Consider the displacement operator Tψ(x)=ψ(x+a). Is T Hermitian?
To get you started, what's the definition of a Hermitian operator?
here is the definition: <f│Ag>=<Af│g> always if A is Hermition.
Don't know how to start.
Suppose the wave function of the state |g> is ##\psi_g(x)## and the wave function of state |f> is ##\psi_f(x)##. Then what does the above definition of an operator A being Hermitian say if you translate the inner product into an integral of wave functions, using
##\langle a | b \rangle = \int dx \psi_a(x)^* \psi_b(x)##
and plug in A = T?
Is this what you are suggesting?
If so I don't know how to proceed.
It does not look like this would give <Tf│g> but don't know why.
If you think they aren't equal, perhaps you can find an explicit counterexample?
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