Why tensor product?

Question

Let $A$ an $B$ be two discrete observables (like spins). When exactly and why we have to consider their tensor product when talking about the mutual observation of the corresponding phenomena?

Answer 1

For describing the behavior of a particle we usually write the Schrodinger equation, or Dirac equation, or other. All the possible solutions of the equation form a space of functions with certain properties, and we name such a space of functions Hilbert space. Now, if we have two particles, $1$ and $2$, which don't interact, behave as if none of them is aware if the presence of the other. The behavior of the two particles can be described, by anyone of the functions in the Hilbert space $\mathcal H_1$ for the particle $1$, and anyone of the functions in $\mathcal H_2$ for the particle $2$. We denote this situation by saying that the Hilbert space of the two particles is $\mathcal H = \mathcal H_1 \otimes \mathcal H_2$. This is what means this notation.

So, picking a state $|a\rangle$ from $\mathcal H_1$ and a state $|b\rangle$ from $\mathcal H_2$, the state $|a\rangle \otimes |b\rangle$, which as explained above belongs to $\mathcal H_1 \otimes \mathcal H_2$, can be very well a state describing the behavior of the pair of particles $1$ and $2$.

However we may have a more complicated situation, name entanglement. An entanglement has no classical equivalent, it is a topic of which you have to read separately. In short, without interacting with one another by any classical field, the particle aren't though independent. If you heard of the spin singlet, the state of the two fermions is

$$|S\rangle = \frac {|\uparrow\rangle_1 \otimes |\downarrow\rangle_2 - |\uparrow\rangle_2 \otimes |\downarrow\rangle_1}{\sqrt {2}}.$$

Well, in order to economize the subscripts $1$ and $2$ we use to write first the particle $1$ and second the particle $2$,

$$|S\rangle = \frac {|\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle \otimes |\uparrow\rangle}{\sqrt {2}}.$$

You see the state $|S\rangle$ still belongs to the space $\mathcal H_1 \otimes \mathcal H_2$, but it is no more the simple tensor product of one element from $\mathcal H_1$ with one element from $\mathcal H_2$, but a superposition of two such products.

Well, I hope that this time the situation is more clear.