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it is well known that a general operator from SU(2) can be effectively represented as a composition of at most three "one-parameter" ones. For instance in the form $\exp(i\, \alpha\, \sigma_z) \exp(i\, \beta\, \sigma_x) \exp(i\, \gamma\, \sigma_z)$ where $\alpha, \beta, \gamma$ are real parameters, $\sigma_x, \sigma_z$ are the Pauli x and z.

The question is the following: how would this generalize to SU(d) , $d > 2$ ? I suppose at least $d^2-1$ "one-parameter" operators are necessary (and, out of dimensionality argument also, at least locally, sufficient). But how does one write a decomposition like this effectively?

[I am most interested in $d=3$ right now, but the appetite might grow :-)]

luksen
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user160320
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1 Answers1

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Figured it out with kind help from a colleague. A matrix from SU(d) can be reduced to a diagonal unitary using at most $d(d-1)/2$ Givens rotations. Each Givens rotation is defined by two rotation parameters and a scalar, but we will let all the scalars get absorbed into the residual diagonal unitary. Thus we need $d(d-1)$ one-parameter factors for reduction to diagonal. Because the residual diagonal is unitary, it splits into $d-1$ ( not $d$) one-parameter diagonal factors.

Now $d(d-1)+(d-1) = d^2-1$ which is what was claimed.

user160320
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