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  Scalar products in the spinor helicity formalism

+ 2 like - 0 dislike

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
asked Dec 6, 2014 in Theoretical Physics by soliton (110 points) [ no revision ]
retagged Feb 1, 2015 by dimension10

1 Answer

+ 3 like - 0 dislike

Soliton is correct, there is a factor of two coming in when the indices are raised or lowered with the Levi-Civita tensors and then a contraction is made. Here is how it appears.

Consider a four-vector as a Lorentz spinor, \begin{equation} 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] \ . \end{equation} 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, \begin{equation} X^{\dot{A}B}\frac{\partial\det(X)}{\partial X^{\dot{A}B}}=X^{\dot{A}B}X_{B\dot{A}}=2\det(X) \end{equation} 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. \begin{equation} X_{B\dot{A}}=\epsilon_{\dot{A}\dot{C}}\epsilon_{BD}X^{\dot{C}D} \ . \end{equation}

This post imported from StackExchange Physics at 2014-12-09 15:07 (UTC), posted by SE-user Stephen Blake
answered Dec 6, 2014 by Stephen Blake (70 points) [ no revision ]

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