# How to make this matrix unitary?

I have the following matrix from quantum mechanics, $$X = a_0 + \sigma\cdot\mathbf{a}$$, where $$\sigma$$ are the usual Pauli matrices. I can expand this into a matrix form of $$X$$, $$X = \begin{bmatrix} a_0 + a_z & a_x - ia_y\\ a_x + ia_y & a_0 - a_z \end{bmatrix}.$$ But, I want to see/find how to make this part of the unitary group.

I know it is at least Hermitian since $$X^\dagger = X$$, but when I test to see if it is unitary, $$X^\dagger X = XX^\dagger = 1$$, I always find constraints on the terms in $$X$$, but they never make sense. For instance, the off-diagonals becomes $$(a_0 + a_z)^2 + a_y^2 + a_z^2$$ must equal one, but this doesn't seem right to me.

Also, another part of what I want to do is to show that if I have another matrix $$Y = b_0 + \sigma\cdot\mathbf{b}$$ (exactly the same at $$X$$ but with different coefficients), I know that IF $$X$$ is truly unitary with some specific contraints, then if $$X,Y\in G$$ ($$G$$ being the unitary group) then their product must be in $$G$$ as well (this is basic closure in group theory). When I do this, I find that I must require $$a_yb_x - a_xb_y = 0$$ for the product to have the same structure of $$X$$ and $$Y$$.

I really just need someone (or more then one amazing person) to double check my reasoning is correct. The first part just doesn't seem "nice" enough to be correct, or I am just wrong, that's always possible.

EDIT: maybe this would be much easier if I simply did this without matrices, but I am unsure since $$X^\dagger = X$$ then $$X^\dagger X = (a_0 + \sigma\cdot\mathbf{a})(a_0 + \sigma\cdot\mathbf{a}) = a_0^2 + (\sigma\cdot\mathbf{a})^2 + 2a_0\sigma\cdot\mathbf{a}$$ and then I only need to figure out how to make $$2a_0\sigma\cdot\mathbf{a} = 1$$ (identity matrix).

• The condition $\det X=1$ is $a_0^2-\mathbf{a}\cdot\mathbf{a}=1$.
– J.G.
Sep 15, 2022 at 18:13
• @J.G. would this restrict $X$ enough to make it unitary, since $X^\dagger = X^{-1} = (\text{det}(X))^{-1}[[X_4, -X_2],[-X_3, X_1]]$? Sep 15, 2022 at 18:16

Let me review the mainstream representation of the conventional SU(2) group element, instead, $$U =a_0 + i \sigma\cdot\mathbf{a} =\begin{bmatrix} a_0 + ia_z & ia_x + a_y\\ ia_x -a_y & a_0 - ia_z \end{bmatrix}, ~~~~\leadsto \\ U U^\dagger =(a_0^2+\mathbf{a}^2) 1\!\! 1, \qquad \det U = (a_0^2+\mathbf{a}^2) ,$$ now with real a s. It is then evident U is simple unitary for $$a_0^2+\mathbf{a}^2 =1$$.
Conventionally, one parameterizes $$a_0=\cos\theta$$ and $$\mathbf{a}=\sin\theta ~~ \mathbf{n}$$, where n is a unit vector.
• You may then convince yourself that $$U= e^{i\theta ~~ \mathbf{n}\cdot \sigma },$$ and that the product of two such unitary matrices, with differing θs and n s is also unitary with new composite θs and n, the magic of Rodrigues/Gibbs formulas.
• Ok, I believe this is all that I am looking for, since I was simply looking for the requirements that make $X$ unitary. So, as long as the determinant is equal to 1, we are done. Then, to answer the second part of my own question, then $\text{det}(XY) = \text{det}(X)\text{det}(Y) = 1$ must be true for $XY$ to be part of the unitary group? Sep 15, 2022 at 20:00
• Thank you! I was using Mathematica to compute the entirety of $XY$ to try to find the constraint on each component, and did not use much group theory (as I should have), which makes it $\textit{much}$ easier. Thank you! Sep 15, 2022 at 20:05
• the problem I am running into is that yours and mine differ by a factor of $i$ on the $a_z$ component. This makes the determinant requirement get an extra factor of $2a_0a_z$. So to $\textbf{make}$ my matrix part of the unitary group, I need to have $2a_0a_z =0$ which is annoying, and doesn't give anything useful so, I guess that's that. Sep 15, 2022 at 21:28