes \frac{1}{2}\right)\oplus \left(0\otimes \frac{1}{2}\right) \\ &=\left(\frac{5}{2}\oplus\frac{3}{2}\right)\oplus \left(\frac{3}{2} \oplus\frac{1}{2}\right)\oplus \left(\frac{1}{2}\right) \\ &=\frac{5}{2}\oplus\frac{3}{2}\oplus\frac{3}{2}\oplus\frac{1}{2}\oplus\frac{1}{2}. \end{align}

Comments:

  1. $\oplus$ and $\otimes$ are commutative.
  2. $0\otimes\frac{1}{2}$ makes $0$ disappear on the line before the last one... I don't understand that. It also happens with $0\otimes 1$, but that feels wrong. Can I not get a total spin of 0 in that case?
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  • $\begingroup$ Your notation is such that $n$ denotes the representation of $\mathrm{SU}(2)$ labeled by the half-integer spin/Casimir value $n$? $\endgroup$ – ACuriousMind May 10 '15 at 15:56
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/183139/2451 $\endgroup$ – Qmechanic May 10 '15 at 15:57
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    $\begingroup$ ACuriousMind - please bear with me, as (as you know :-), I'm just a student in a QM class... I am not sure I understand your point: I thought I would just try my hand at this kind of computation with a system of 2 particles of spin 1 and 1 particle of spin 1/2. Is that somehow illegal? $\endgroup$ – Frank May 10 '15 at 16:14
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Your final result looks right to me. Everything should be half-integers.

A basic rule of combining two quantized angular momenta is that the quantum number of the resultant can be anywhere between the sum of the original quantum numbers and the absolute value of the difference of them, in integer steps. Consider $\ell_1$ = 1 combining with $\ell_2$=3. The quantum number, $L$, of the combination can be between |3-1|=2 and 3+1=4, stepping by 1, or $L = 2, 3, \text{ or } 4$.

So if you have only one particle with a half-integer quantum number, the combination must be a half-integer quantum number. There is no way to get 0.

Also, with quantum numbers of 1 and 0, the result would be between |1-0| and 1+0, which is 1.

The reason for this rule is based in the mathematics of preserving angular momentum wave functions. If you study more QM, you'll eventually get to rotation group theory, 3j symbols, Clebsch-Gordon coefficients, etc, which will all help explain the rule. But the rule works.

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$\newcommand{\ket}[1]{\left| #1 \right>}$ Note that $ l=0$ has only one state $m=0$. Therefore the tensor product of $l=1$ and $l=0$ can be written as:

$$ (l=1)\otimes (l=0) = \left\{ \begin{array} &\ket{l= 1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \end{array} \right\}=(l=1) $$ As you have probably noticed this is just the spectrum of $l=1$ state hence you just consider it as $l=1$ triplet.

Edit: For some reason thought you asked $(l=1) \otimes (l=0)$ but you can equally well apply it to $ \left(l= \frac{1}{2} \right) \otimes (l=0)$

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To dramatize the point of the above correct answers, let me rewrite your multiplets the way mathematicians, or SU(3), SU(5),..., mavens, write them, through the actual dimensionality 2l+1 of the multiplets, in boldface.

So then, l=1/2 is a doublet, 2; 1 is a triplet, 3; 3/2 is a quartet, 4; and 5/2 a sextet, 6; and 0 a singlet, 1, the source of your confusion.

Then your Clebsch-Gordan series for the Kronecker product is expressible as just $$ {\bf 3 } \otimes {\bf 3 } \otimes {\bf 2 }= {\bf 6 } \oplus {\bf 4 } \oplus {\bf 4 } \oplus {\bf 2 } \oplus {\bf 2 } ~, $$ which manifestly matches at the level of elementary school arithmetic the dimensionality of the vector space of the entire reducible representation.

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  • $\begingroup$ This is a good idea for SU(2) because the dimension of a rep uniquely specifies the rep. For SU(N>2) there are often different reps with the same dimension as exemplified in SU(3) by people distinguishing the two octets by subscripts ($\bf{}8_A$ and $\bf{8_S}$). One complete way of labelling the different reps of the same dimension is by using the upper Gelfand operator pattern. The lower Gelfand operator pattern specifies the state in the rep. (cds.cern.ch/record/345437/files/CM-P00057105.pdf) $\endgroup$ – Gary Godfrey Feb 26 '16 at 20:01

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