I am going through chapter 34 of Srednicki and having doubts about Weyl Spinors. He defines \begin{equation}[\psi_a(x)]^\dagger=\psi^\dagger_{\dot{a}}(x)\tag{1}\end{equation} where $\psi_a(x)$ is a left handed Weyl Spinor and $\psi^\dagger_{\dot{a}}(x)$ is a right handed Weyl spinor. In my head, I am thinking of $\psi_a(x)$ as a column matrix that is complex conjugated and transposed to get the $\psi^\dagger_{\dot{a}}(x)$ which is a row matrix. Further he defines $$\psi^a=\epsilon^{ab}\psi_b\tag{2}$$ and says that a similar definition is true for $$\psi^{\dagger\dot{a}}=\epsilon^{\dot{a}\dot{b}}\psi_{\dot{b}}^{\dagger}\tag{3}.$$ But this shouldn't be true as $\psi_{\dot{b}}^{\dagger}$ is a row matrix and the actual expression should be $$\psi^{\dagger\dot{a}}=\psi_{\dot{b}}^{\dagger}\epsilon^{\dot{b}\dot{a}}.\tag{4}$$ I arrive at this by taking the adjoint of the equation 2 as \begin{aligned} \psi^{\dagger\dot{a}}&= (\epsilon^{ab}\psi_b)^\dagger\\ &=\psi^{\dagger}_{\dot{b}}(\epsilon^{ab})^\dagger\\ &=\psi^{\dagger}_{\dot{b}}\epsilon^{\dot{b}\dot{a}} \end{aligned} where daggering the product of two matrices reverses their order, daggering of $\epsilon$ switches the indices and puts a dot on the indices but as all the components of $\epsilon$ itself are real, no complex conjugation symbol of that. So which of them is correct? Equation 3 or equation 4?

  • $\begingroup$ If I remember correctly the order does not matter, since what you have there is merely some matrix elements labelled with $\dot{a}$ or $\dot{a}\dot{b}$ etc... The order indeed matters if you write down the corresponding matrices, for instance if $\epsilon$ is the matrix corresponding to the levi-civita symbol, then $\epsilon=\pmatrix{0 & 1 \\ -1 & 0}$ (or maybe up to minus signs, I do not remember). But it's $(\dot{a},\dot{b})=(0,1)$ element, for instance commutes with the $\dot{b}$ element of $\psi^{\dagger}$, so Eq. (3) is correct as it stands $\endgroup$
    – schris38
    Sep 11, 2023 at 16:24
  • 1
    $\begingroup$ Thanks this is something that I missed thinking. This helps. One thing that I would like to ask here is how do we interpret the $\psi^\dagger_{\dot{a}}$ when we add it to the $\psi_a$ field and make it into a majorana field? $\psi^\dagger_{\dot{a}}$ is no longer a row matrix, right? $\endgroup$ Sep 11, 2023 at 19:47
  • $\begingroup$ I do not remember that, sorry... Let someone else take the lead :) $\endgroup$
    – schris38
    Sep 12, 2023 at 4:17

1 Answer 1


Let's say you have already quantized the theory, then in $(1)$ you have to think about $[\psi_a(x)]^\dagger$ as the Hermitian adjoint operator of $\psi_a(x)$ acting on the Hilbert space, not as the conjugate transpose matrix of the column vector $$ \begin{pmatrix} \psi_1(x) \\ \psi_2(x) \end{pmatrix}. $$ So eq. $(1)$ defines 2 objects, $(\psi^\dagger)_{\dot{1}}$ and $(\psi^\dagger)_{\dot{2}}$, separately (but obviously they transform together under Poincaré action), only after that you can rearrange these 2 things as you like, for example in a column or row vector.

Then $(4)$ is not more true, because the $\epsilon^{ab}$ are just coefficients you use to obtain the objects $\psi^{(\cdot)}$ with upper indices from the ones $\psi_{(\cdot)}$ with lower indices. More precisely, since taking the Hermitian adjoint is an antilinear operation and $\epsilon$ is real, the correct computation leads to what we expect: $$ (\psi^a)^\dagger = (\epsilon^{ab} \psi_b)^\dagger = (\epsilon^{ab})^*(\psi_b)^\dagger = \epsilon^{\dot{a}\dot{b}}(\psi^\dagger)_{\dot{b}} = (\psi^\dagger)^{\dot{a}} $$ so using the conjugate transpose matrix $(\epsilon^{ab})^\dagger$ is wrong because here $\dagger$ really means taking the Hermitian adjoint operator.

Finally you can rearrange $\psi_{(\cdot)}$ and $(\psi^\dagger)^{(\cdot)}$ in the Majorana spinor: $$ \Psi = \begin{pmatrix} \psi_1 \\ \psi_2 \\ (\psi^\dagger)^{\dot{1}} \\ (\psi^\dagger)^{\dot{2}} \end{pmatrix} $$ which simply means constructing a column vector whose entries are the spinor components you defined earlier, here there's no need to represent $\psi$ or $\psi^\dagger$ separately as row or column matrices.


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