0
$\begingroup$

Let us begin in a two-dimensional Euclidean plane. The vector is e.g. $\vec{V}(x,y)$ It is often useful – but in this case, it's just a mathematical trick that doesn't make the complex numbers "fundamental" – to combine the components into a complex number, $z=x+iy$. The two-dimensional rotations in $SO(2)$ are fully determined by the angle $δ$. And the matrix acting on the vector $\vec{V}(x,y)$ $M= \begin{bmatrix}+\cosδ & −\sinδ\\+\sinδ & +\cosδ\end{bmatrix}$ may be fully replaced by the complex coefficient $e^{iδ}$ that multiplies our complex coordinate $z$. Instead of $z$, however, we could have dealt with its power $z^p$. The rotation could be described by the complex transformation $z^p→z^{′p},z^{′p}=e^{ipδ}z^p$. In this context, the complex number $e^{ipδ}$ plays the role of the "transformation matrix" that rescales the one and only component of our tensor-spinor-whatever, $z^p$. In our overly trivial two-dimensional context, the coefficient or exponent $p$ may be anything you want. But the value $p=1/2$ may be identified with the spinors in two dimensions. The spinor in two dimensions may be represented as $\sqrt{z}$ where $z=x+iy$ encodes a vector; so the spinor $z^{1/2}$ is literally the square root of a vector (translated to a complex number) in this case.

So I got all this from a blog that tries to explain spinors which unfortunately I don't remember. I don't understand why the $p$ in $e^{ipδ}$ has to be there since any complex number can be rotated without the need for that coefficient. Is there some rule for rotating complex numbers that are raised to higher powers $p=2,...,n$ or am i missing something?

$\endgroup$
2
$\begingroup$

If $z \mapsto \mathrm{e}^{\mathrm{i}\delta}$, then $z^p \mapsto (\mathrm{e}^{\mathrm{i}\delta} z)^p = \mathrm{e}^{\mathrm{i}\delta p} z^p$.

What we are actually looking at is that the "transformation of the $p$-th power of $z$" is a way to speak of the representation of the circle group $\mathrm{U}(1)$ labeled by $p$.

$\endgroup$
  • $\begingroup$ Is $p$ a generator like in the $SO$ groups or does it change with the powers? $\endgroup$ – pkjag Aug 9 '14 at 20:53
  • 1
    $\begingroup$ @pkjag: The generator of $\mathrm{U}(1) \equiv \mathrm{SO}(2)$ is $\mathrm{i}$, the $p$ is a number labeling the representation by $\mathrm{i} \mapsto p\mathrm{i}$. $\endgroup$ – ACuriousMind Aug 9 '14 at 21:01
  • $\begingroup$ I still don't understand how you get $i$ from differentiating the $M$ matrix components and evaluating the derivatives at zero? $\endgroup$ – pkjag Aug 9 '14 at 21:11
  • 1
    $\begingroup$ @pkjag: $\partial_\delta \mathrm{e}^{\mathrm{i}\delta} = \mathrm{i}\mathrm{e}^{\mathrm{i}\delta} = \mathrm{i}$ @ $\delta = 0$ for the $\mathrm{U}(1)$. For the matrix, i.e. $\mathrm{SO}(2)$, use that $\mathrm{i} = \sin(0) + \mathrm{i}\cos(0)$. $\endgroup$ – ACuriousMind Aug 9 '14 at 21:24

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.