Is the Displacement Operator Tψ(x)=ψ(x+a) Hermitian?

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Discussion Overview

The discussion centers on the properties of the displacement operator defined as Tψ(x)=ψ(x+a) and whether it is Hermitian. Participants explore the definitions and implications of Hermitian operators in the context of quantum mechanics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Post 1 introduces the question of whether the displacement operator T is Hermitian.
  • Post 2 asks for the definition of a Hermitian operator to guide the discussion.
  • Post 3 provides a definition of a Hermitian operator, but expresses uncertainty about how to apply it.
  • Post 4 elaborates on the definition by suggesting a translation of the inner product into an integral form, using wave functions.
  • Post 5 questions whether the expression for the inner product leads to and expresses confusion about the equality.
  • Post 6 confirms the expression from Post 5 and suggests finding a counterexample if the two inner products are not equal.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the displacement operator is Hermitian, and the discussion remains unresolved with ongoing questions and explorations of the definitions involved.

Contextual Notes

There are limitations in the discussion regarding the application of the Hermitian operator definition, as participants express uncertainty about the mathematical steps and implications of their expressions.

Rude
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Consider the displacement operator Tψ(x)=ψ(x+a). Is T Hermitian?
 
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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.
 
Rude said:
here is the definition: <f│Ag>=<Af│g> always if A is Hermition.

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?

<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 said:
Is this what you are suggesting?

<f│Tg>=∫dxΨ(x)*ψ(x+a)

Yup.

Rude said:
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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