2
$\begingroup$

I am wondering if anyone could explain to me either why my method is not valid or point out where I have made an algebraic slip. I have been looking at this for a long time, to no avail.

Let $\{\cdot , \cdot\}$ be a Poisson bracket.

I wish to show that for angular momentum $M=x\times p$, $$\{M_i,M_j\}=\varepsilon_{ijk} M_k$$

I have previously shown that $\{x_i,M_j\}=\varepsilon_{jik} x_k$ and $\{p_i,M_j\}=\varepsilon_{jik} p_k$

So I proceed as follows:

$$\{M_i,M_j\}=\varepsilon_{iab} \{x_a p_b,M_j\}$$ $$= \varepsilon_{iab} (x_a\{p_b,M_j\}+\{x_a,M_j\}p_b)$$ $$= \varepsilon_{iab} (x_a\varepsilon_{jbk} p_k+\varepsilon_{jak} x_kp_b)$$ $$=(\delta_{ik}\delta_{aj}-\delta_{ij}\delta_{ak})x_a p_k+ (\delta_{bk}\delta_{ij}-\delta_{bj}\delta_{ik})x_k p_b$$ $$=x_jp_i-x_kp_k+x_bp_b-x_ip_j$$ $$=x_jp_i-x_ip_j$$ $$=\varepsilon_{kji}M_k$$

which is minus the result I should be getting.

(I know that I could have expanded both $M_i,M_j$ and prove it that way, but I don't see any reason why this method should not work.)

Thank you.

  • 1
    Your method and your algebra are both correct. Your starting point isn't right. In fact, $x$ and $p$ and $M$ are all _vectors_ (which is presumably what you're trying to show). For any vector $V$, you should have $\{M_i, V_j\} = \varepsilon_{ijk} V_{k}$, or alternately $\{V_i, M_j\} = -\{M_j, V_i\} = -\varepsilon_{jik} V_{k} = \varepsilon_{ijk} V_{k}$. So your Poisson brackets of $x$ and $p$ with $M$ both have the wrong sign.2012-07-19
  • 0
    @mjqxxxx: Thanks for spotting that!2012-07-19

0 Answers 0