1
$\begingroup$

I am trying to understand a simple demonstration in Ashok Das' Lectures in QFT. He does the following on p. 134

$$[P_\mu,M_{\nu\lambda}]=[\partial_\mu ,x_\nu\partial_\lambda-x_\lambda\partial_\nu]=\partial_\mu(x_\nu\partial_\lambda-x_\lambda\partial_\nu)-(x_\nu\partial_\lambda-x_\lambda\partial_\nu)\partial_\mu\tag{4.37}$$

So far everything is fine, just replaced the expressions for $P_\mu$ and $M_{\nu\lambda}$. However, the next step is where I have trouble. I calculate

$$=\eta_{\mu\nu}\partial_\lambda-\eta_{\mu\lambda}\partial_\nu-x_\nu\partial_\lambda\partial_\mu+x_\lambda\partial_\nu\partial_\mu.$$

I understand where the $\eta$ come from in the first few terms, but according to the explanation, the 3rd and four terms must vanish. However, for them to vanish, it would have to mean that $x$ and $\partial$ commute, and I am not sure why that would be the case. If they commute, wouldn't that change the definition of $M_{\nu\lambda}$? After all, it's terms would commute and maybe even cancel out! I know that there is something here that I am understanding wrong, but I'm not sure what it is.

$\endgroup$
  • $\begingroup$ I'm afraid that of all the books that I have looked into, Das' was the only one that went into detail into the calculations, all others so far just give you the result. $\endgroup$ – Nick Heumann Sep 5 '19 at 23:15
  • 1
    $\begingroup$ This calculation is just really sloppy, he’s misapplying the product rule. Just act with both sides on a test function and you’ll see that your two unwanted terms actually each appear twice, and cancel each other. $\endgroup$ – knzhou Sep 6 '19 at 4:42
2
$\begingroup$

@knzhou in comments has I think the right answer.

After expanding out $$ \begin{align} [P_a,M_{bc}]&=[\partial_a ,x_b\partial_c-x_c\partial_b]\\ &=\partial_a (x_b \partial_c-x_c\partial_b)-(x_b\partial_c-x_c\partial_b)\partial_a \end{align} $$ it is not formally correct to apply that first $\partial_a$ to only the $x_{b,c}$ terms (which should generate those $\eta_{a\{b,c\}}$ terms); if there were a test function in the mix this would miss several terms that apply $\partial_a$ directly to the test function, due to the product rule. So assuming that the product rule still applies here you instead would get $$ \begin{align} [P_a,M_{bc}]&= (\partial_a x_b) \partial_c- (\partial_a x_c)\partial_b + x_b \partial_a \partial_c -x_c \partial_a \partial_b - x_b\partial_c\partial_a+x_c\partial_b\partial_a\\ &=\eta_{ab}\partial_c - \eta_{ac}\partial_b - x_c [\partial_a, \partial_b] - x_b [\partial_c, \partial_a]\\ &=\eta_{ab}\partial_c - \eta_{ac}\partial_b \end{align} $$ and those two terms vanish because we get to deal with only those nice functions for which partials commute.

Especially if these are plain transcriptions of lectures it is possible that he simply got a bit hurried while writing and he knew what the answer was and he didn’t see his mistake so he simply rattled off a quick term to the effect of “whoops those two terms are not supposed to still be there, they vanish” and simply wrote down the right result. I’ve done that before in recitation sections.

$\endgroup$
1
$\begingroup$
  1. The core of OP's problem seems to be the notational ambiguity in what parts/objects that a differential operator acts on, cf. e.g. my Phys.SE answer here. In the operator eq. (4.37), the main point is that the differential operators act to the right on all explicitly & implicitly written objects according to Leibniz product rule. Recall in particular that a differential operator is ultimately supposed to act on a function (in order to produce a result of the operation).

  2. Alternatively. one may use the operator rule $$ [A,BC]~=~ [A,B]C+ B[A,C], $$ and the fundamental commutators $$ [\partial_{\mu},x_{\nu}]~=~\eta_{\mu\nu}, \qquad [\partial_{\mu},\partial_{\nu}]~=~0, $$ to correctly reduce the left-hand side of eq. (4.37). It seems that that was what Das had in mind. (It should be emphasized that the presentation in Das' textbook is correct in its present form.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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