Thursday, 11 January 2018

homework and exercises - Construction of Pauli Matrices


How can we construct the Pauli matrices starting from σi=[abcd]

by using the conditions σ2i=1,
[σx,σy]=2iσz,
and so on?



Answer



Since each σi is a scalar multiple of a Lie bracket of other finite matrices, each σi must be traceless. So straight away we know:


σi=(abca)


and σ2i=id then yields a2+bc=1.


The eigenvalues of any matrix of the form in (1) with a2+bc=1 are ±a2+bc=±1. Therefore, for any set of matrices we find fulfilling all the given relationships, we can do a similarity transformation on the whole set and thus (1) diagonalize any member of the set we choose whilst (2) keeping all the required relationships intact. Exercise: Prove that the given relationships (Lie brackets and σ2i=id) are indeed invariant under any similarity transformation.


Thus, without loss of generalness, we can always choose one of the set to be:


σz=(1001)


So now work out the Lie bracket of σz and σx=(axbxcxax): result must be 2iσy and so we get:



σy=(0ibxicx0)


But given σ2y=1 we get bxcx=1 whence ax=0 (since a2x+bxcx=1). So our remaining two matrices are of the forms:


σx=(0bx1bx0)

σy=(0ibxibx0)


and the remaining commutation relationships then give you the unknown constant bx.


Once you have found bx, we know from our comments above that any set of matrices fulfilling the required commutation relationships and σ2i=id is gotten from this particular set (the "standard" Pauli matrices) by a similarity transformation.


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