# Scalar products in the spinor helicity formalism

In A. Zee's book Quantum Field Theory in a Nutshell (2nd edition), Chapter N.2, page 486, the momentum $p$ is written as a $2\times 2$ matrix:

$$p_{\alpha\dot{\alpha}} = p_{\mu} (\sigma^{\mu})_{\alpha\dot{\alpha}} = (p_0I - p_i\sigma^i)_{\alpha\dot{\alpha}} = \begin{pmatrix} (p^0 - p^3) && -(p^1 - ip^2) \\ -(p^1 + ip^2) && -(p^0 + p^3) \end{pmatrix}_{\alpha\dot{\alpha}}$$ Given two vectors $p$ and $q$, their scalar product is given by $$p\cdot q = \varepsilon^{\alpha\beta}\varepsilon^{\dot{\alpha}\dot{\beta}} p_{\alpha\dot{\alpha}}q_{\beta\dot{\beta}}$$ In E. Witten's article arXiv:hep-th/0312171, the same formula can also be found above Eq.(2.7) in page 5. However, I checked explicitly that it might be not valid $$\begin{split} \varepsilon^{\alpha\beta}\varepsilon^{\dot{\alpha}\dot{\beta}} p_{\alpha\dot{\alpha}}q_{\beta\dot{\beta}} &= \varepsilon^{12}\varepsilon^{\dot{1}\dot{2}}p_{1\dot{1}}q_{2\dot{2}} + \varepsilon^{12}\varepsilon^{\dot{2}\dot{1}}p_{1\dot{2}}q_{2\dot{1}} + \varepsilon^{21}\varepsilon^{\dot{1}\dot{2}}p_{2\dot{1}}q_{1\dot{2}} + \varepsilon^{21}\varepsilon^{\dot{2}\dot{1}}p_{2\dot{2}}q_{1\dot{1}} \\ &= 2(p^0q^0 - p^1q^1 - p^2q^2 - p^3q^3) \end{split}$$ which differs with the above formula in a factor of $2$. This is only a simple exercise, but I don't know whether they use a different summation convention. And why there is a factor of 2 difference? Thanks a lot!

• So... just to be clear, what exactly are you asking? Why there is a factor of 2 difference? (Phrases like "Any clarifications will be appreciated" usually make a question less clear.) Dec 6 '14 at 6:38
• @DavidZ Yes. Why there is a factor of 2 difference and whether they use a different summation convention or not Dec 6 '14 at 6:41
• Oh, I meant to suggest you edit the question to clarify that. Dec 6 '14 at 6:43

Consider a four-vector as a Lorentz spinor, $$x^{\mu}\rightarrow X^{\dot{A}B}= \left[ \begin{array}{cc} x^{0}+x^{3} & x^{1}-ix^{2} \\ x^{1}+ix^{2} & x^{0}-x^{3} \end{array} \right] \ .$$ The natural way to get an invariant is to take the determinant and this agrees with the scalar product of a four vector with no factors of two coming in $x^{\mu}x_{\mu}=\det(X)$. The determinant is a homogeneous polynomial of degree two in the variables $X^{\dot{A}B}$ so by Euler's theorem on homogeneous functions, $$X^{\dot{A}B}\frac{\partial\det(X)}{\partial X^{\dot{A}B}}=X^{\dot{A}B}X_{B\dot{A}}=2\det(X)$$ where the spinor with lowered indices $X_{B\dot{A}}$ is the cofactor in the expansion of the determinant. This is how the factor of two appears. By differentiating the determinant, the cofactor is given by the standard formula for lowering the spinor indices. $$X_{B\dot{A}}=\epsilon_{\dot{A}\dot{C}}\epsilon_{BD}X^{\dot{C}D} \ .$$