Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Stimulated by some Physics backgrounds, consider the following two sets of matrices.

Notations and definitions:Let $A,B$ be two complex $n\times n$ matrices, then $\left [ A,B \right ]\overset{\underset{\mathrm{def}}{}}{=}AB-BA$ and $ e^A\overset{\underset{\mathrm{def}}{}}{=} \sum_{m=0}^{\infty }\frac{A^m}{m!}. $

Let $M_1,M_2,M_3$ be three $n\times n$ Hermitian matrices, and they satisfy $[M_1,M_2]=iM_3,[M_2,M_3]=iM_1,[M_3,M_1]=iM_2$ identities, where $i=\sqrt{-1}$.

Now define a set $X$ of unitary matrices:$X=\left \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ and another set $Y$ of unitary matrices:$Y=\left \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} :\alpha,\beta,\gamma \in \mathbb{R}\right \},$ where $i=\sqrt{-1}$ ( Note that the number indices of the Hermitian matrices $M$ are different in $X$ and $Y$ ).

And my questions are as follows:

(1) Is $X=Y$ ?

(2) If (1) is true, is the set $X$ a group ?

(3) If both (1) and (2) are true, is $X\cong SU(2)$ true ?

Supplements: For concreteness, let's take a look at the following simple example. Consider the Physically called spin-$\frac{1}{2}$ "Pauli" matrices $M_1=\frac{1}{2}\bigl(\begin{smallmatrix} 0& 1\\ 1&0 \end{smallmatrix}\bigr),M_2=\frac{1}{2}\bigl(\begin{smallmatrix} 0& -i\\ i&0 \end{smallmatrix}\bigr)$ and $M_3=\frac{1}{2}\bigl(\begin{smallmatrix} 1& 0\\ 0&-1 \end{smallmatrix}\bigr),$ and it's easy to find that $M_1^2+M_2^2+M_3^2=\frac{1}{2}(\frac{1}{2}+1)\mathbb{I}$, where $\mathbb{I}$ is a $2\times2$ identity matrix.

In the above example, direct calculation of matrices $e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}$ in $X$ shows that $X=SU(2)$ (then $X$ is a group), and it's also easy to verify that $Y\subseteq SU(2)$. So now the question is, is $SU(2)\subseteq Y$ too ?

Thanks in advance.

share|improve this question
2  
In the definition of X there is $M_3$ twice. Is this an error or not? :) –  Riccardo May 3 '13 at 21:06
1  
I would imagine that any answer would depend heavily on $M_1, M_2$ and $M_3$, which you have not elaborated on. Could be wrong though. I don't immediately grasp what those commutation conditions say. –  rschwieb May 3 '13 at 21:20
1  
@ Ric Ped, Dear Ped, $M_3$ definitely appears twice, it's not an error. –  K-boy May 3 '13 at 21:33
2  
@ Ric Ped, in fact, if you choose $M_1=\bigl(\begin{smallmatrix} 0 &1 \\ 1&0 \end{smallmatrix}\bigr),M_2=\bigl(\begin{smallmatrix} 0 &-i \\ i&0 \end{smallmatrix}\bigr),M_3=\bigl(\begin{smallmatrix} 1 &0 \\ 0&-1 \end{smallmatrix}\bigr)$, then $X$ is just the Euler's angles $(\alpha,\beta ,\gamma) $ representation of group $SU(2).$ –  K-boy May 3 '13 at 21:39
2  
Those relations don't hold for the Euler's angles matrices, instead $[M_1,M_2]=2iM_3$ $[M_2,M_3]=2iM_1$ $[M_3,M_1]=2iM_2$. Also for those matrices $\sum M_i^2=3\mathbb{I}$, where you say it should be $\frac{3}{4}\mathbb{I}$. Are you missing a factor of $2$ somewhere? –  Alexander Gruber May 3 '13 at 22:48

1 Answer 1

up vote 1 down vote accepted

Comments to the question (v3):

(We will from now on assume that the $n\times n$ matrices $M_1$, $M_2$ and $M_3$ are linearly independent, and that $n\geq 2$.)

OP defines two sets:

$$\tag{1} X_n~:=~ \{ e^{i\alpha M_3}e^{i\beta M_2}e^{i\gamma M_3}\in {\rm Mat}_{n\times n}(\mathbb{C}) \mid \alpha,\beta,\gamma \in \mathbb{R} \}, $$

$$\tag{2} Y_n~:=~ \{ e^{i(\alpha M_1+\beta M_2+\gamma M_3)} \in {\rm Mat}_{n\times n}(\mathbb{C}) \mid\alpha,\beta,\gamma \in \mathbb{R} \}.$$

The set $Y_n$ is the image $Y_n=\rho(SU(2))$ of an $n$-dimensional $SU(2)$ group representation $\rho:SU(2) \to GL(n,\mathbb{C})$, also known in physics as a spin $\frac{n-1}{2}$ representation. In particular, the set $Y$ is itself a group.

$$\tag{3} Y_n~\cong~\left\{ \begin{array}{rcl} SU(2)/ \mathbb{Z}_2 ~\cong~SO(3) &\text{for}&n&\text{odd}, \\ SU(2) &\text{for}&n&\text{even}.\end{array} \right. $$

The set $X_n\subseteq Y_n$ is a subset of $Y_n$; due, among other things, to the Baker-Campbell-Hausdorff formula.

In fact, the set $X_n$ is a generalized Euler angle realization of $SU(2)$, as mentioned by OP himself. One may check by inspection that the $X_n$ hit every element in $Y_n$, so that $X_n=Y_n$.

share|improve this answer
    
@ Qmechanic, thanks for your wonderful comment. So do you mean that set $Y$ is always "larger" than set $X$ ? Is this conclusion also true for single spin-$\frac{1}{2}$ ? In spin-$\frac{1}{2}$ example, is $Y$ a group ? And is only the set $X$ the spin-rotation group for spin-$\frac{1}{2}$ while $Y$ not ? –  K-boy May 5 '13 at 17:33
    
@ Qmechanic, In spin-$\frac{1}{2}$ example, it's easy to show that $Y$ is a subset of group $SU(2)$(then a subset of $X$), if what you say "$X$ is a subset of $Y$" is simultaneously true, then we arrive at $X=Y=SU(2)$ for spin-$\frac{1}{2}$, right? –  K-boy May 6 '13 at 8:13
1  
I updated the answer. –  Qmechanic May 7 '13 at 15:20
    
@ Qmechanic, thank you very much. To conclude your explanations, for spin-$\frac{n-1}{2}$** , we have $X_n=Y_n\cong SO(3)$ for $n=odd$ and $X_n=Y_n\cong SU(2)$ for $n=even$ . So $X_n$ and $Y_n$ are in fact the same one called **spin-rotation group for spin-$\frac{n-1}{2}$. Am I right ? –  K-boy May 7 '13 at 19:44
    
@ Qmechanic, and for a general angular momentum $M_1,M_2,M_3$ instead of spin , is $X_n$ still identical to $Y_n$ and $X_n$ or $Y_n$ still the rotation group associated with $M_1,M_2,M_3$ ? Thanks a lot. –  K-boy May 10 '13 at 11:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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