Writing the Klein-Gordon in terms of the gamma matrices

Hello I have a quick question regarding the Dirac gamma matrices or Dirac equation and the Klein-Gordon equation. Recall that the Klein-Gordon is given by the following $$$$\left(\Box + m^2\right)\Psi=0$$$$ where $$\Box$$ is the D'Alembert operator. Then, we have the Dirac equation written in terms of the gamma matrices (or covariant form) $$$$\left( i\gamma^{\mu}\partial_{\mu}-m\right)\Psi=0$$$$ I multiplied both sides of the Dirac equation by the $$\beta$$ matrix, such that I may be able to reduce the gamma matrices, $$\gamma^{\mu}$$, to the Dirac matrices, $$\alpha^{i}$$, after which I $$\textbf{square}$$ both sides of the equation, from which we have the Klein-Gordon equation after imposing certain restrictions on the alpha and beta matrices, i.e. $$$$\left(-i \alpha_{x} \frac{\partial}{\partial x}-i \alpha_{y} \frac{\partial}{\partial y}-i \alpha_{z} \frac{\partial}{\partial z}+\beta m\right)\left(-i \alpha_{x} \frac{\partial}{\partial x}-i \alpha_{y} \frac{\partial}{\partial y}-i \alpha_{z} \frac{\partial}{\partial z}+\beta m\right) \psi=-\frac{\partial^{2} \psi}{\partial t^{2}}$$$$ \begin{aligned} \implies -\frac{\partial^{2} \psi}{\partial t^{2}}=&-\alpha_{x}^{2} \frac{\partial^{2} \psi}{\partial x^{2}}-\alpha_{y}^{2} \frac{\partial^{2} \psi}{\partial y^{2}}-\alpha_{z}^{2} \frac{\partial^{2} \psi}{\partial z^{2}}+\beta^{2} m^{2} \psi \\ &-\left(\alpha_{x} \alpha_{y}+\alpha_{y} \alpha_{x}\right) \frac{\partial^{2} \psi}{\partial x \partial y}-\left(\alpha_{y} \alpha_{z}+\alpha_{z} \alpha_{y}\right) \frac{\partial^{2} \psi}{\partial y \partial z}-\left(\alpha_{z} \alpha_{x}+\alpha_{x} \alpha_{z}\right) \frac{\partial^{2} \psi}{\partial z \partial x} \\ &-\left(\alpha_{x} \beta+\beta \alpha_{x}\right) m \frac{\partial \psi}{\partial x}-\left(\alpha_{y} \beta+\beta \alpha_{y}\right) m \frac{\partial \psi}{\partial y}-\left(\alpha_{z} \beta+\beta \alpha_{z}\right) m \frac{\partial \psi}{\partial z} \end{aligned} such that, \begin{aligned} \alpha_{x}^{2}=\alpha_{y}^{2}=\alpha_{z}^{2}=\beta^{2} &=1 \\ \alpha_{j} \beta+\beta \alpha_{j} &=0 \\ \alpha_{j} \alpha_{k}+\alpha_{k} \alpha_{j} &=0 \quad(j \neq k) \end{aligned}

Now, this works swell for the normal Dirac equation, but what about the Dirac equation which may have additional terms, i.e. the Dirac equation in curved spacetime, in which we have the spin connection, as well as other terms, i.e. the Levi-Civita term. This proves extremely inefficient, as multiplying both sides by the $$\beta$$ matrix, then squaring both sides gives an enormous amount of terms. Is there any other way in which perhaps we can derive the Klein-Gordon using gamma matrices, or perhaps derive the Klein-Gordon from the Dirac? Thank you for your help

• Free Dirac and conjugated Dirac solutions are simultaneously KG solutions. This should no longer be true if there is gravitational interaction with spin. Apr 10, 2020 at 12:55

Simply hit the Dirac equation with the complex conjugate of $$(i\gamma^\mu \partial_\mu-m)$$. So \begin{align} (i\gamma^\mu \partial_\mu-m)\Psi &= 0 \\ -(i\gamma^\nu \partial_\nu+m)(i\gamma^\mu \partial_\mu-m)\Psi &= 0 \\ (\gamma^\mu\gamma^\nu\partial_\mu\partial_\nu+m^2)\Psi &= 0 \end{align} Since both $$\mu$$ and $$\nu$$ are dummy indices it doesn't hurt to do a dumb move and rename them, so you get \begin{align} \left(\frac{1}{2}\big(\gamma^\mu\gamma^\nu\partial_\mu\partial_\nu+\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu\big)+m^2\right)\Psi &= 0. \end{align} But now observe that $$\partial_\mu\partial_\nu = \partial_\nu\partial_\mu$$ so \begin{align} \left(\frac{1}{2}\big(\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu\big)\partial_\mu\partial_\nu+m^2\right)\Psi &= 0. \end{align} Finally the gamma matrices obey the Clifford algebra $$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$$, so the last equation becomes simply \begin{align} \left(\frac{1}{2}2\;\eta^{\mu\nu}\partial_\mu\partial_\nu+m^2\right)\Psi &= 0 \\ \left(\Box+m^2\right)\Psi &= 0, \end{align} which is the Klein-Gordon that you wanted.

• Thank you so much this is great! But as I mentioned above this works swell for the free Dirac, but when one applies curvature, I don’t believe the curvature displays analogous properties to the gamma matrices and their Clifford Algebra. I shall give it a try nonetheless. Again, thank you for your suggestion! Apr 10, 2020 at 16:01
• Your actual question was: "Is there any other way in which perhaps we can derive the Klein-Gordon using gamma matrices, or perhaps derive the Klein-Gordon from the Dirac?". And that's what I did. For curved space, I believe that combining this derivation with @mike stone's curved space Dirac operator should get you to the curved space Klein-Gordon Apr 10, 2020 at 18:47
• I see. So then substituting in the operator given by mike stone into the derivation you provided above, I should arrive at what I am looking for? Or is the operator squared which mike stone provided already what I’m looking for? Apr 10, 2020 at 20:08

For the curved space Dirac operator $${D}= \gamma^a e^\mu_a \left(\partial_\mu + \textstyle{\frac 12} \sigma^{bc}\,\omega_{bc\mu}\right)= \gamma^a D_a\equiv \gamma^\mu D_\mu.$$ we have the simple looking Lichnerowicz formula $$D^2 = \nabla^2 - \frac 14 R,$$ where $$\nabla^2\equiv\frac 1{\sqrt{g}} D_\mu \sqrt{g}g^{\mu\nu} D_\nu$$ is the "rough" or spin-connection Laplacian on spinors and $$R$$is the scalar curvature. Although named for Andre Lichnerowicz who published it in 1963 (A. Lichnerowicz, Spineurs harmonique, Compt. Rend. Acad. Sci. Paris, Ser. A 257, (1963) 7-9), the formula appears as the very last equation in a paper by Erwin Schroedinger writen some 30 years earlier (E. Schroedinger, Diracsches Elektron im Schwerefeld I, Sitzungsber. Preuss. Akad. Wiss., Phys. Math. Kl. {11}, (1932) 105-128.). As you say, there are a lot of terms and quite a bit of manipulation (23 pages in Schroedinger's paper) to get the Schroedinger-Lichnerowicz formula!

(Actually it's not that bad. You can do in a couple pf pages using the Bianchi identities.)

• Could you confirm if the Laplacian that you described above, i.e. the "rough" or spin-connection Laplacian is the Laplace-Beltrami operator applied to a scalar function $f$, i.e. $\Delta f=\frac{1}{\sqrt{|g|}} \partial_{i}\left(\sqrt{|g|} g^{i j} \partial_{j} f\right)$ Apr 11, 2020 at 7:05
• No. It cannot be the scalar laplacian. The $D_\mu$ contain the spin connection, as they must if the laplacian is to be covariant under changes of coordinates or frames. What forming $D^2$ does is to get rid of the $\gamma^\mu$ that pre-multiply $D$. Apr 11, 2020 at 12:24
• I see. So then the $D_{\mu}$ in the Laplacian above is the operator that you defined above when you wrote, $\gamma^{\mu}D_{\mu}$ which involves the spin connection? Or is it simply a partial derivative or covariant derivative? From what you have said, I believe it is the operator you defined above involving the spin connection, but just want to make sure. Apr 12, 2020 at 20:27
• Hence, it would only be the section inside of the parentheses, since we used the tetrad to change the indices of the gamma matrix, or am I incorrect? Apr 12, 2020 at 20:58
• You are correct: $D_\mu=\partial_\mu+\frac 12 \sigma^{bc}\omega_{bc\mu}$ Apr 12, 2020 at 21:03