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? $\endgroup$ Nov 22, 2015 at 19:32
  • $\begingroup$ Yes I'm familiar with the gamma-matrices. So you do you evaluate complex conjugation and transpose? $\endgroup$ Nov 22, 2015 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? $\endgroup$ Nov 22, 2015 at 19:40
  • $\begingroup$ $s=\sin{\frac{\theta}{2}}$ and $c=\cos{\frac{\theta}{2}}$...what would c* and s* become? $\endgroup$ Nov 22, 2015 at 19:45
  • $\begingroup$ Are we looking at $c^{*},s^{*}=c,s$ purely because there's no complex part to the expression... $\endgroup$ Nov 22, 2015 at 19:50

1 Answer 1


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. $\endgroup$ Nov 22, 2015 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.