# Why scale tensor products like that?

Studying the sum of angular momentum in quantum mechanics, tensor products were introduced to us to get the general system from the individual states. There's a property that says: $$(av)\otimes w = a(v \otimes w)= v\otimes (aw)$$. We will call it (1). But I don't get it's physical meaning. It's as if I scale one vector, but not the other, and the vector of the total system scales by that factor¿?.

For example, for two particles with spin 1/2, I have the state $$|1/2,m1> \otimes |1/2,m2>$$ for $$m_i=1/2,-1/2$$. If I measure the z component of the spin of the first particle ($$S_1$$), I get the value $$\hbar m_1$$, with $$|1/2,m1>$$ an eigenvector of that operator. Hence, we have that: $$(S_{1,z} \otimes 1) (|1/2,m1> \otimes |1/2,m2>)=(\hbar m_1 |1/2,m1>) \otimes |1/2,m2>$$ and by the property (1) that's just equal to: $$\hbar m_1( |1/2,m1> \otimes |1/2,m2>)$$. Which seems to imply that $$|1/2,m1> \otimes |1/2,m2>$$ is an eigenvector of $$S_{1,z} \otimes 1$$ with eigenvalue $$\hbar m_1$$ (the measurement of state 1 is scaling the total system by that factor). Or what's worse, by the third equality in (1), that: $$(S_{1,z} \otimes 1) (|1/2,m1> \otimes |1/2,m2>)= |1/2,m1> \otimes (\hbar m_1 |1/2,m2>)$$ (that the measurement in the first state scales the second one by that factor ¿¿¿???). But that's not the case in reality. The measurement in state 1 doesn't change things that way.

I don't get what's going on, I don't understand much tensor products. If you can shed some light on this. Thank you

• This would be so much easier to read if you formatted the $m_{1,2}$ with subscripts, $\frac{1}{2}$ as proper fractions and the angle brackets using $\langle,\rangle$ (\langle, \rangle). – jacob1729 Mar 27 at 17:57
• "But that's not the case in reality." - if I've read what you've written correctly, all the things you've said are true. Why do you think they are "not the case in reality"? – ACuriousMind Mar 27 at 17:57

This takes care of the additivity of tensors. Your question is about the internal scaling of tensors where $$au \otimes v = u \otimes av = a(u \otimes v)$$.
Again this is a geometric property, it means that we can recale a tensor whilst keeping its geometric area the same. If you draw a 2-tensor, that is a 2-vector as two arrows so that they form a parallelogram, you'll see that rescaling one side by $$a$$ means that you must rescale the other side by $$1/a$$, and hence the geometric area remains the same. I recommend drawing out the diagram as it's quite clear then.