Quantum tensor product closer to Kronecker product?

Question

Coming more from a computer science background, I never really studied tensor products, covariant/contravariant tensors etc. So until now, I was seeing the "tensor product" operation mostly as (what appears to be) a Kronecker product between the matrix representation (in some fixed basis) of my vector/linear operator, i.e. if I have two vectors/matrices

$$A = \begin{pmatrix}a_{11} & a_{12} & \cdots \\ a_{21} & a_{22} & \cdots \\ \vdots\end{pmatrix}$$ $$B = \begin{pmatrix}b_{11} & b_{12} & \cdots \\ b_{21} & b_{22} & \cdots \\ \vdots\end{pmatrix}$$

Then: $$A \otimes B = \begin{pmatrix}a_{11}B & a_{12}B & \cdots \\ a_{21}B & a_{22}B & \cdots \\ \vdots\end{pmatrix} $$ i.e. $$A \otimes B = \begin{pmatrix} a_{11}b_{11} & a_{11} b_{12} & \cdots & a_{12}b_{11} & a_{12}b_{12} & \cdots \\ a_{11} b_{21} & a_{11} b_{22} & \cdots & a_{12}b_{21} & a_{12}b_{22} & \cdots \\ \vdots & \vdots & & \vdots & \vdots \\ a_{21}b_{11} & a_{21} b_{12} & \cdots & a_{22}b_{11} & a_{22}b_{12} & \cdots \\ a_{21} b_{21} & a_{21} b_{22} & \cdots & a_{22}b_{21} & a_{22}b_{22} & \cdots \\ \vdots & \vdots & & \vdots & \vdots \end{pmatrix} $$

In particular, if we consider $|0\rangle = \begin{pmatrix}1\\0\end{pmatrix}$ and $|1\rangle = \begin{pmatrix}0\\1\end{pmatrix}$, then $|0\rangle \otimes |1\rangle = \begin{pmatrix}0\\0\\0\\1\end{pmatrix}$, i.e. $|0\rangle \otimes |1\rangle$ is a vector.

Now, if I look at the tensor product webpage of wikipedia, they seem to define $v \otimes w \colon= v w^T$, i.e. $v \otimes w$ is a matrix (ok, the matrix is just a reshape of the vector obtained by the Kronecher product so both of them are isomorphic, but in term of "type" isn't it a bit strange to define it like that?). But on the other hand, when $v$ and $w$ are matrices, we are back to the Kronecher product.

So here is my question: why do they define the tensor product like that for vector? Is there different "kinds" of tensors? How are they linked with tensors used in physics?

Answer 1

The tensor product of two objects with $m$ and $n$ components is an object with $mn$ components that consists of the pairwise products of the components of the inputs. The Kronecker product and the $v w^T$ product (vector outer product) are both tensor products by that definition.

The reason for first defining a tensor product by matrix multiplication in the Wikipedia article is probably pedagogical: most readers will be familiar with matrix multiplication, and with that example under their belt they might have an easier time with the Kronecker product.

The tensor product of objects with $k$ and $\ell$ indices should naturally have $k+\ell$ indices. The vector outer product combines two 1-index vectors into a 2-index matrix, but the Kronecker product has to encode its 2+2 input indices into 2 output indices because matrices only support two indices. For that reason I'd say that the outer product is the more natural tensor product on vectors.

In the explicit-index tensor notation used in general relativity and particle physics, there is no operation of "matrix multiplication" and you instead always write it out explicitly. Instead of $M=NP$ you write $M_{ac} = \sum_b N_{ab} P_{bc}$, or just $M_{ac} = N_{ab} P_{bc}$ where the doubling of $b$ signals that it should be summed over. There's also no Kronecker product; you instead just generalize the outer product and use $k+\ell$ indices. Instead of $|ψ'\rangle = (I\otimes H\otimes I)|ψ\rangle$ you'd write $ψ'_{ab'c} = H_{b'b} ψ_{abc}$. Of course, the disadvantage of this system is you have to write out a lot of indices. (And, in quantum computing, the fact that you'd be the only one using it.)

Answer 2

The tensor product in linear algebra is defined for linear maps. When these linear maps are represented as matrices, you can use the Kronecker product to calculate it. https://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_linear_maps

Answer 3

In quantum computing, we compute the tensor product exactly as you specify. This includes the case of taking a tensor product with two vectors.

Now, it is also true that $|\psi\rangle\langle\phi|$ can be thought of as a tensor product between $|\psi\rangle$ and $\langle\phi|$. But in QC we would never define the tensor product between $|\psi\rangle$ and $|\phi\rangle$ to be that.