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The properties of the projection operators are defined as:

$$P_+ = \frac{1}{2}(1+\gamma^5)$$ $$P_- = \frac{1}{2}(1-\gamma^5)$$

where $\gamma^5 = -i\gamma^0\gamma^1\gamma^2\gamma^3$

and their key properties are that $P_+^2 = P_+, P_+P_- = 1, P_-^2 = P_-$.

But since $\gamma^0$ has the same property as $\gamma^5$ that ${\gamma^0}^2=1$, if we replaced $\gamma^5$ with $\gamma^0$ in the definition of the projection operators, the newly defined projection operators would also satisfy all the key properties. And it's not like ${\gamma^0}^2=1$ is basis dependent; to show that ${\gamma^5}^2=1$ we actually need to make use of that fact. Why do we then bother to define $\gamma^5$ at all if we could use $\gamma^0$ to define the projection operators?

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  • $\begingroup$ Have you heard of chirality? $P_{\pm}$ are of course not the only possible projectors but they are those projecting a spinor onto its left/right chiral components. $\endgroup$
    – user154997
    Commented Jun 3, 2017 at 23:33
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    $\begingroup$ $\gamma^5$ anticommutes with gamma matrices, $\gamma^0$ does not. $\endgroup$
    – Kosm
    Commented Jun 3, 2017 at 23:45

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Why do we then bother to define $\gamma^5$ at all if we could use $\gamma^0$ to define the projection operators?

The key observation is that chirality should be preserved under the Lorentz transformation. Hence the $X$ in the chiral projection $$ P_{\pm} = \frac{1}{2}(1 \pm X) $$ should commute with 6 Lorentz algebra elements $$ \gamma^{\mu}\gamma^{\nu}. $$

Out of all the combinations (15 of them: 4 vectors $\gamma^{\mu}$, 6 bivectors $\gamma^{\mu}\gamma^{\nu}$, 4 pseudovectors $\gamma^{5}\gamma^{\mu}$, 1 pseudoscalar $\gamma^{5}$) of gamma operators, only the pseudoscalar $$ \gamma^{0}\gamma^{1}\gamma^{2}\gamma^{2} $$ commutes with ALL 6 Lorentz (antisymmetric) bivectors $\gamma^{\mu}\gamma^{\nu}$, while $$ \gamma^{0} $$ does NOT.

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  • $\begingroup$ Notice that if you define $\tilde \gamma ^0=\gamma ^5$ and $\tilde \gamma ^5=\gamma ^0$, then $\tilde \gamma ^5 $ commutes with $\tilde \gamma ^\mu \tilde \gamma ^\nu$ $\endgroup$
    – pppqqq
    Commented Oct 2, 2018 at 20:30
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    $\begingroup$ They are isomorphic, conditional on redefining Lorentz operators as $\tilde \gamma ^\mu \tilde \gamma ^\nu$. Then we are back to square one (since $\tilde \gamma ^5 \sim \tilde \gamma ^0 \tilde \gamma ^1\tilde \gamma ^2\tilde \gamma ^3$). Aren't we? $\endgroup$
    – MadMax
    Commented Oct 2, 2018 at 20:39
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To understand the importance of $\gamma^5$ you need to understand chirality first. Let me describe briefly. Chirality is a property of asymmetry, an object/system is called chiral if it is distinguishable from its mirror image. In other words the object can not be superposed on its mirror image just by rotations. Chirality defines the handedness (right/left handed) of the object.

Now come to $\gamma^5$, this matrix is the generator of chiral transformations of spinors$e^{-i\gamma^5\alpha}$. So this matrix can be used to check the handedness of spinors. This kind of matrix is possible in Clifford algebra in only even space time dimensions. So in 3+1 D we defince $\gamma^5$ as you did.

Why not $\gamma^0$? Just because it is not the generator of chiral transformations. (I think) It has nothing to do with chirality.

Projection operators are the operators which separate left/right chiral parts of Dirac spinor. These operators would be constructed with $\gamma^5$. Hope it helps. You can discuss it further.

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