Compute #hatP^2# and is #hatP# idempotent?

Let #|psi_n># be the eigenstates of some Hermitian operator, #hatO#

Consider the operator #hatP = 1/2 (|ψ_m><ψ_m| + |ψ_n><ψ_n| )#

1 Answer
Truong-Son N.
Feb 17, 2018

Assuming you really mean

#hatP = 1/2 (| psi_m >><< psi_m | + |psi_n >> << psi_n |)#,

then since #hatP ne hatO#, the #|psi_m >># and #|psi_n >># need not be eigenstates of #hatP#. We have:

#hatP^2 = 1/4 (| psi_m >><< psi_m | + |psi_n >> << psi_n |)^2#

#= 1/4 (| psi_m >><< psi_m | + |psi_n >> << psi_n |)(| psi_m >><< psi_m | + |psi_n >> << psi_n |)#

#= 1/4 (|psi_m >> << psi_m | psi_m >> << psi_m| + |psi_m >> << psi_m | psi_n >> << psi_n| + |psi_n >> << psi_n | psi_m >> << psi_m| + |psi_n >> << psi_n | psi_n >> << psi_n |)#

By default, proper eigenstates of Hermitian operators are conventionally such that the #| psi_m >># and #| psi_n >># are orthonormal, so

#= 1/4 [##color(green)(|psi_m >> color(black)(cancel(color(green)(<< psi_m | psi_m >>))^(1)) color(green)(<< psi_m| + color(purple)(|psi_m >> color(black)(cancel(color(purple)(<< psi_m | psi_n >>))^(0)) color(purple)(<< psi_n| + color(orange)(|psi_n >> color(black)(cancel(color(orange)(<< psi_n | psi_m >>))^(0)) color(orange)(<< psi_n| + color(red)(|psi_n >> color(black)(cancel(color(red)(<< psi_n | psi_n >>))^(1)) color(red)(<< psi_n |# #]#

#= 1/4 (| psi_m >><< psi_m | + |psi_n >> << psi_n |)#

#= 1/2hatP ne hatP#

Even if the eigenstates are not normalized, each eigenvector must be orthogonal for distinct eigenvalues. Orthonormal states are the best-case scenario for simplifying this, so no matter how you spin it, #hatP# is not idempotent.

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