Trying to Make a (p[slash])^2 operator - is this right?

[bjnartowt]

Homework Statement

Find out what p[slash]p[slash] is (Feynman slash-notation), because Maple doesn't like it when you feed it p[slash]p[slash], and let it uber-"FOIL" out the (four non-commuting terms) x (four non-commuting terms), where the "x" denotes plain Jane matrix-multiplication.

Homework Equations

$$\begin{array}{c}<br /> [{\partial _\mu },{\gamma _\nu }] \equiv 0 \\ <br /> {\gamma _0}^2 = - {\gamma _i}^2 \equiv {\bf{I}} \\ <br /> \end{array}$$

The Attempt at a Solution

$$\begin{array}{c}<br /> {p_{{\rm{slash}}}}{p_{{\rm{slash}}}} = - ({\gamma _\mu }{\partial ^\mu })({\gamma _\nu }{\partial ^\nu }) \\ <br /> = - \left( {{\gamma _0}{\partial ^0} - \vec \gamma \bullet \vec \partial } \right)\left( {{\gamma _0}{\partial ^0} - \vec \gamma \bullet \vec \partial } \right) \\ <br /> = - \left( {{\gamma _0}^2{{({\partial ^0})}^2} + (\vec \gamma \bullet \vec \gamma ){\nabla ^2} - 2({\gamma _0}{\partial ^0})(\vec \gamma \bullet \vec \partial )} \right) \\ <br /> {p_{{\rm{slash}}}}{p_{{\rm{slash}}}} = - \left( {({\bf{I}}){{({\partial ^0})}^2} + ( - 3{\bf{I}}){\nabla ^2} - 2({\gamma _0}{\partial ^0})(\vec \gamma \bullet \vec \partial )} \right) \\ <br /> \end{array}$$

 

[bjnartowt]

sorry! LaTeX is bad.

 

[bjnartowt]

sorry! LaTeX is bad.

Um...writing text below the thread-start actually fixed the problem

 

[chrispb]

No, it is not right. To get the right answer, begin by noting that pslash^2 is symmetric in d^mu and d^nu.

There is one more relevant equation that the gammas satisfy that you need to solve this, as well.

 

[bjnartowt]

pslash^2 is symmetric in d^mu and d^nu.

Hi chrispb, thank you for stopping to help us answer this question.

May I ask: does "symmetric" mean pslash^2 is equal to its own transpose?

 

[chrispb]

No, I just mean $${\gamma _\mu }{\partial ^\mu }{\gamma _\nu }{\partial ^\nu }={\gamma _\mu }{\gamma _\nu }{\partial ^\mu }{\partial ^\nu }={\gamma _\mu }{\gamma _\nu }{\partial ^\nu }{\partial ^\mu }$$.

What this means is you can write $${\partial ^\mu }{\partial ^\nu }=\frac{1}{2}({\partial ^\mu }{\partial ^\nu }+{\partial ^\nu }{\partial ^\mu })$$ free of charge.