I'm am sure that I must be missing something very simple, so apologies in advance.

Considering the Lorentz transformation $\Lambda$ of a spinor fields, for the plane-wave solution $u(p)$, I cannot for the life of me agree why

(1) $$ u^s(\Lambda^{-1} {p'}) = \Lambda_{\frac{1}{2}} u^s(p') $$


$$ p' = \Lambda p $$

This is in Peskin & Schroeder, pg 59, just above equation (3.110).

I have tried to get this a dozen times, to no avail.

I know that, for a scalar field, under a Lorentz transformation $\Lambda$ we get, as per Peskin & Schroeder, pg 36, equation (3.2)

$$ \phi(x) \rightarrow \Lambda \phi(x) = \phi'(x) = \phi(\Lambda^{-1} x) $$

This makes sense to me as "the transformed field at the transformed point in spacetime should be the same as the un-transformed field at the untransformed-transformed point in spacetime".

So trying to do that with inverse transformations, now using $\Lambda_{\frac{1}{2}}$ for a spinor plane-wave solution, I get

$$ \Lambda u(p) = u (\Lambda^{-1} p) $$

and applying an inverse transformation would give

$$ \Lambda^{-1} \Lambda u(p) = \Lambda^{-1} u (\Lambda^{-1} p) $$


$$ u(\Lambda^{-1} \Lambda p) = u' (\Lambda^{-1} p) $$


$$ u(\Lambda^{-1} p') = u( [\Lambda^{-1}]^{-1} \Lambda^{-1} p) $$


$$ u(\Lambda^{-1} p') = u(p) $$

that is,

$$ u(p) = u(p) $$

So it's consistent alright, but not of much use!

Can anyone show me what I'm missing to derive equation (1) above.


It is confusing and frankly disgusting to talk about transformations of points in spacetime. This results from abusing that in Minkowski space the tangent space can be identified with spacetime itself; the result is that one confuses coordinates and points in spacetimes. This is a terrible habit and should be avoided.

The proper statement is that under a change of tetrad (orthonormal frame) by $\Lambda$ every field transforms according to some representation of (the covering of) the Lorentz group. In formulas $$\phi(p) \mapsto \phi'(p) = \phi(p)$$ $$A_\mu(p) \mapsto A_\mu'(p) = L^\nu_\mu A_\nu(p)$$ $$\psi(p) \mapsto \psi'(p) = \Lambda_{1/2} \psi(p) $$ I have explicitly the same point on both transformed and untransformed fields because the transformation is pointwise. Here $p$ is a point in spacetime. Unprimed and primed observers disagree on the components of a field at a point, but it is the same point, you should not compare fields at primed points and unprimed points.

Now for general spacetimes coordinate changes and frame changes are entirely independent. In fact you can find a coordinate frame that is also a tetrad if and only if spacetime is flat, that is Minkowski space. Basically this says that there exists a global inertial system iff spacetime is flat. Then in Minkowski space and Minkowski space only, if $x^\mu$ and $y^\mu$ are coordinates for inertial systems, the corresponding tetrads $\{ dx^\mu\}$ and $\{ dy^\mu\}$ are everywhere related by a Lorentz transformation. Conversely you can compose any inertial coordinate function with a Lorentz transformation to get a new inertial coordinate function, $$x^\mu = L^\mu_\nu y^\nu$$ such that $$dx^\mu = L^\mu_{\nu}dx^\nu.$$

Then for the scalar field, $$\phi(p) = \phi((x^{-1})^\mu(x^\mu)) \mapsto \phi'(p) = \phi((y^{-1})^\mu(y^\mu)) = \phi((y^{-1})^\mu({L^{-1}}^\mu_\nu x^\mu))$$ This is where this horrible idea of transforming points comes from. As you can see it's entirely about coordinates. You should not confuse points and coordinates, alas, it is all too easy to do so in Minkowski space.

  • $\begingroup$ Could you please elaborate on the difference between "point" and "coordinate"? E.g. where does $p=x^{-1}x$ come from? $\endgroup$ – ersbygre1 May 27 at 6:45

I struggled for this answer for quite some time. When I turned to 46 of Peskin and Schroeder, I found a plausible explanation. On page 46, Peskin and Schroeder, write the following equation.

$u(p) = exp[-\frac{1}{2} \eta \begin{pmatrix}\sigma^3&0\\0&-\sigma^3\end{pmatrix}] u(p_0)$

The exponential form can be generalized $\Lambda_\frac{1}{2}$ from the following equation no. 3.30:

$\Lambda_\frac{1}{2} = exp(-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu})$

Hence that Equation can be now be rewritten for a general transformation as:

$ u(p) = \Lambda_\frac{1}{2} u(p_0) $

Taking inverse transformations on both sides and taking $p_0 = \Lambda^{-1} p$, we can see the equation come out as:

$ u(\Lambda^{-1} p) = \Lambda_\frac{1}{2}^{-1} u(p)$

The equation $ u^{s}(\Lambda^{-1} p) = \Lambda_\frac{1}{2}^{-1} u^{s}(p)$ is the generalized version of the above equation.


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.