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).

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

1 Answer 1


I think requiring Hermiticity for X, and hence real, instead of pure imaginary, a dooms you.

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.
  • $\begingroup$ 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? $\endgroup$
    – MathZilla
    Sep 15, 2022 at 20:00
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    $\begingroup$ Yes and yes. The determinant of the product of two unimodular matrices is also 1. The idea is the unitary matrix has a hermitian and an antihermitean part., necessarily.... $\endgroup$ Sep 15, 2022 at 20:01
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    $\begingroup$ 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! $\endgroup$
    – MathZilla
    Sep 15, 2022 at 20:05
  • $\begingroup$ 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. $\endgroup$
    – MathZilla
    Sep 15, 2022 at 21:28
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    $\begingroup$ No, your a is my i a for all three components, not just the z component.... Remember, your a must be pure imaginary! $\endgroup$ Sep 15, 2022 at 21:31

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