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I'm trying to obtain the commutation relation between the Pauli-Lubanski vector operator and the generators of the Lorentz Group:

$$[W^\mu,P_\sigma]=[\frac{1}{2}\epsilon^{\mu\nu\lambda\rho} P_\nu M_\lambda\rho,P_\sigma]\\ \hspace{2.3cm}= \frac{1}{2}\epsilon^{\mu\nu\lambda\rho}[ P_\nu M_{\lambda\rho},P_\sigma]\\ \hspace{3.9cm}=\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}P_\nu[ -\eta_{\lambda\sigma}P_\rho+\eta_{\rho\sigma}P_\lambda]\\ \hspace{5.cm}=\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\rho\sigma}P_\nu P_\lambda-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho $$

Now, I know a priori that this commutator is zero and I'm trying to change indices accordingly. For instance, in the first term I want to rename the dummy indices $\lambda$ to $\rho$ and vice-versa. This will permute the respective last indices in the Levi-Civita tensor. Since I'm only renaming indices, I assume that I don't have to put a minus sign when doing so: $$=\frac{1}{2}\epsilon^{\mu\nu\rho\lambda}\eta_{\lambda\sigma}P_\nu P_\rho-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho$$

But now, to get the same form for the Levi-Civita tensor in both terms I permute the last two indices of that tensor in the first term, taking into account that it is an antisymmetric tensor: $$=-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho-\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho=-\epsilon^{\mu\nu\lambda\rho}\eta_{\lambda\sigma}P_\nu P_\rho$$ , which is not $0$.

Have I gone wrong somewhere?

If I do the same for the second term instead of the first I get the same result but positive. Since the results must be equal, the only possibility for something of the form $+\text{final result}=-\text{final result}$ is for $\text{final result}=0$. Does it make sense?

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    $\begingroup$ Everything is zero, every term is separately zero, including the thing you said was not zero, because e.g. in $\varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho}$ the $\varepsilon^{\mu \nu \rho \sigma}$ is anti-symmetric in $\nu$ and $\rho$ while $P_{\nu} P_{\rho}$ is symmetric in $\nu$ and $\rho$. $\endgroup$
    – bolbteppa
    Commented Mar 17, 2020 at 20:56
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    $\begingroup$ Note $\varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} = \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} + \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} = \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} - \frac{1}{2} \varepsilon^{\mu \rho \nu \sigma} P_{\nu} P_{\rho} = \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} - \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\rho} P_{\nu} = \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} - \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} = 0$ $\endgroup$
    – bolbteppa
    Commented Mar 17, 2020 at 20:59
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    $\begingroup$ Does this answer your question? Commutator of the Pauli-Lubanski with a vector $P^{\mu}$ $\endgroup$ Commented Mar 17, 2020 at 22:22
  • $\begingroup$ @bolbteppa Consider writing this into an answer. It's more easily searchable, permanent and its quality can be assessed by up/downvotes. Even if you think it is trivial, your comment is a complete answer to the question that OP asked. $\endgroup$
    – MannyC
    Commented Mar 17, 2020 at 23:16
  • $\begingroup$ Thank you very much for your answers @bolbteppa! I still have one more question regarding the renaming of dummy indices. If I have $\epsilon^{\mu\nu\rho\sigma}P_\rho P_\sigma$ and rename $\rho$ to $\sigma$ and $\sigma$ to $\rho$, is the result $\epsilon^{\mu\nu\sigma\rho}P_\sigma P_\rho$ or $-\epsilon^{\mu\nu\sigma\rho}P_\sigma P_\rho$? $\endgroup$
    – RicardoP
    Commented Mar 18, 2020 at 12:07

1 Answer 1

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Every term of the form $$\varepsilon^{\mu \nu \rho \sigma} P_{\rho} P_{\sigma}$$ in your calculation is zero, including the thing you said was not zero, because e.g. in $\varepsilon^{\mu \nu \rho \sigma} P_{\rho} P_{\sigma}$ the $\varepsilon^{\mu \nu \rho \sigma}$ is anti-symmetric in $\rho$ and $\sigma$, $\varepsilon^{\mu \nu \rho \sigma} = - \varepsilon^{\mu \nu \sigma \rho}$, while $P_{\rho} P_{\sigma}$ is symmetric in ν and ρ, $P_{\rho} P_{\sigma} = P_{\sigma} P_{\rho}$.

The simplest way to show that this is zero is to prove that $A = - A$ so that $2A = 0$. For simplicity we consider the two-dimensional analogue $$\varepsilon^{\mu \nu} P_{\mu} P_{\nu}$$ where $\mu, \nu = 0,1$. I want to show that $\varepsilon^{\mu \nu} P_{\mu} P_{\nu} = - \varepsilon^{\mu \nu} P_{\mu} P_{\nu}$. The calculation is as follows: \begin{align} \varepsilon^{\mu \nu} P_{\mu} P_{\nu} &= + \varepsilon^{\nu \mu} P_{\nu} P_{\mu} \ \ (1) \\ &= - \varepsilon^{\mu \nu} P_{\nu} P_{\mu} \ \ (2) \\ &= - \varepsilon^{\mu \nu} P_{\mu} P_{\nu} \ \ (3) \end{align} In line $(1)$ I used the fact that I can just re-label dummy indices whatever way I want, since they are dummy indices, and here I want to have them written in reverse order so I can later invoke anti-symmetry on $\varepsilon^{\mu \nu}$ and symmetry on $P_{\mu} P_{\nu}$. To see very explicitly why I can re-label dummy indices, just write it out: \begin{align} \varepsilon^{\mu \nu} P_{\mu} P_{\nu} &= \varepsilon^{0 \nu} P_{0} P_{\nu} + \varepsilon^{1 \nu} P_{1} P_{\nu} \\ &= (\varepsilon^{00} P_{0} P_{0} + \varepsilon^{0 1} P_{0} P_{1}) + (\varepsilon^{1 0} P_{1} P_{0} + \varepsilon^{1 1} P_{1} P_{1}) \\ &= (\varepsilon^{0\mu} P_{0} P_{\mu}) + (\varepsilon^{1 \mu} P_{1} P_{\mu}) \\ &= \varepsilon^{\nu \mu} P_{\nu} P_{\mu}. \end{align} Note I have done absolutely nothing but write it out so that there are no dummy indices, then collect the terms up with dummy indices again, but now using a different labelling.

In going from $(1)$ to $(2)$ I used the anti-symmetry of $\varepsilon^{\mu \nu}$ and in going from $(2)$ to $(3)$ I used the symmetry of $P_{\mu} P_{\nu}$. Now I have $A = - A$ so that $2A = 0$.

Another way to prove this result is to write it in the form $A = \frac{1}{2} A + \frac{1}{2} A = \frac{1}{2}A - \frac{1}{2} A = 0$ which is just a longer way of doing the above calculation, and clearly uses the above calculation in going from the $+$ to the $-$, but it is often used (e.g. to derive the angular momentum operators in special relativity/classical mechanics etc...) so it's good to be aware: \begin{align} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} &= \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} + \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} \ \ (1) \\ &= \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} - \frac{1}{2} \varepsilon^{\mu \rho \nu \sigma} P_{\nu} P_{\rho} \ \ (2) \\ &= \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} - \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\rho} P_{\nu} \ \ (3) \\ &= \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} - \frac{1}{2} \varepsilon^{\mu \nu \rho \sigma} P_{\nu} P_{\rho} \ \ (4) \\ &= 0. \end{align} In line $(1)$ I know that $\varepsilon^{\mu \nu \rho \sigma}$ is anti-symmetric while $P_{\nu} P_{\rho}$ is symmetric so that the whole thing is immediately zero, and I want to show this explicitly by turning it into something like $A = \frac{1}{2} A + \frac{1}{2} A = \frac{1}{2} A - \frac{1}{2} A = 0$, so I introduce the $1/2$ just to get two copies of it which I expect will cancel one another. In going $(1)$ to to $(2)$ I just used the anti-symmetry of $\varepsilon^{\mu \nu \rho \sigma}$ to write one of them with a $-$ sign. In going from $(2)$ to $(3)$ I then re-labelled the dummy indices so that I would have $\varepsilon^{\mu \nu \rho \sigma}$ in both terms, In going from $(3)$ to $(4)$ I then used commutativity of $P_{\mu}$ and $P_{\nu}$. Note $(4)$ is now in the form $A = \frac{1}{2}A - \frac{1}{2}A = 0$.

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