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dEdt
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"Proof" that all operators are linear
I've "proven" that all operators acting on a Hilbert space are linear. Obviously this isn't true, so there must be a fault in my reasoning somewhere. I having trouble finding it though, and would appreciate input by someone who can.
Let [itex]|\psi\rangle = \alpha |a\rangle+\beta|b\rangle[/itex], [itex]A[/itex] be an arbitrary operator, and [itex]|\phi\rangle[/itex] be an arbitrary vector.
[tex]\langle\phi|A|\psi\rangle =(\langle\phi|A)|\psi\rangle=(\langle\phi|A) (\alpha |a\rangle+\beta|b\rangle)=\alpha (\langle\phi|A)|a\rangle+\beta(\langle\phi|A) |b\rangle= \alpha \langle\phi|(A|a\rangle)+\beta\langle\phi|(A |b\rangle)= \langle\phi| (\alpha A|a\rangle +\beta A|b\rangle).[/tex]
So
[tex]\langle \phi |(A|\psi \rangle)=\langle\phi| (\alpha A|a\rangle +\beta A|b\rangle)[/tex]
for all [itex]\langle \phi |[/itex]. Therefore
[tex]A(\alpha |a\rangle+\beta|b\rangle)= \alpha A|a\rangle +\beta A|b\rangle.[/tex]
I've "proven" that all operators acting on a Hilbert space are linear. Obviously this isn't true, so there must be a fault in my reasoning somewhere. I having trouble finding it though, and would appreciate input by someone who can.
Let [itex]|\psi\rangle = \alpha |a\rangle+\beta|b\rangle[/itex], [itex]A[/itex] be an arbitrary operator, and [itex]|\phi\rangle[/itex] be an arbitrary vector.
[tex]\langle\phi|A|\psi\rangle =(\langle\phi|A)|\psi\rangle=(\langle\phi|A) (\alpha |a\rangle+\beta|b\rangle)=\alpha (\langle\phi|A)|a\rangle+\beta(\langle\phi|A) |b\rangle= \alpha \langle\phi|(A|a\rangle)+\beta\langle\phi|(A |b\rangle)= \langle\phi| (\alpha A|a\rangle +\beta A|b\rangle).[/tex]
So
[tex]\langle \phi |(A|\psi \rangle)=\langle\phi| (\alpha A|a\rangle +\beta A|b\rangle)[/tex]
for all [itex]\langle \phi |[/itex]. Therefore
[tex]A(\alpha |a\rangle+\beta|b\rangle)= \alpha A|a\rangle +\beta A|b\rangle.[/tex]