I have a simple question, I probably misunderstood some minor thing, because this should be trivial. My question is regarding the book Peskin & Schroeder 1995 Introduction to Quantum Field Theory, at page 59.

Question briefly

It seems to me, that there is a contradiction between (3.106) and (3.110). Let's say in coordinate system $A$ I have a vector $p$, and in coordinate system $B$ the same vector looks like $\Lambda p$, then a field should be transformed as $U(\Lambda)\psi(x)U^{-1}(\Lambda) = R(\Lambda) \psi(\Lambda^{-1}x)$, not as in the book $R(\Lambda^{-1}) \psi(\Lambda x)$.

Question detailed

In Peskin & Schroeder 1995 Introduction to Quantum Field Theory, at page 59 (3.106), it says, that $$ \tag{3.106} |\mathbf{p}, s \rangle = \sqrt{2 E_\mathbf{p}} a_\mathbf{p}^{s \dagger} |0\rangle. $$

Now, according to the book, if we look this in an other coordinate system, this state will look like $|\mathbf{\Lambda}\mathbf{p}, s \rangle$. And so, we can derive, that

$$ \tag{3.109} U(\Lambda) a_\mathbf{p}^{s} U^{-1}(\Lambda) = \frac{E_{\mathbf{\Lambda}\mathbf{p}}} {E_{\mathbf{p}}} a_{\mathbf{\Lambda}a_\mathbf{p}}^{s}. $$

Now, our convention here was that that any vector will look like in that other system as $v' = \Lambda v$. Let's say I have a scalar field $\phi(x)$, this looks like $\phi(\Lambda^{-1}x)$ in the other coordinate system. The same applies to the fermion field, this should be transformed as

$$ U(\Lambda)\psi(x)U^{-1}(\Lambda) = R(\Lambda) \psi(\Lambda^{-1}x), $$

but in the book it's quiet the opposite, it says there,

$$ \tag{3.110} U(\Lambda)\psi(x)U^{-1}(\Lambda) = R(\Lambda^{-1}) \psi(\Lambda x). $$

What am I missing?

Related: Lorentz transformations for scalar fields in QFT --- Peskin and Schroder

  • $\begingroup$ The point is P&S defined $\Lambda | p \rangle = | \Lambda p \rangle$. If they did it as $|\Lambda p \rangle = | \Lambda^{-1} p \rangle$, then your assumption is correct. Both of the two ways work correctly and it is just a conventional problem I think. $\endgroup$
    – Keyflux
    Apr 7, 2022 at 17:18

1 Answer 1


One can check these kind of formulae by seeing if they respect the goup property. If $$ U(\Lambda)_1\psi U(\Lambda_1)^{-1}= R(\Lambda_1^{-1})\psi $$ then $$ U(\Lambda_2)U(\Lambda)_1\psi U(\Lambda_1)^{-1}U(\Lambda_2)^{-1}= U(\Lambda_2)R(\Lambda_1^{-1})\psi U(\Lambda_2)^{-1}\\ = R(\Lambda_1^{-1}) U(\Lambda_2) \psi U(\Lambda_2)^{-1}\\ \hbox{(operators $U$ commute with arrays of numbers $R$)}\\ = R(\Lambda_1^{-1})R(\Lambda_2^{-1})\psi\\ = R(\Lambda_1^{-1}\Lambda_2^{-1})\psi\\ = R((\Lambda_2 \Lambda_1)^{-1})\psi $$ which is coincides with the action of $$ U(\Lambda_2)U(\Lambda_1)=U(\Lambda_2\Lambda_1). $$ So P&S are correct. Your formula would lead to the R"s being backwards. (I did not look at the \Lambda x), but you can check this too.

Note that the wavefunction corresponding to a state $|{\bf p}\rangle$ is $$ \phi_{\bf p}(x) = \langle 0|\psi(x)|{\bf p}\rangle $$ so that the wavefunction for $U(\Lambda) |{\bf p}\rangle$ is $$ \langle 0|\psi(x)U(\Lambda)|{\bf p}\rangle\\ = \langle 0|U(\Lambda)U(\Lambda)^{-1}\psi(x) U(\Lambda)|{\bf p}\rangle\\ = \langle 0|U(\Lambda)^{-1}\psi(x)U(\Lambda)|{\bf p}\rangle $$

which has the $\Lambda$'s inverted as you want them to be.

  • $\begingroup$ Aha, this is interesting. I know, that P&S are correct, I just don't really understand why. I would like to reconcile the two operations that I presented. It doesn't have to be QFT. A classical vector field transforms as $f(x) \rightarrow \Lambda f(\Lambda^{-1} x)$, if for example a solution to a plane wave solution transforms as $exp(-ixp) \rightarrow exp(-i \Lambda p x)$. I will think about it more, you answer is an other good insight. $\endgroup$
    – Gabor
    Jul 20, 2020 at 11:15
  • $\begingroup$ I've added a bit more that might help. $\endgroup$
    – mike stone
    Jul 20, 2020 at 13:32

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