- #36
atyy
Science Advisor
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Fredrik said:That's a projection operator, so it only has eigenvalues 0 and 1.
Note that if ##P^2=P## and ##Px=\lambda x## where ##\lambda## is a complex number and ##\|x\|\neq 0##, we have ##P^2x=P(Px)=P(\lambda x)=\lambda Px=\lambda^2x##, and also ##P^2x=Px=\lambda x##. So ##\lambda(\lambda-1)x=0##, and this implies that ##\lambda## is 0 or 1.
Ok, thanks. I have to say that it seems very unintuitive in the case of the wave function, but perhaps it's just very hard to measure such an observable, although it exists. (It seems quite intuitive to me for spin).