Value of orbital angular momentum for two particles

[Decimal]

Hello,

I encountered the following statement in my lecture notes and there is a couple of things I don't understand:"Let's consider two particles with spins $s_1 = \frac{1}{2}$ and $s_2 = 1$ with a spherically symmetric interaction potential. Assume these two particles are in a two particle state with orbital quantum number $l=2$. Now a measurement of $L^2$ will always give the value $6 \hbar^2$."

First, I assume this means that the total orbital quantum number of the two particles is equal to 2? If both particles were to carry $l=2$ this would result in a total quantum number $l_{tot}=4$ right? This would mean the measurement of $L^2$ should give $20\hbar^2$. Again, please correct me if I am wrong.

Also I thought when adding up angular momenta there would always be multiple possible values. So let's say we add up two $l=2$ particles then the total orbital angular momenta would be $l_{tot} = 4,3,2,1,0$. Then one would also find multiple values for $L^2$ right? Yet apparently there is only one value, so what am I not understanding?

I feel like I am missing something, so any help would be greatly appreciated! Thanks!

 

[PeterDonis]

I assume this means that the total orbital quantum number of the two particles is equal to 2?

Yes. More precisely, it means that the total orbital quantum number of the two particle state is equal to 2. That doesn't mean that either particle has a well-defined orbital quantum number by itself (see below).

I thought when adding up angular momenta there would always be multiple possible values.

If you have two particles that are individually in states that both have $l = 2$, then there can be multiple values of the total $l$ of the two particles together, because there can be multiple two-particle states that have $l = 2$ for each particle individually. But that's not what your lecture notes are describing. They are describing a single two-particle state that has $l = 2$ for the two-particle state. The notes are not telling you anything definite about $l$ for each particle individually (and if you think about it you will see that neither particle can have a definite value of $l$ by itself; they must be entangled so that their individual orbital angular momenta are correlated, with each individual particle being in a superposition of different values of $l$).

 

[Decimal]

Ah yes, I was trying to think of the individual angular momenta as having some value (like both particles having $l=1$). What you said makes a lot more sense indeed. Thank you!