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 $\langle y|y\rangle=1$.

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    $\begingroup$ Isn't this rather the definition of what $U\otimes V$ is? $\endgroup$ – Norbert Schuch Oct 21 '18 at 21:21
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    $\begingroup$ It's not the definition, it's a property of the left Kronecker product. $\endgroup$ – user1271772 Oct 21 '18 at 21:32
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    $\begingroup$ It’s a property that one requires that leads to the tensor product being the correct thing to do mathematically $\endgroup$ – DaftWullie Oct 22 '18 at 5:42
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    $\begingroup$ I'd say the tensor product is defined by the way it acts on product states (or maybe more precisely, a product basis), and then extended by linearity. $\endgroup$ – Norbert Schuch Oct 22 '18 at 7:51
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    $\begingroup$ @ahelwer You define "tensor product" through "tensoring together". That's almost tautological. What does "tensoring together" mean? $\endgroup$ – Norbert Schuch Oct 23 '18 at 22:17

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.


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.

  • $\begingroup$ Could this argument not be very circular? If I'm free to pick any basis (by which I assume you mean orthonormal) for each of the two Hilbert spaces, why don't I just pick bases that contain $|x\rangle$ and $|y\rangle$ as elements? Then it's just a trivial statement of (1) and you're not really proving anything. $\endgroup$ – DaftWullie Oct 22 '18 at 6:44
  • $\begingroup$ Whether you use the ket notation or not, we are dealing with vectors anyway. We can just adapt after. The proof would be too long anyway here and I prefer he has a few elements to vizualise. $\endgroup$ – cnada Oct 22 '18 at 7:15
  • $\begingroup$ My comment was nothing to do with the use of Dirac notation! $\endgroup$ – DaftWullie Oct 22 '18 at 9:11
  • $\begingroup$ Then I am sorry I did not fully understand your comment. $\endgroup$ – cnada Oct 22 '18 at 9:12
  • $\begingroup$ You say "let $\{e_i\}$ be a basis for the space..". I'm pointing out that if you choose a specific basis, where $e_1=x$ (and something similar for the second space, using $y$), your statement of equation (1) for $i=j=1$ is exactly what you're being asked to prove, and you're just stating it. $\endgroup$ – DaftWullie Oct 22 '18 at 9:24

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

enter image description here

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.

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    $\begingroup$ Linking to Wikipedia is good, but you should also cite the relevant content in your answer. That way if the Wikipedia page changes or gets deleted over the next ten years, the answer is still useful. $\endgroup$ – Craig Gidney Oct 23 '18 at 22:17
  • $\begingroup$ Mr. Gidney, I have seen other people link to Wikipedia. $\endgroup$ – user2898 Oct 23 '18 at 22:47
  • $\begingroup$ Linking to Wikipedia is great, but see meta.stackexchange.com/questions/8231/… . Answers should stand on their own even if the reader can't follow the link. In this case your answer doesn't include a proof of the property, it just says that yes the property is a thing and then delegates to wikipedia. $\endgroup$ – Craig Gidney Oct 23 '18 at 23:54
  • $\begingroup$ @user2898 Welcome to Quantum Computing SE. As Craig Gidney mentions, link-only answers are discouraged on this site. Also, images are not acceptable as a substitute for text as they are not "searchable". They are only suitable for diagrams or illustrations. Please type out the material in the image as text using MathJax. Thank you. $\endgroup$ – Sanchayan Dutta Oct 24 '18 at 0:00

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