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Consider two spin systems $\mathcal{H}_1$ and $\mathcal{H}_2$, each with spin operator $L_1$ and $L_2$, respectively.

What is the difference between $L_1+L_2$ and $L_1 \cdot L_2$?

As far as I understand, both act component-wise on the tensor product space $\mathcal{H}_1\otimes\mathcal{H}_2$ so shouldn't they really be the same?

Qmechanic
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test123
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  • $L=L_1+L_2$ is diagonal in the product basis, i.e., it describes no coupling - just the total angular momentum. On the other hand, $L_1\cdot L_2$ is a true coupling. In general, a rule-of-thumb for two operators to be the same, is to have the same matrix representation - this is clearly not the case here. – Roger V. Jul 22 '21 at 09:05
  • @RogerVadim But isn't $(L_1\cdot L_2)h_1\otimes h_2=L_1h_1\otimes L_2h_2$ for all $h_i\in\mathcal{H}_i,i=1,2$? That is basically the source of my confusion. – test123 Jul 22 '21 at 09:26
  • I suggest doing an exercise with two spins 1/2. The product basis is just four states and all the math is easily explicitly done on a page. – Roger V. Jul 22 '21 at 09:33
  • But what is the definition of $L_1\cdot L_2$? That's what I was referring to – test123 Jul 22 '21 at 09:39
  • Usually it is a scalar product of two operators (since angular momentum is a vector). But you have to check what it means in your context. If $+$ and $\cdot$ are not vector operatoions in your case, you would need to clarify the question. – Roger V. Jul 22 '21 at 09:41
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    You might be interested in this. – Tobias Fünke Jul 22 '21 at 09:48
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    @Jakob Thanks! Exactly what I was looking for – test123 Jul 22 '21 at 09:51

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