Spinor rotation

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  • Thread starter gentsagree
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  • #1
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1

Main Question or Discussion Point

Hi,

I am confused on a very basic fact. I can write [itex]\xi = (\xi_{1}, \xi_{2}) [/itex] and a spin rotation matrix as

[tex]
U =
\left( \begin{array}{ccc}
e^{-\frac{i}{2}\phi} & 0 \\
0 & e^{\frac{i}{2}\phi}
\end{array} \right)
[/tex]

A spinor rotates under a [itex]2\pi[/itex] rotation as

[tex]
\xi ' =
\left( \begin{array}{ccc}
e^{-i\pi} & 0 \\
0 & e^{i\pi}
\end{array} \right)
\left( \begin{array}{c}
\xi_{1} \\
\xi_{2}
\end{array} \right)
=
\left( \begin{array}{ccc}
-\xi_{1} \\
\xi_{2}
\end{array} \right)
[/tex]

which is [itex](-\xi_{1}, \xi_{2})[/itex], and not [itex]-\xi[/itex], so only one component changes sign. Is this correct?
 

Answers and Replies

  • #2
vanhees71
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How do you come to this conclusion? Since ##\exp(\mathrm{i} \pi)=\exp(-\mathrm{i} \pi)=-1## your rotation by ##2 \pi## leads to ##\hat{U} \xi=-\xi## as it should be.
 
  • #3
96
1
Ahah, ok, I should not try to do physics today.
 

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