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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!

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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.

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  • $\begingroup$ 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. $\endgroup$
    – zjs
    Commented Jul 1, 2020 at 0:17

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