# Imaginary eigenvalues of gamma matrices

• center o bass
In summary, the conversation discusses a statement in David Tong's notes on QFT where it is claimed that the square of a matrix being positive and negative respectively means real and imaginary eigenvalues. Further discussion involves trying to obtain a specific equation involving exponential expansion and factorizing out gammas, with some difficulties arising due to incorrect expansions. Eventually, the issue is resolved and it is determined that the incorrect expansion was the source of the problem.

#### center o bass

Hi! I'm reading David Tong's notes on QFT and I'm now reading on the chapter on the dirac equation

http://www.damtp.cam.ac.uk/user/tong/qft/four.pdf

and I stumbled across a statement where he claims that

$$(\gamma^0)^2 = 1 \ \ \Rightarrow \text{real eigenvalues}$$

while

$$(\gamma^i)^2 = -1 \ \ \Rightarrow \text{imaginary eigenvalues}.$$

I'm a bit rusty on my linear algebra and just wondered why this is necessarily true. Why does the square of a matrix being positive and negative respectively mean real and imaginary eigenvalues?

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Suppose x is an eigenvector of the matrix A with eigenvalue a, so that Ax = ax. Then A^2 x = a^2 x.

Suppose A^2 x = x for all x (including the eigenvectors of A). Then in view of the above, all eigenvalues of A satisfy a^2 = 1. So the possible eigenvalues are +1 and -1.

Now suppose A^2 x = -x for all x. Then all eigenvalues of A satisfy a^2 = -1. So the possible eigenvalues are +i and -i.

Ah thanks a lot! :)

The_Duck said:
Suppose x is an eigenvector of the matrix A with eigenvalue a, so that Ax = ax. Then A^2 x = a^2 x.

Suppose A^2 x = x for all x (including the eigenvectors of A). Then in view of the above, all eigenvalues of A satisfy a^2 = 1. So the possible eigenvalues are +1 and -1.

Now suppose A^2 x = -x for all x. Then all eigenvalues of A satisfy a^2 = -1. So the possible eigenvalues are +i and -i.

Maybe you could help me with the next thing I'm stuck on? :)

After obtaining that

$$(S^{\mu \nu})^\dagger = \frac{1}4 [\gamma^\mu, \gamma^\nu]^\dagger = - \gamma^0 S^{\mu \nu} \gamma^0$$

D. Tong claims that

$$\exp ( \frac{1}{2} \Omega_{\mu \nu} (S^{\mu \nu})^\dagger ) = \gamma^0 \exp( -\frac{1}{2} \Omega_{\mu \nu} S^{\mu \nu} ) \gamma^0.$$

I tried obtaining this by doing an exponential expansion and factorizing out the gammas, plus the property that
$$(\gamma^0)^2 = 1$$

but I get problems with the even terms in the expansion.

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center o bass said:
I tried obtaining this by doing an exponential expansion and factorizing out the gammas, plus the property that
$$(\gamma^0)^2 = 1$$

but I get problems with the even terms in the expansion.

What problems? Could you write out what you think the expansions of the left and right-hand sides of the equations are?

The_Duck said:
What problems? Could you write out what you think the expansions of the left and right-hand sides of the equations are?

Ofcourse.

$$\exp(\gamma^0 (-\frac{1}2 \Omega_{\mu \nu} S^{\mu \nu}) \gamma^0) = 1 + \gamma^0 A \gamma^0 + \frac{1}{2!} (\gamma^0)^2 A^2 (\gamma^0)^2 + \frac{1}{3!}(\gamma^0)^3 A^3 (\gamma^0)^3 + \ldots = \gamma^0 ( 1 + A + \frac{1}{2!} \gamma^0 A \gamma^0 + \frac{1}{3!} A^3 + \ldots )\gamma^0$$

where i have defined

$$A = -\frac{1}2 \Omega_{\mu \nu} S^{\mu \nu}$$

OK, two things:

1) Your expansion of the LHS isn't right because gamma-0 and A don't necessarily commute. If A and B are not-necessarily commuting matrices or operators, then

exp(AB) = 1 + AB + (1/2!)ABAB + (1/3!)ABABAB + ...

(recall that, for instance, (AB)^2 = (AB)*(AB)). But you seem to be assuming the incorrect form

exp(AB) = 1 + AB + (1/2!)AABB + (1/3!)AAABBB + ...

which is only equivalent to the correct form if A and B commute (because (AB)(AB) != AABB unless A and B commute). (Of course, you need to apply the equivalent statement for the exponential of the product of three matrices, not just two).

2) Why are there some gamma-0's inside the parentheses in your expansion of the RHS?

The_Duck said:
OK, two things:

1) Your expansion of the LHS isn't right because gamma-0 and A don't necessarily commute. If A and B are not-necessarily commuting matrices or operators, then

exp(AB) = 1 + AB + (1/2!)ABAB + (1/3!)ABABAB + ...

(recall that, for instance, (AB)^2 = (AB)*(AB)). But you seem to be assuming the incorrect form

exp(AB) = 1 + AB + (1/2!)AABB + (1/3!)AAABBB + ...

which is only equivalent to the correct form if A and B commute (because (AB)(AB) != AABB unless A and B commute). (Of course, you need to apply the equivalent statement for the exponential of the product of three matrices, not just two).

2) Why are there some gamma-0's inside the parentheses in your expansion of the RHS?

Ah, thank you again. The problem lied in the wrong expansion :) The gamma-0's were there due to this wrong expansion. When i tried to factor out gamma-0's to the left and right of the sum gamma-0's remained at the even terms.