Is the term:

$$γ^μ γ_μ$$ An identity matrix? Since,if we start with both the Dirac equation, $$(iγ^μ ∂_μ-m)Ѱ=0$$ We find that, $$iγ^μ ∂_μ=m$$ If we square both sides, we get, $$-γ^μ γ_μ∂^μ ∂_μ=m^{2}$$ Therefore, $$-γ^μ γ_μ ∂^μ ∂_μ-m^2=0$$ $$γ^μ γ_μ ∂^μ ∂_μ+m^2=0$$ This situation above resembles the Klein-Gordon equation, solely as an operator, as seen below, $$(\Box+m^2 )=0$$ Were $\Box$ denotes the d'Alembert operator.

Therefore, $$γ^μ γ_μ ∂^μ ∂_μ+m^2=(\Box+m^2 )$$

Therefore, $$γ^μ γ_μ ∂^μ ∂_μ=\Box$$ Since the d'Alembert operator, is defined as $$\Box=∂^μ ∂_μ$$ The only way for the above equation to be true is if, $$γ^μ γ_μ $$ equals an identity matrix?

Is my reasoning correct, if it isn't then what is the flaw?

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    $\begingroup$ In an eigenvalue problem $A\mathbf x=\lambda\mathbf x\to (A-\lambda I)\mathbf x=0$ does this necessarily mean that $A-\lambda I=0$? $\endgroup$ – BioPhysicist Nov 22 '19 at 4:31
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    $\begingroup$ en.wikipedia.org/wiki/Gamma_matrices#Miscellaneous_identities $\endgroup$ – G. Smith Nov 22 '19 at 5:01
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    $\begingroup$ There is a conventional explicit choice for the four gamma matrices. You could have tried computing $\gamma^\mu\gamma_\mu$ straightforwardly by matrix multiplication, addition, and subtraction, and discovered that it is not the identity matrix (so you would have realized that you must have made a mistake somewhere). However, it turns out that it is a multiple of the identity matrix. $\endgroup$ – G. Smith Nov 22 '19 at 5:05

You are using $\mu$ as an index too many times. When you square $\gamma^\mu \partial_\mu$, you have to use a different index for the two factors. Here is a correct derivation:

Starting with $(i \gamma^\mu \partial_\mu - m) \psi = 0$, apply $-(i \gamma^\nu \partial_\nu + m)$ from the left: $$ 0 = -(i \gamma^\nu \partial_\nu + m)(i \gamma^\mu \partial_\mu - m) \psi = (\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu + m^2) \psi. $$ Now, since $\partial_\mu \partial_\nu = \partial_\nu \partial_\mu$, i.e. it is symmetric, only the symmetric part $$ \gamma^{(\mu} \gamma^{\nu)} = \frac{\{\gamma^\mu, \gamma^\nu\}}{2} = \frac{\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu}{2} $$ will survive: $$ \gamma^\mu \gamma^\nu \partial_\mu \partial_\nu = \gamma^{(\mu} \gamma^{\nu)} \partial_\mu \partial_\nu. $$ But, the defining feature of the $\gamma$ matrices is that $\gamma^{(\mu} \gamma^{\nu)} = \eta^{\mu\nu}$. So you get $$ (\eta^{\mu\nu} \partial_\mu \partial_\nu + m^2) \psi = 0, $$ i.e. $$ (\partial^\mu \partial_\mu + m^2) \psi = 0, $$ the Klein-Gordon equation.

As for what $\gamma^\mu \gamma_\mu$ is, $$ \gamma^\mu \gamma_\mu = \gamma^\mu \gamma^\nu \eta_{\mu\nu} = \gamma^{(\mu} \gamma^{\nu)} \eta_{\mu\nu} = \eta^{\mu\nu} \eta_{\mu\nu} = \delta^\mu_\mu = d = 4 $$ if you are working in a four-dimensional spacetime. Here, they key step is using that $\eta_{\mu\nu}$ is symmetric.

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