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Rude
Recent content by Rude
R
Is the Displacement Operator Tψ(x)=ψ(x+a) Hermitian?
Is this what you are suggesting? <f│Tg>=∫dxΨ(x)*ψ(x+a) If so I don't know how to proceed. It does not look like this would give <Tf│g> but don't know why.
Rude
Post #5
Mar 16, 2013
Forum:
Quantum Physics
R
Is the Displacement Operator Tψ(x)=ψ(x+a) Hermitian?
here is the definition: <f│Ag>=<Af│g> always if A is Hermition. Don't know how to start.
Rude
Post #3
Mar 15, 2013
Forum:
Quantum Physics
R
Is the Displacement Operator Tψ(x)=ψ(x+a) Hermitian?
Consider the displacement operator Tψ(x)=ψ(x+a). Is T Hermitian?
Rude
Thread
Mar 15, 2013
Tags
Displacement
Operator
Replies: 5
Forum:
Quantum Physics
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