Simple proof that $(U \otimes V)(|x\rangle \otimes |y\rangle) = U|x\rangle \otimes V|y\rangle$?

Question

This transformation comes up a lot during symbolic manipulation of quantum operations on state vectors. It's the reason why, for instance, $(X\otimes \mathbb{I}_2)|00\rangle = |10\rangle$ - it lets us operate on a single qbit by tensoring a unitary operation $U$ with identity operators where $U$ is at the same position of significance as the qbit to which we want to apply $U$.

I've been trying to write out a proof of why this transformation works, but I lack good notation for representing and reasoning about tensored matrices and vectors - it becomes very clunky very quickly. Is there a simple way to prove this transformation holds, or a convenient notation for representing tensored matrices/vectors?

Assume $U$ is a square complex unitary matrix of size $n$, $V$ a square complex unitary matrix of size $m$, $|x\rangle$ an $n$-element complex column vector where $\langle x|x\rangle=1$, and $|y\rangle$ an $m$-element complex column vector where $\langle y|y\rangle=1$.

Answer 1

I will give you a few elements for the demonstration on real vectors which you can extend to complex.

Let {$e_i$} be the standard basis for the space where $U (n*n)$ is defined . Let {$e_j$} be the standard basis for the space where $V (m*m)$ is defined.

First, it is a property that the basis {$e_i \otimes e_j$} is a basis for the n*m-matrices space.

$ U \otimes V $ is a linear mapping on the space and we have that : $$ U \otimes V (e_i \otimes e_j) = (U e_i) \otimes(V e_j) (1) $$

A remark to be given is that in linear algebra, when $W$ is linear and $W e_i$ is known, W is uniquely determined. As {$e_i \otimes e_j$} is a basis for the linear mapping $ U \otimes V $, it will be unique for the definition (1).

In particular,
$$ U \otimes V (x \otimes y) = (U x) \otimes(V y) $$

Indeed : $$ U \otimes V (x \otimes y) = U \otimes V (\sum_i x_i e_i \otimes \sum_j y_j e_j) $$ $$ = U \otimes V (\sum_{i,j} x_i y_j (e_i \otimes e_j)) $$ $$ = \sum_{i,j} x_i y_j U \otimes V (e_i \otimes e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) \otimes(V e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) (V e_j)^T $$ $$ = \sum_{i} x_i (U e_i) \sum_{j} y_j (V e_j)^T $$ $$ = U (\sum_{i} x_i e_i) (V (\sum_{j} y_je_j))^T $$ $$ = U x (V y)^T $$ $$ = U x \otimes V y $$

You can look at that PDF if it makes it more clear.

Answer 2

If we write $$ U=\sum_{i,j}U_{ij}|i\rangle\langle j|\quad V=\sum_{kl}V_{kl}|k\rangle\langle l|, $$ and $$ |x\rangle=\sum_jx_j|j\rangle\quad |y\rangle=\sum_ly_l|l\rangle, $$ then we can evaluate both sides of the equation $$ (U\otimes V)(|x\rangle\otimes|y\rangle)=(U|x\rangle)\otimes(V|y\rangle) $$ using the definition of the tensor product as $$ U\otimes V=\sum_{ijkl}U_{ij}V_{kl}|ik\rangle\langle jl|. $$

So, the left-hand side is \begin{align*} (U\otimes V)(|x\rangle\otimes|y\rangle)&=\left(\sum_{ijkl}U_{ij}V_{kl}|ik\rangle\langle jl|\right)\left(\sum_{jl}x_jy_l|jl\rangle\right) \\ &=\sum_{ijkl}U_{ij}x_jV_{kl}y_l|ik\rangle. \end{align*} Similarly, the right-hand side is \begin{align*} (U|x\rangle)\otimes(V|y\rangle)&=\left(\sum_{ij}U_{ij}x_j|i\rangle\right)\otimes\left(\sum_{kl}V_{kl}y_l|k\rangle\right) \\ &=\sum_{ijkl}U_{ij}x_jV_{kl}y_l|ik\rangle \end{align*} The two are the same.

You may worry that there's a little bit of trickery going on with the kets, that contained within the "definition" of the tensor product is already hiding an implicit use o f the tensor product because I'm going from $|i\rangle\otimes|k\rangle$ to $|ik\rangle$, and that makes the definition rather circular. However, remember that the text in a ket is just a label, so you can really think about what I'm doing as defining a new composite label in some different Hilbert space.

Answer 3

What you have in the title of your question is known as the "mixed-product property":

and you can get this plus many other properties from the Kronecker product wikipedia page. The other answers have shown why the mixed-product property holds true for the left Kronecker product.