I have the following helicity spinor:

$$ u_R=\sqrt{E} \begin{pmatrix} c \\ se^{i\phi} \\ c \\ se^{i\phi} \end{pmatrix} $$

We take $s=\sin\theta/2$ and $c=\cos\theta/2$. Also, $E$ and $\phi$ are real.

What would $\bar{u}_R$ correspond to? This isn't a homework question, but it would help me show something. I just want to know what this bar means? Is it a case of flipping signs of $i\phi$ to $-i\phi$, or is there something more?

Are there any rules for the conversion. Please keep your answer as simple as possible. I'm a bit of a quantum mechanics n00b (you could probably tell me my level of question!).

  • $\begingroup$ Could you show me an example method to calculate the hermitian of something like $u_{R}$ above? $\endgroup$ – DarthPlagueis Nov 22 '15 at 19:32
  • $\begingroup$ Yes I'm familiar with the gamma-matrices. So you do you evaluate complex conjugation and transpose? $\endgroup$ – DarthPlagueis Nov 22 '15 at 19:39
  • $\begingroup$ Or rather, sorry, how would I work out the complex conjugate of the above expression. That is what I think I need to do...so, what would $u_{R}^{*}$ become? $\endgroup$ – DarthPlagueis Nov 22 '15 at 19:40
  • $\begingroup$ $s=\sin{\frac{\theta}{2}}$ and $c=\cos{\frac{\theta}{2}}$...what would c* and s* become? $\endgroup$ – DarthPlagueis Nov 22 '15 at 19:45
  • $\begingroup$ Are we looking at $c^{*},s^{*}=c,s$ purely because there's no complex part to the expression... $\endgroup$ – DarthPlagueis Nov 22 '15 at 19:50

Let $a$ be any spinor; then, by definition $\bar a\equiv a^\dagger \gamma^0$, where $\dagger$ stands for hermitian conjugation (transpose+complex conjugation: $a^\dagger=(a^T)^*$), and $\gamma^0$ is one of the Dirac matrices.

With this in mind, the steps are as follows:

first, we transpose the spinor: $$ u^T=\sqrt{E}\begin{pmatrix} c & s\mathrm e^{i\phi} & c & s\mathrm e^{i\phi} \end{pmatrix} $$

next, complex conjugation (real quantities dont change, and $i\to-i$): $$ u^\dagger=(u^T)^*=\sqrt{E}\begin{pmatrix} c & s\mathrm e^{-i\phi} & c & s\mathrm e^{-i\phi} \end{pmatrix} $$

finally, the zero gamma matrix is $$ \gamma^0=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{pmatrix} $$

thus $$ \bar u=u^\dagger\gamma^0=\sqrt{E}\begin{pmatrix} c & s\mathrm e^{-i\phi} & -c & -s\mathrm e^{-i\phi} \end{pmatrix} $$

Hope this answers your question.

  • 2
    $\begingroup$ Thank you very much for this answer, I believe I am begin to understand the mathematics of this. $\endgroup$ – DarthPlagueis Nov 22 '15 at 20:07

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