0
$\begingroup$

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).

$\endgroup$
2
  • 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

3
$\begingroup$

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.
$\endgroup$
6
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.