General rule of thumb for tensor products.

Question

When would I want to use a tensor product? For example, say I am trying to derive "something" but, to get to the right derivation, I should use the tensor product.

The question comes from quantum computing, but I'm more interested in understanding the context of when to use it. If someone could answer this by a) focusing more of "what it does", and b) by providing a very simple example that showcases the intuition, then very much appreciated.

Also, I have the definition of the tensor product on hand, but--if I didn't make it obvious enough--I don't understand the bigger picture of its utility.

Thanks again.

Answer 1

The motivation for the tensor product from quantum states is quite straightforward in the finite-dimensional case.

1. Let $X$ be a finite set, say $X = \{x_1, x_2\}$. We want to talk about $x_1$ and $x_2$ as being the two observable states of some particle. (Say $x_1$ is spin-up, and $x_2$ is spin-down or something).
2. However, the system can be in a $\mathbb{C}$-linear combination of those states, so we should really be writing down a state as a linear function $f: X \to \mathbb{C}$. Let $F(X, \mathbb{C})$ denote the vector space of all functions $X \to \mathbb{C}$.
3. Now introduce another particle, with observables $Y = \{y_1, y_2, y_3\}$. A quantum state for this system is a function belonging to $F(Y, \mathbb{C})$.

Now, if we want to talk about these particles together, their set of observables should be $$X \times Y = \{(x_1, y_1), (x_1, y_2), (x_1, y_3), (x_2, y_1), (x_2, y_2), (x_2, y_3)\}$$ and so each quantum state should be a vector belonging to $F(X \times Y, \mathbb{C})$. Now we can ask: how are the vector spaces $F(X, \mathbb{C})$, $F(Y, \mathbb{C})$, and $F(X \times Y, \mathbb{C})$ related? The answer is the tensor product: $$F(X \times Y, \mathbb{C}) \cong F(X, \mathbb{C}) \otimes_{\mathbb{C}} F(Y, \mathbb{C})$$

Note that there is a way to take $f \in F(X, \mathbb{C})$ and $g \in F(Y, \mathbb{C})$ and produce a function $(f \otimes g) \in F(X \times Y\mathbb)$, by defining $$ (f \otimes g)(x, y) = f(x) g(y).$$ We can produce all such "separable" states in this way. However, there are obviously functions on $X \times Y$ that do not arise in such a way, for example the function $h: X \times Y \to \mathbb{C}$ defined by $h(x_1, y_1) = h(x_2, y_2) = 1$ and all else 0. This function $h$ is not a separable state (it is an entangled state), but it is a linear combination of separable states. (Every function in $F(X \times Y, \mathbb{C})$ is a linear combination of separable states).

So, one way to view the tensor product is as the "product" operation once you take linear combinations on some base set. There are many other ways to view the tensor product, this is just one.