Least Action

Nontrivializing triviality..and vice versa.

A lemma on a rather frequently encountered set of Hermitian matrices

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I’m documenting a small ‘lemma’ that I think is worth mentioning. Suppose M^{i} are 4 Hermitian matrices (i = 1, \ldots, 4) satisfying

M^{i}M^{j} + M^{j}M^{i} = 2\delta^{ij}I

where I denotes the identity matrix. Then these matrices have eigenvalues \pm 1, are traceless and are necessarily of even dimension.

For i = j, the anticommutator above gives (M^{i})^2 = I. So, for any eigenvector X and eigenvalue \lambda, we have

M^{i}X = \lambda X \implies (M^{i})^2 X = \lambda^2 X

or equivalently \lambda = \pm 1.

Traceless-ness has a neat proof:

Suppose j \neq i. Then

tr(M^{i}) = tr(M^{i}(M^{j})^2) = tr(M^{i}M^{j}M^{j}) = tr(M^{j}M^{i}M^{j})

where the last equality follows from tr(ABC) = tr(CAB). But M^{j}M^{i} = -M^{i}M^{j}, so

tr(M^{i}) = -tr(M^{i}M^{j}M^{j}) = -tr(M^{i}) \implies tr(M^{i}) = 0.

Finally, suppose the numbers of +1 eigenvalues and -1 eigenvalues are a and b respectively. The dimension of the matrix is then a + b. Since the trace equals the sum of eigenvalues, we have

a - b = 0

So, the dimension is a + b = 2a = 2b, which is clearly always an even number.

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Written by Vivek

September 19, 2010 at 20:33

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