I am reading Schwartz's "QFT and the standard model". On pg 13 he gives the Lorentz transform of a rotation around the x-axis:

$ \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos \theta _x & \sin \theta _x \\ 0 & 0 & -\sin \theta _x & \cos \theta _x \\ \end{array} \right) $

For a boost along the x-axis he gives:

$ \left( \begin{array}{cccc} \cosh \beta _x & \sinh \beta _x & 0 & 0 \\ \sinh \beta _x & \cosh \beta _x & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $

I believe this corresponds to the Lorentz transform in the form $\Lambda^\alpha{}_\beta$. I believe the first index is the row and the second is the column. But at the bottom of the pg he has $ V^\mu=\Lambda_\nu^\mu V^\nu. $ as though the order of the indices on $\Lambda$ doesn't matter. But surely they do matter as $\Lambda^\alpha{}_\beta\not=\Lambda_\beta{}^\alpha$ in general?

  • 1
    $\begingroup$ This question seems to be about a typographic detail in a specific book. $\endgroup$
    – zzz
    Commented Jun 6, 2015 at 2:51
  • $\begingroup$ @bechira I think it would be interesting to know as its otherwise such a good qft textbook. I think the answer is by $\Lambda^\mu_\nu$ he means $\Lambda^\mu{}_\nu$. $\endgroup$
    – Virgo
    Commented Jun 6, 2015 at 3:59

1 Answer 1


Short answer

His notation is not ambiguous because the expression $$V^{'\mu} \equiv \Lambda^\mu_\nu V^\nu$$ can only mean sum along the $\nu$ component. Since $\Lambda$ is a representation of the Lorentz group, it is a linear operator, hence it can only act on a vector by the usual way that matrices act on vectors. Hence the above is unambiguous.

Longer answer

I'll explain why in general this notation is unambiguous.

The convention is the following: the upper index denotes a vector index and the lower index denotes a dual vector index and. That is, for a general rank $(p, q)$ tensor, one would usually write

$$ T^{\mu_1...\mu_p}_{\nu_1...\nu_q} \partial_{\mu_1} \otimes ... \otimes \partial_{\mu_p} \otimes dx^{\nu_1}... dx^{\nu_q} $$

where $\{\partial_\mu\}$ denotes a basis for your vector space and ${dx^\nu}$ denotes a basis for its dual vector space. (The notation I chose is from differential geometry, which is certainly useful when you define a field theory in over a general spacetime manifold).

As a special case, a linear transformation is a rank $(1,1)$ tensor. Namely $\Lambda^\mu_\nu$ unambiguously denotes components of the tensor:

$$\Lambda^\mu_\nu ~\partial_\mu \otimes dx^\nu$$

Acting on a vector $V \equiv V^\tau \partial_\tau$, we get:

$$\Lambda[V] \\ = \Lambda^\mu_\nu ~\partial_\mu \otimes dx^\nu [V^\tau \partial_\tau] \\ = \Lambda^\mu_\nu V^\tau (\partial_\tau dx^\nu) ~\partial_\mu \\ = \Lambda^\mu_\nu V^\tau \delta_\tau^\nu ~\partial_\mu \\ = \Lambda^\mu_\nu V^\nu ~\partial_\mu $$

Observe no horizontal padding in the indices are needed to make these manipulations unambiguous.

  • $\begingroup$ but $\Lambda^\mu_{\ \nu}$ is not a tensor, right? $\endgroup$
    – Nahc
    Commented Jan 21, 2017 at 16:39
  • $\begingroup$ @Chan Why not? It's a well-defined multi linear operator in any basis. $\endgroup$
    – zzz
    Commented Jan 22, 2017 at 19:51

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