Can anybody confirm the following Feynman slash identity?

[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.

 

[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.

 

[moss]

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.

I have corrected the typo.
Lets say that I don't 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?

 

[fzero]

I have corrected the typo.
Lets say that I don't 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?

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.

 

[moss]

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.

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.

 

[moss]

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.

Okay, I have done the explicit calculation and its a single very long scalar term. I don't get why are you writing the identity matrix? product of 2 slashes will give a scalar.

Now about your 2nd eq. above, let's 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.

 

[fzero]

Okay, I have done the explicit calculation and its a single very long scalar term. I don't get why are you writing the identity matrix? product of 2 slashes will give a scalar.

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.

Now about your 2nd eq. above, let's 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.

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.

 

[moss]

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.

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 that's it.

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

 

[fzero]

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 that's it.

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

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.