- #1
einai
- 27
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Another quantum question...I feel dumb .
A Hermitian operator A has the spectral decomposition
A = Σ an|n><n| (summation in n)
where A|n> = an|n> (the an's are eigenvalues of A, and |n>'s the eigenstates).
So, how can I find the spectral decomposition of the inverse of A so that
AA-1 = A-1A = 1?
My intuition would be A = Σ (1/an)|n><n| (summation in n), since 1 = Σ |k><k| (summation in k), but I don't think it's that easy.
Thanks in advance!
A Hermitian operator A has the spectral decomposition
A = Σ an|n><n| (summation in n)
where A|n> = an|n> (the an's are eigenvalues of A, and |n>'s the eigenstates).
So, how can I find the spectral decomposition of the inverse of A so that
AA-1 = A-1A = 1?
My intuition would be A = Σ (1/an)|n><n| (summation in n), since 1 = Σ |k><k| (summation in k), but I don't think it's that easy.
Thanks in advance!