The three generators of $su(2)$ satisfy the commutation relations

$$ [J_0 , J_\pm] = J_\pm , \quad [J_+, J_- ] = +2J_0 .$$

The three generators of $su(1,1)$ satisfy the commutation relations

$$ [K_0 , K_\pm] = K_\pm , \quad [K_+, K_- ] = -2K_0 .$$

Now, let us define

$$ K_0 = J_0, \; K_+ = J_+,\; K_- = - J_-. $$

It is apparent that so defined $K$'s satisfy the $su(1,1)$ algebra! Does this mean that $su(1,1)$ is actually equivalent to $su(2)$?

Where is the argument wrong?

  • $\begingroup$ What exactly do you mean by $su(1,1)$? Is it the same as $so(1,1)$ by any chance? $\endgroup$ – childofsaturn May 11 '15 at 15:22
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    $\begingroup$ @childofsaturn - no, SU and SO are never the same. $SU(m,n)$ is pseudounitary, i.e. complex matrices with $\det M=1$ obeying $MGM^\dagger=M$ where $G$ is diagonal with $m$ times $+1$ and $n$ times $-1$. $SU(1,1)$ ends up isomorphic to $SL(2,R)$ or also $SO(2,1)$. $\endgroup$ – Luboš Motl May 11 '15 at 16:12

You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients.

A general matrix in the $SU(2)$ group is written as $$ M = \exp[ i( \alpha J_+ + \bar\alpha J_- + \gamma J_0 )] $$ where $\alpha\in {\mathbb C}$ and $\gamma\in {\mathbb R}$ while the general matrix in $SU(1,1)$ is given by $$ M' = \exp [ i( \alpha_+ J_+ + \alpha_- J_- + \beta J_0 ] $$ where $\alpha_+,\alpha_-,\beta\in {\mathbb R}$ are three different real numbers.

To summarize, for $SU(2)$, the coefficients in front of $J_\pm$ are complex numbers conjugate to each other, while for $SU(1,1)$, they are two independent real numbers. (And I apologize that I am not sure whether the $i$ should be omitted in the exponent of $SU(1,1)$ only according to your convention. Probably.)

If you allow all three coefficients in front of $J_\pm,J_0$ to be three independent complex numbers, you will obtain the complexification of the group. And as Qmechanic also wrote, the complexification of both $SU(2)$ and $SU(1,1)$ is indeed the same, namely $SL(2,{\mathbb C})$.

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    $\begingroup$ so $su(2)$ and $su(1,1)$ are not determined by their commutation relations only? What are the extra conditions? $\endgroup$ – kaiser May 11 '15 at 19:57
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    $\begingroup$ @kaiser: They are determined by their commutation relations, but the $J_\pm$ are not elements of the real Lie algebra $\mathfrak{su}(2)$ (which is obvious from the $\mathrm{i}$ in the definition $J_\pm = J_1 \pm \mathrm{i} J_2$, so what you wrote down are not the commutation relations of $\mathfrak{su}(2)$, but of its complexification, as Qmechanics writes. $\endgroup$ – ACuriousMind May 11 '15 at 20:56
  • $\begingroup$ I suspect that $M' = \exp [ i( \alpha_+ J_+ + \alpha_- J_- + \beta J_0 ]$ should be $M' = \exp [ i( \alpha J_+ - \bar{\alpha} J_- + \beta J_0 ]$ with $\alpha$ complex. $\endgroup$ – kaiser May 12 '15 at 7:43
  • $\begingroup$ It's actually possible, @kaiser. Could someone please settle this question? For example by an explicit form of the matrices in terms of Pauli matrices etc. It isn't quite needed to explain why the groups are different but it would be nice to fix errors in "redundant" formulae, too. $\endgroup$ – Luboš Motl May 12 '15 at 9:51

The ladder operators do belong to the real Lie algebras$^1$ $$\begin{align} su(1,1)~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid m^{\dagger}\sigma_3=-\sigma_3m,~ {\rm tr}(m)=0\}\cr ~=~&{\rm span}_{\mathbb{R}}\{ \sigma_1, \sigma_2, i\sigma_3 \}\cr ~\cong~sl(2,\mathbb{R}) ~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid {\rm tr}(m)=0\}\cr ~=~&{\rm span}_{\mathbb{R}}\{ \sigma_1, i\sigma_2, \sigma_3 \}\cr ~=~&{\rm span}_{\mathbb{R}}\{ \sigma_+, \sigma_-, \sigma_3 \}\cr ~\cong~ so(2,1)~:=~&\{m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^{t}\eta =- \eta m\}, \end{align}$$ $$ \sigma_{\pm}~:=~\frac{\sigma_1\pm i \sigma_2}{2}, \qquad \eta~=~{\rm diag}(1,1,-1), $$ but they do not belong to the real Lie algebras $$\begin{align} su(2)~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid m^{\dagger}=-m,~ {\rm tr}(m)=0\}\cr ~=~&{\rm span}_{\mathbb{R}}\{ i\sigma_1, i\sigma_2, i\sigma_3 \} \cr ~\cong~ so(3)~:=~&\{m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^{t}=-m\}. \end{align}$$ All the above 5 real Lie algebras have complexifications isomorphic to $$\begin{align}sl(2,\mathbb{C})~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}(m)=0\}\cr ~=~&{\rm span}_{\mathbb{C}}\{ \sigma_1, \sigma_2, \sigma_3 \}.\end{align}$$


$^1$ Here we follow the mathematical definition of a real Lie algebra. Be aware that in much of the physics literature, the definition of a real Lie algebra is multiplied with a conventional extra factor of the imaginary unit $i$, cf. footnote 1 in my Phys.SE answer here.

  • $\begingroup$ what do you mean by the ladder operators? $\endgroup$ – kaiser May 11 '15 at 19:51
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    $\begingroup$ $J_{\pm}$ and $K_{\pm}$. Note in particular that OP's first equation does not display the generators of $su(2)$. $\endgroup$ – Qmechanic May 11 '15 at 20:44

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