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We have the angular momentum quantum numbers $(j,l,s)$ representing total angular momentum, orbital angular momentum and spin respectively.

I am solving a problem where we try to show that if you have two spin-1 particles then $l+s$ is even. I didn't understand how to relate these numbers together to be able to show this, so I tried to look at some Wikipedia pages, such as this one. I saw a relation which I thought may help solve the problem: $j=|l\pm s|$. However, I don't understand this equation - which value do you take, $+$ or $-$?

Question:

Does this help solve my problem? If so can someone explain the $\pm$ part?
If not, could someone guide me towards something that will help me solve the problem? An equation or relation perhaps, or a hint?

Thanks.

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  • $\begingroup$ In $l+s$, are you referring to the actual quantum numbers or the angular momentum vectors? Is $l$ related to a single particle or the combined orbital a.m. of the pair? Same question for $s$. $\endgroup$
    – Bill N
    Nov 8 '17 at 2:19
  • $\begingroup$ Also, on the page you link, it's talking about electrons, so $s=1/2$. The equation for $j$ on that page won't work for $s=1$. It should be $l+s\le j \le |l-s|$. $\endgroup$
    – Bill N
    Nov 8 '17 at 2:22
  • $\begingroup$ @BillN $j,l,s$ correspond to the quantum numbers in relation to the operators $J^2,L^2,S^2$. $\endgroup$
    – John Doe
    Nov 8 '17 at 2:33
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When coupling two systems, one with angular momentum $\ell_1$ the other with angular momentum $\ell_2$, the possible values of the net angular momentum $J$ are those in the set $\{\ell_1+\ell_2,\ell_1+\ell_2-1,\ell_1+\ell_2-2,\ldots, \vert \ell_1-\ell_2\vert\}$, i.e. they range from the sum to the absolute value of the difference of $\ell_1$ and $\ell_2$.

In the simplest example of application of this rule, $\ell_1=\ell_2=1/2$ so the possible values of such a two-spin system would be $J=0$ and $J=1$.

In general all possible values in the range will appear, although there might be some symmetry requirements that will eliminate some of the final $J$ values. In the specific case of $\ell_1=\ell_2=1$, one can show using Clebsch-Gordan methods that the $J=2$ and $J=0$ states are symmetric under permutations of the particle labels, whereas $J=1$ is antisymmetric. If the particles are indistinguishable bosons, this automatically eliminates the antisymmetric part $J=1$ in the list of possible $J$.

You might be interested by this answer or this answer.

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  • $\begingroup$ Oh ok, so the relation $j=|l\pm s|$ simply says "j can take the values between these two values". And yes, I had previously shown these were symmetric / antisymmetric. Thanks! $\endgroup$
    – John Doe
    Nov 8 '17 at 2:37
  • $\begingroup$ @JohnDoe usually this is written as $l+s\le j\le \vert l-s\vert$ so I don’t quite know why one would use $\vert l\pm s\vert$. Barring this the range as given is correct. $\endgroup$ Nov 8 '17 at 3:12
  • $\begingroup$ It is my honor to refer to one of my answers, especially when it comes from users like you. $\endgroup$
    – Frobenius
    Nov 10 '17 at 0:07
  • $\begingroup$ @Frobenius I'm not quite sure what you mean but I hold that your answer to a similar question was quite detailed and constructive, and deserves to be read. $\endgroup$ Nov 10 '17 at 0:41
  • $\begingroup$ By my comment above I thank you for your reference to my answer. I do not mean anything else. $\endgroup$
    – Frobenius
    Nov 10 '17 at 5:55

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