# Representations of the Poincaré group: question in a proof

• I

## Main Question or Discussion Point

Hello!

On page 51 where he want to invert
$$\Lambda^{\mu}_{\nu} = \tfrac{1}{2} \text{tr}( \bar{\sigma}^{\mu}A \sigma_{\nu} A^{\dagger})$$
the person says we may use
$$\sigma_{\nu} A^{\dagger} \bar{\sigma}^{\nu} = 2 \text{tr}(A^{\dagger})I.$$
to do that ... how do you prove this formula?

I have one method which is not good at all, and he does not even prove it so it may be very simple and so I hope you can see it and explain

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dextercioby
Homework Helper
What method do you have?

Hello!

The method I have to invert the formula is to work from
$$M' = AMA^{\dagger} = A x^{\nu} \sigma_{\nu} A^{\dagger} = x^{\nu} A \sigma_{\nu} A^{\dagger} = x'^{\mu} \sigma_{\mu} = \Lambda^{\mu}_{\nu} x^{\nu} \sigma_{\mu} = x^{\nu} \Lambda^{\mu}_{\nu} \sigma_{\mu} \mapsto \Lambda^{\mu}_{\nu} \sigma_{\mu} = A \sigma_{\nu} A^{\dagger}$$
and use the identity
$$\sigma_{\nu cd} \bar{\sigma}^{\nu}_{eb} = 2 \delta_{cb} \delta_{de}.$$
With it i get
$$\Lambda^{\mu}_{\nu} \sigma_{\mu} \bar{\sigma}^{\nu} = A \sigma_{\nu} A^{\dagger} \bar{\sigma}^{\nu} = (A \sigma_{\nu} A^{\dagger} \bar{\sigma}^{\nu})_{ab} = A_{ac} \sigma_{\nu cd} A^{\dagger}_{de} \bar{\sigma}^{\nu}_{eb} = A_{ac} \sigma_{\nu cd} \bar{\sigma}^{\nu}_{eb} A^{\dagger}_{de} = A_{ac} 2 \delta_{cb} \delta_{de} A^{\dagger}_{de} = 2 A_{ab} A^{\dagger}_{dd} = 2 A \mathrm{tr}(A^{\dagger}).$$
But this identity
$$\sigma_{\nu cd} \bar{\sigma}^{\nu}_{eb} = 2 \delta_{cb} \delta_{de}$$
I don't understand properly. How would it be predicted and explained?

But this is a bit different from what I asked in my first question, and that way seems easier. So now we have maybe two ways to fix!

dextercioby
Homework Helper
[...]
$$\sigma_{\nu} A^{\dagger} \bar{\sigma}^{\nu} = 2 \text{tr}(A^{\dagger})I.$$
to do that ... how do you prove this formula?[...]
Why won't you expand the lhs, assuming ##A^{\dagger}## is (a b\\c d)?

Yes!

I take
$$\sigma_0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \sigma_2 = \begin{bmatrix} 1 & -i \\ i & 0 \end{bmatrix}, \sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & - 1 \end{bmatrix}$$
and set
$$\overline{\sigma}_{\mu} = (I,-\sigma_i)$$
so that I now have (with ##\eta^{\mu \nu} = (1,-1,-1,-1)##)
$$\overline{\sigma}^{\mu} = \eta^{\mu \nu} \overline{\sigma}_{\nu} = (I,\sigma_i) = \sigma_{\mu}$$
but then when I write out ##A##
\begin{align}
\sigma_{\mu} A \bar{\sigma}^{\mu} &= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \bar{\sigma}^{\mu} \\
&= \sigma_{\mu}[a \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} ]\bar{\sigma}^{\mu}
\end{align}
perhaps I should invoke my ##\sigma_{\mu}## expressions
\begin{align}
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} &= \frac{1}{2}(\sigma_0 + \sigma_3), \\
\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} &= \frac{1}{2}(\sigma_1 + i \sigma_2), \\
\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} &= \frac{1}{2}(\sigma_1 - i \sigma_2), \\
\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} &= \frac{1}{2}(\sigma_0 - \sigma_3)
\end{align}
which lets me say
\begin{align}
\sigma_{\mu} A \bar{\sigma}^{\mu} &= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \bar{\sigma}^{\mu} \\
&= \sigma_{\mu}[a \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} ]\bar{\sigma}^{\mu} \\
&= \sigma_{\mu}[a \frac{1}{2}(\sigma_0 + \sigma_3) + b \frac{1}{2}(\sigma_1 + i \sigma_2) + c \frac{1}{2}(\sigma_1 - i \sigma_2) + d \frac{1}{2}(\sigma_0 - \sigma_3) ]\bar{\sigma}^{\mu} \\
&= \frac{1}{2} (a+d)\sigma_0 \sigma_{\mu}\bar{\sigma}^{\mu} + 0 \\
&= \frac{1}{2} (a+d)4 \sigma_0 \\
&= 2 \text{tr}(A)I
\end{align}
because the following
\begin{align}
\frac{1}{2}b\sigma_{\mu} \sigma_1\bar{\sigma}^{\mu} &= \frac{1}{2}b[\sigma_{0} \sigma_1\bar{\sigma}^{0} + \sigma_{1} \sigma_1\bar{\sigma}^{1} + \sigma_{2} \sigma_1\bar{\sigma}^{2} + \sigma_{3} \sigma_1\bar{\sigma}^{3}] = \frac{1}{2}b[ \sigma_1 + \bar{\sigma}^{1} - \sigma_1 \sigma_{2} \bar{\sigma}^{2} - \sigma_1 \sigma_{3} \bar{\sigma}^{3}] \\
&= \frac{1}{2}b[ \sigma_1 + \sigma - \sigma_1 - \sigma_1 ] = 0
\end{align}
holds, also for ##\sigma_2,\sigma_3##.

Can it be seen in a quicker fashion like he can see it?

Last edited:
samalkhaiat
$$\sigma_{\nu} A^{\dagger} \bar{\sigma}^{\nu} = 2 \text{tr}(A^{\dagger})I.$$
... how do you prove this formula?
Take the ${}_\alpha {}^{\beta}$ entries of $(2 \ \mbox{Tr}(A^{\dagger}) \ I_{2})$: $$( 2 \ \mbox{Tr}(A^{\dagger}) \ I_{2} )_{\alpha}{}^{\beta} = 2 \ \mbox{Tr}(A^{\dagger}) \ \delta_{\alpha}{}^{\beta} . \ \ \ \ \ \ (1)$$ Now substitute $$\mbox{Tr}(A^{\dagger}) = (A^{\dagger})^{\bar{\alpha}}{}_{\bar{\alpha}} = ( A^{\dagger} )^{\bar{\alpha}}{}_{\bar{\beta}} \ \delta^{\bar{\beta}}{}_{\bar{\alpha}} ,$$ in (1) and use the identity $$\left(\sigma_{\mu} \right)_{\alpha \bar{\alpha}} \left(\bar{\sigma}^{\mu} \right)^{\bar{\beta}\beta} = 2 \ \delta_{\alpha}{}^{\beta} \ \delta^{\bar{\beta}}{}_{\bar{\alpha}} .$$ This gives you

$$( 2 \ \mbox{Tr}(A^{\dagger}) \ I_{2} )_{\alpha}{}^{\beta} = ( \sigma_{\mu} )_{\alpha \bar{\alpha}} ( A^{\dagger} )^{\bar{\alpha}}{}_{\bar{\beta}} \left( \bar{\sigma}^{\mu} \right)^{\bar{\beta}\beta} = ( \sigma_{\mu} \ A^{\dagger} \ \bar{\sigma}^{\mu} )_{\alpha}{}^{\beta} .$$

vanhees71
Ummm... what????

That is quite a statement, would prefer being ignored with this unfair attitude.

I already posted a proof using the steps you used:

But this identity
$$\sigma_{\nu cd} \bar{\sigma}^{\nu}_{eb} = 2 \delta_{cb} \delta_{de}$$
I don't understand properly. How would it be predicted and explained?

But this is a bit different from what I asked in my first question, and that way seems easier. So now we have maybe two ways to fix!
and mentioned my issue with it, I see it is you decided not to interact with peoples posts

vanhees71
Gold Member
2019 Award
I think given this posting #7 we should suggest to Greg to introduce a "dislike button"

samalkhaiat
Ummm... what????

That is quite a statement, would prefer being ignored with this unfair attitude.
1) That statement was referring to

2) Clearly, #6 is an answer to the question you posed in #1.

3) The following two identities can be proved by inspection, i.e. by considering all possible values of the spinor indices,

$$\eta_{\mu\nu} \ (\sigma^{\mu})_{\alpha \bar{\alpha}} \ (\sigma^{\nu})_{\beta \bar{\beta}} = 2 \ \epsilon_{\alpha \beta} \ \epsilon_{\bar{\alpha}\bar{\beta}} , \ \ \ \ \ \ (1)$$ $$\eta_{\mu\nu} \ (\sigma^{\mu})_{\alpha \bar{\alpha}} \ (\bar{\sigma}^{\nu})^{\bar{\eta}\eta} = 2 \delta_{\alpha}{}^{\eta} \ \delta^{\bar{\eta}}{}_{\bar{\alpha}} . \ \ \ \ \ \ (2)$$

And once you prove any one of them, the second one follow from the following spinor-metric relation $$(\bar{\sigma}^{\mu})^{\bar{\eta}\eta} = - \epsilon^{\eta\beta} \ (\sigma^{\mu})_{\beta \bar{\beta}} \ \epsilon^{\bar{\beta}\bar{\eta}} , \ \ \ \ \ (3)$$ where the spinor metric and its inverse are given by

$$\epsilon_{\alpha \beta} = \epsilon_{\bar{\alpha}\bar{\beta}} = - \epsilon^{\alpha \beta} = - \epsilon^{\bar{\alpha}\bar{\beta}} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} ,$$ $$\epsilon_{\alpha \beta}\ \epsilon^{\beta \eta} = \delta_{\alpha}{}^{\eta} , \ \ \epsilon_{\bar{\alpha}\bar{\beta}}\ \epsilon^{\bar{\beta}\bar{\eta}} = \delta^{\bar{\eta}}{}_{\bar{\alpha}} . \ \ \ \ (4)$$

Proving (1): expand the LHS as $$\mbox{LHS}(1) = \sigma^{0}_{\alpha \bar{\alpha}} \ \sigma^{0}_{\beta \bar{\beta}} - \sigma^{1}_{\alpha \bar{\alpha}} \ \sigma^{1}_{\beta \bar{\beta}} - \sigma^{2}_{\alpha \bar{\alpha}} \ \sigma^{2}_{\beta \bar{\beta}} - \sigma^{3}_{\alpha \bar{\alpha}} \ \sigma^{3}_{\beta \bar{\beta}} .$$ Now consider, for example, the case $\alpha = \bar{\beta} = 1, \ \beta = \bar{\alpha} = 2$: $$\mbox{LHS}(1) = (0)(0) - (1)(1) - (-i)(i) - (0)(0) = -2 ,$$ $$\mbox{RHS}(1) = 2 \epsilon_{12}\epsilon_{21} = 2(-1)(1) = -2 .$$ Now, take another possible values, say $\alpha = \bar{\alpha} = \bar{\beta} = 1 , \ \beta = 2$:

$$\mbox{LHS}(1) = (1)(0) - (0)(1) - (0)(-i) - (1)(0) = 0 ,$$ $$\mbox{RHS}(1) = 2 \epsilon_{12}\epsilon_{11} = 2(-1)(0) = 0.$$ And you can check that for all other possible combination values of the spinor indices, you always have $\mbox{LHS}(1) = \mbox{RHS}(1)$.
Now that we proved the identity (1), we can easily prove the identity (2): Substitute (3) in the LHS of (2), then using (1) and (4) leads you to the RHS of (2).

dextercioby and vanhees71
The notes I posted, on page 51, seemed to believe that the proof should be so extremely obvious it was not worth proving.

The only proof I was aware of was the one I posted in post #3, as samalkhaiat nicely just explicitly explained them. I believe I tracked down the original paper which was this same proof by inspection, and I believed I understood it, until days later when I completely forgot it, so I posted a question hoping there was a simpler way as the notes seemed to say.

If you prove equation 2.15 of the notes, which we all know, then the proof is immediate directly from this with almost no work. This way also lets you do the ##SO(3)## case very quickly, and the expression is more complicated, as you see on the line between equation 2.23 and 2.24.