Commutator Confusion

[MathNerd]

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[matt grime]

Why are the diagional entries of the commutator zero? It has trace zero ([little white lie about to occur] and trace isn't defined for infinite matrices, before you bring that up).

work the commutator of the two by two matrices A and B, where A has a 1 in the top right, B has a 1 in the bottom left and all other entries are zero.

[MathNerd]

Okay so Tr( [A,B] ) = 0 , for the sqaure matrices of nxn dimension A and B...

Heisenberg Matrix Mechanics

[ X , P_X ] = i \hbar I

The trace of the left hand side is zero, while the trace of the right hand side is definitely not zero. So how is this suppose to work?

[matt grime]

Heisnberg's matrix mechanics is an infinite dimensional theory isn't it? So that proviso that infinite dimensional matrices don't have traces applies.

The trace is the sum of the diagonal entries, or equivalently the sum of the eigenvalues. What if there were an uncountable basis - how would you sum that? Even if there were a countable basis it still doesn't make sense for ALL matrices (the ones it does are called compact - these operators aren't even bounded never mind compact).