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If I express the Dirac equation in the form of

$$i\hbar \frac{\partial}{\partial t} \psi_a(x) = \left(-i\hbar c(\alpha^j)_{ab}\partial _j + mc^2(\beta)_{ab}\right)\psi_b(x),$$

with the constraints

$$\{\alpha^j,\alpha^k\}_{ab} = 2\delta^{jk}\delta_{ab}, \qquad \{\alpha^j,\beta\}_{ab}=0, \qquad (\beta^2)_{ab} = \delta_{ab},$$

since the Gamma matrices are traceless and their eigenvalues are either +1 or -1, we can show that the Gamma matrices are of even dimension.

My question is: how can I show $2 \times 2$ matrices are not enough?

Emilio Pisanty
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Kevin Kwok
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  • See Greiner, Relativistic quantum mechanics. – BoundaryGraviton Oct 20 '16 at 08:25
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    Possible duplicates: http://physics.stackexchange.com/q/53318/2451 , http://physics.stackexchange.com/q/259514/2451 , http://physics.stackexchange.com/q/97770/2451 , http://physics.stackexchange.com/q/41282/2451 and links therein. – Qmechanic Oct 20 '16 at 08:52
  • @BoundaryGraviton I've checked his book. In page 78, he simply states that because only three anticommuting matrices exists for N = 2, namely Pauli matrices. What he omitted is exactly what I want to show-why there are only three? – Kevin Kwok Oct 20 '16 at 10:17

1 Answers1

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You need four linearly independent matrices. Write down the representation:

a + ib   c + id
e + if   g + ih

The traceless condition gives $a = -g$, $h=-b$. Now we have

a + ib   c + id
e + if  -a - ib

This has six unknowns (six possible independent matrices).

Then the eigenvalue conditions for the four matrices give four equations in six unknowns, which means that there is no linearly independent solution with $2\times2$ matrices, since that would require four degrees of freedom, but we only have two left.

Incidentally this question is often an exercise in books on quantum field theory or relativistic quantum mechanics.

Suzu Hirose
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    Thank you! This simple reasoning is what I expected. Actually, I am reading a quantum field theory book. – Kevin Kwok Oct 20 '16 at 11:11
  • @KevinKwok - yes it is usually in those kinds of books. I had a look at the new edition of Mandl & Shaw (2010) and can't find it there so I seem to have been mistaken. – Suzu Hirose Oct 20 '16 at 11:52