# Can anybody confirm the following Feynman slash identity?

1. Oct 29, 2015

### moss

I want to compute the following when the 4-vectors are already given i.e $$x^{\mu},y^{\mu}$$ are given and are orthonormal ( x, y are complex vectors);

\begin{eqnarray}

\left(/\negmedspace\negmedspace x/\negmedspace\negmedspace y\right)^{2} & = & /\negmedspace\negmedspace x/\negmedspace\negmedspace y/\negmedspace\negmedspace x/\negmedspace\negmedspace y\\

& = & -/\negmedspace\negmedspace x/\negmedspace\negmedspace y/\negmedspace\negmedspace y/\negmedspace\negmedspace x\\

& = & -/\negmedspace\negmedspace x/\negmedspace\negmedspace x\\

& = & -1\\

\Rightarrow/\negmedspace\negmedspace x/\negmedspace\negmedspace y & = & i

\end{eqnarray}

Is the above computation right?
I only want to compute the last equation and want to confirm the above.

Last edited: Oct 29, 2015
2. Oct 29, 2015

### fzero

Apart from the typo in the first line, you've shown that $({\not x}{\not y})^2 = - I_4$, the identity matrix. However, you cannot conclude that ${\not x}{\not y} = i I_4$ from this, since there are an infinite number of matrices $M$ that satisfy $M^2 = - I_4$.

I don't think ${\not x}{\not y}$ simplifies very much, though it's trace does.

3. Oct 29, 2015

### moss

I have corrected the typo.
Lets say that I dont know the square of x and y and that x.x =x^2 and y.y = y^2.
I use the fact that a-slash x a-slash = a^2. what would I get by the above computation?

4. Oct 29, 2015

### fzero

From the anticommutation rule for Dirac matrices we have
$${\not x}{\not y} + {\not y}{\not x} = 2 x\cdot y I_4,$$
so
$$({\not x}{\not y})^2 = - x^2 y^2 I_4+ 2 (x\cdot y) {\not x}{\not y}.$$
Since ${\not x}{\not y}$ doesn't simplify further, we can't do much more with this. However, we can always compute ${\not x}{\not y}$ in a chosen basis of Dirac matrices. Perhaps that would be useful to you.

5. Oct 29, 2015

### moss

Thanks, I also though to compute explicitly with the Dirac matrices but I though to try some shortcut.
I will update this thread after the explicit try.

6. Oct 29, 2015

### moss

Okay, I have done the explicit calculation and its a single very long scalar term. I dont get why are you writing the identity matrix? product of 2 slashes will give a scalar.

Now about your 2nd eq. above, lets say x & y are orthogonal so the 2nd term of the 2nd eq. is zero. One is then left with -x^2y^2 and its a scalar since x, y = vectors. Now if you further impose the condition of x^2 = 1 and y^2=1 then here you go.........your 2nd eq. will then give ' i '.

In my original qs. i stated that x and y are orthonormal and if you use the orthonormality then your quoted eq. will also give ' i '.
This identity in my original qs. is okay so far because it save a lot of computation. My idea is only valid if one knows the vectors x & y .

Thanks to you because with your argument I came to the right conclusion.

7. Oct 29, 2015

### fzero

By definition, ${\not x} = x^\mu \gamma_\mu$. Since $\gamma_\mu$ are matrices, this expression is a matrix $({\not x})_{ab} = x^\mu (\gamma_\mu)_{ab}$. We can take products as much as we want, but we have to take matrix products and get matrices back out. It is possible to get a scalar by taking a trace or determinant and sometimes there are nice identities for those.

Your identity is incorrect for the same reason I gave originally. I would suggest trying to compute it explicitly if you are still having trouble seeing why.

8. Oct 30, 2015

### moss

Dear fzero, its not only my identity but according to your 2nd eq. above, my original suggestion is correct. why its correct? it is correct because I know the vectors x & y and I know that they are orthonormal. I just used your 2nd relation and thats it.

I still have to check why you have that identity matrix? maybe because the metric tensor is accompanied with it.

Last edited: Oct 30, 2015
9. Oct 30, 2015

### fzero

Yes, there is an identity matrix assumed in the commutation relation for Dirac matrices:
$$(\{\gamma^\mu,\gamma^\nu\})_{ab} = 2 \eta^{\mu\nu} (I_4)_{ab}.$$
See e.g., wikipedia.