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Based on that notes (equation 54): https://warwick.ac.uk/fac/sci/physics/staff/academic/boyd/stuff/dirac.pdf

I was reading about the wave functions and I have a question about the notation. You can define the wave function like:

$\psi(p) = e^{-ix_{\mu}p^{\mu}}u(p)$

My question is: Why there is no dependence in the $\mu$ on the left hand side? The dependence in the wave function is only taking care $u(p)$.

UPDATE:

My question is because I am doing something more advanced and I have this same question, but for the wave function related to the matrix elements and fermions representation:

I have the follow definition for the pseudoscalar current:

$P^{ij}_5 (\vec{x}, t) = \sum_{y} \bar{\psi}^{i} (\vec{x}+\vec{y}, t) \gamma^5 \psi^{j} (\vec{y}, t)$

and then the wave function is defined as:

$\Psi_{P}(\vec{x}) = <\Omega|P^{ij}_5(\vec{x}, 0)| P>$

i and j are flavour indices.

then the same question: why there is no dependence in the LHS of the equation? Should I apply some theory or some convention?

Paper with the information: http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/22/036/22036311.pdf

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  • $\begingroup$ The IJ in the bra/ket are free indices. They should be carried in the PSI in the LHS, too. $\endgroup$
    – DanielC
    Commented Jan 31, 2018 at 16:09
  • $\begingroup$ @DanielC thanks. Also the index of the gamma matrix (dirac matrix) in this case \gamma_5 or for the vector case \gamma_\mu should also appear in the LHS? I mean do you think people hide it to be more easy to write or there is another explanation? Also last question, I don't know if you will be able to answer but for this wave function they say it is "gauge fixed". Do you know what it means? Thanks a lot $\endgroup$
    – Gabriela
    Commented Jan 31, 2018 at 16:48
  • $\begingroup$ Might be useful to include context of the $\Psi_P(x)$ equation (author, publication or text title, etc) $\endgroup$
    – Kyle Kanos
    Commented Feb 1, 2018 at 0:39
  • $\begingroup$ @KyleKanos this is the link for the paper: iaea.org/inis/collection/NCLCollectionStore/_Public/22/036/… $\endgroup$
    – Gabriela
    Commented Feb 1, 2018 at 10:12

2 Answers 2

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It seems that the notes you are using have used Einstein's summation convention - this means that when you have an index appearing both as a subscript and a superscript, it is summed over, i.e. $$x_\mu p^\mu := \Sigma_\mu x_\mu p^\mu $$ Thus since the index is summed over, there is no dependence on the LHS.

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  • $\begingroup$ Thanks for your answer, but I have update the question, considering my goal. $\endgroup$
    – Gabriela
    Commented Jan 31, 2018 at 14:42
  • $\begingroup$ It seems to me that the Pij are matrix elements, so the bra should provide (say) the other i term over which to sum and the ket should provide the j. So again since you are summing over then there is no dependence on the LHS. $\endgroup$
    – Sreekar Voleti
    Commented Feb 1, 2018 at 1:37
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The equation above makes use of the Einstein's summation convention. According to this convention, all the repeated indices get summed over. Eq: $$x^\mu y_\mu \equiv \sum_{\mu=0} x^0y_0 + x^1y_1 + ...$$ Therefore, there is no dependence on $\mu$ on the LHS. The index $\mu$ is called a dummy index.

Generally, Greek symbols are used to denote the space-time variables ($\mu = 0,1,2,3$) with appropriate metric definition and Latin symbols for space variables ($i=1,2,3$). However, many people use it differently.

The upper index $x^{\mu}$ refers to the contravariant quantity and a lower index $x_{\nu}$ refers to a covariant quantity with a metric connecting the two as $$x^{\mu}=g^{\mu\nu}x_{\nu}$$

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  • $\begingroup$ thanks for your answer. I have update the question. But thanks a lot, because I think I can get some ideas for the next step. $\endgroup$
    – Gabriela
    Commented Jan 31, 2018 at 14:42
  • $\begingroup$ I guess, you refer again to the indices $i$ and $j$. In this (second) example Einstein's summation convention does not apply, because there are no repeated indices. Actually from your description it is not clear what $i$ and $j$ mean. So please tell us what they mean, respectively to what they refer. $\endgroup$
    – Frederic Thomas
    Commented Jan 31, 2018 at 15:14
  • $\begingroup$ @FredericThomas we have i and j different. And they are the flavour indices. $\endgroup$
    – Gabriela
    Commented Jan 31, 2018 at 16:17

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