Why scale tensor products like that?

Question

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 $\frac{1}{2}$, I have the state $ |\frac{1}{2},m_1\rangle \otimes |\frac{1}{2},m_2\rangle $ for $m_i=\frac{1}{2},-\frac{1}{2}$. If I measure the $z$ component of the spin of the first particle ($S_1$), I get the value $\hbar m_1$, with $|\frac{1}{2},m1\rangle$ an eigenvector of that operator. Hence, we have that: $ (S_{1,z} \otimes 1) (|\frac{1}{2},m_1\rangle \otimes |\frac{1}{2},m_2\rangle)=(\hbar m_1 |\frac{1}{2},m_1\rangle) \otimes |\frac{1}{2},m_2\rangle $ and by the property (1) that's just equal to: $ \hbar m_1( |\frac{1}{2},m_1\rangle \otimes |\frac{1}{2},m_2\rangle)$. Which seems to imply that $|\frac{1}{2},m_1\rangle \otimes |\frac{1}{2},m_2\rangle$ 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) (|\frac{1}{2},m_1\rangle \otimes |\frac{1}{2},m_2\rangle)= |\frac{1}{2},m_1\rangle \otimes (\hbar m_1 |\frac{1}{2},m_2\rangle) $ (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.

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 '21 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 '21 at 17:57
@ACuriousMind because by measuring the first state, I'm altering the second one scaling it? I don't get it. Isn't it supposed that they are independent? I just don't see how this property manifest in reality, or in the measurement process — Jorge Ortiz, May 23 '21 at 08:16

score 1 · Answer 1 · answered Mar 27 '21 at 18:21

This is because of geometry.

Tensors actually generalise vectors to higher dimensional arrows. Whilst a vector is a single arrow, a 2-tensor, which was can also call a 2-vector, is defined by two arrows and so on.

This means, that whilst there is only one way of adding two vectors - the usual parallelogram law - there are 2 ways of adding 2-tensors and 5 ways of adding 5-tensors and so on. And these additions are just generalisations of the vector additivity law. It's quite clear if you draw out the second and third cases - the higher cases obviously can't be drawn in our 3d world.

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.

Tensors are very geometric notions, although typically they aren't introduced in this fashion.

Why scale tensor products like that?

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