# Tensor of Hamilton quaternions with C

I'm currently reading Voight's Quaternion Algebras which at 2.4.22 discusses $$\mathbb{H}\otimes_\mathbb{R}\mathbb{C}$$:

"The matrix representation of $$\mathbb{H}$$ in section 2.4 extends to a matrix representation of $$\mathbb{H}\otimes_\mathbb{R}\mathbb{C}$$, and this representation and its connection to unitary matrices is still used widely in quantum mechanics."

My understanding of tensors is pretty weak so I'm confused on a couple points here and am hoping for some clarification. I know that single-qubit gates are in $$PU(2)$$. Is that all that the tensor product is? I can see some resemblance as tensoring by $$\mathbb{C}$$ should quotient away any (complex) scalar matrices (and of course the quaternions are matrices by the natural identification). Is the tensor over $$\mathbb{R}$$ just because we view the quaternions and complex numbers as real algebras/vector spaces? Thanks!

Is that all that the tensor product is?

Is ... is what all that the tensor product is? The previous sentence about single-qubit gates doesn't mention tensor products, so what does "that" refer to in the highlighted sentence?

If $$A$$ and $$B$$ are real algebras, then $$A\otimes B$$ is spanned by elements of the form $$a\otimes b$$, subject to the distributive property and scalar multiplication. The $$\mathbb{R}$$ in the subscript of $$\otimes$$ means only real scalars are allowed to pass across the $$\otimes$$ symbol, as in $$(\lambda a)\otimes b=a\otimes(\lambda b)=\lambda(a\otimes b)$$ for $$\lambda\in\mathbb{R}$$.

In the case of $$\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}$$, yes we are considering $$\mathbb{H}$$ and $$\mathbb{C}$$ as real algebras.

If we view $$\mathbb{H}$$ as a right $$\mathbb{C}$$-vector space, then we can multiply it by scalars from $$\mathbb{H}$$ on the left and by scalars from $$\mathbb{C}$$ on the right - these actions commute with each other because $$\mathbb{H}$$ is associative - which makes $$\mathbb{H}$$ a module over the algebra $$\mathbb{H}\otimes\mathbb{C}$$ as a right $$\mathbb{C}$$-vector space, so we have a $$\mathbb{R}$$- algebra homomorphism

$$\mathbb{H}\otimes\mathbb{C}\to\mathrm{End}_{\mathbb{C}}(\mathbb{H})$$

Note $$\mathbb{H}\cong\mathbb{C}^2$$ as a $$\mathbb{C}$$-vector space so $$\mathrm{End}_{\mathbb{C}}(\mathbb{H})\cong M_2(\mathbb{C})$$. We can check the above is an isomorphism; pick basis elements $$\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$$ and $$\{1,i\}$$ of $$\mathbb{H}$$ and $$\mathbb{C}$$ to form basis elements $$a\otimes b$$ of $$\mathbb{H}\otimes\mathbb{C}$$, then check the corresponding matrices in $$M_2(\mathbb{C})$$ are linearly independent. The homomorphism (turning tensor into matrices) is a bit tricky because we're combining left/right actions...

Here's how to turn $$\mathbf{j}\otimes i$$ into a $$2\times 2$$ complex matrix. First, for $$\mathbb{H}\cong\mathbb{C}^2$$ as right $$\mathbb{C}$$-vector spaces, we'll use $$\{1,\mathbf{j}\}$$ as a basis corresponding to $$(1,0)$$ and $$(0,1)$$. Then we can define $$(a\otimes b)x:=axb$$ (you could also define $$ax\overline{b}$$ instead, the conjugation being useful for ensuring we get a left module in general, but it won't matter here because $$\mathbb{C}$$ is commutative). So we compute

$$(\mathbf{j}\otimes i)(1) ~=~ \mathbf{j}i ~=~ 1(0+0i)+\mathbf{j}(0+1i)$$ $$(\mathbf{j}\otimes i)(\mathbf{j}) ~=~ \mathbf{jj}i ~=~ 1(0-1i)+\mathbf{j}(0+0i)$$

So the matrix is

$$\mathbf{j}\otimes i \quad\longleftrightarrow\quad \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$$

You mean use a different convention for how $$\mathbb{H}$$ is a $$\mathbb{H}\otimes\mathbb{C}$$-module in order to be more consistent with the notes you're using.

• Thank you very much! I guess I was trying to figure out how $\mathbb{H}\otimes_\mathbb{R}\mathbb{C}$ "looks like" a more familiar object/space, and your answer makes the correspondence very clear.
– zjs
Commented Jul 1, 2020 at 0:17