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

1. Jul 6, 2010

### bjnartowt

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
$$\begin{array}{c} [{\partial _\mu },{\gamma _\nu }] \equiv 0 \\ {\gamma _0}^2 = - {\gamma _i}^2 \equiv {\bf{I}} \\ \end{array}$$

3. The attempt at a solution

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

Last edited: Jul 6, 2010
2. Jul 6, 2010

3. Jul 6, 2010

### bjnartowt

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

4. Jul 7, 2010

### 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.

5. Jul 8, 2010

### bjnartowt

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?

6. Jul 8, 2010

### 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.