2
$\begingroup$

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!).

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • 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

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.