# Scalar products in the spinor helicity formalism

+ 2 like - 0 dislike
93 views

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!

This post imported from StackExchange Physics at 2014-12-09 15:07 (UTC), posted by SE-user soliton
retagged Feb 1, 2015

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. $$X_{B\dot{A}}=\epsilon_{\dot{A}\dot{C}}\epsilon_{BD}X^{\dot{C}D} \ .$$
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysic$\varnothing$OverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.