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I Position operator

  1. Mar 11, 2017 #1
    Would the action of the position operator on a wave function ##\psi(x)## look like this?

    $$\psi(x) \ =\ <x|\psi>$$ $${\bf \hat x}<x|\psi>$$

    Question 2: the position operator can act only on the wave function?
     
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  3. Mar 11, 2017 #2

    hilbert2

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    A vector ##|\psi>## is an abstract state and the ##<x|\psi >## is a wave function. An operator like ##\mathbf{x}## acts on the state ##|\psi >##, and the wavefunction of the state that has been acted on by the operator is ##<x|\mathbf{x}|\psi >##.
     
  4. Mar 11, 2017 #3
    Thanks Hilbert2. So we have the freedom to move the operator ##\hat x## from ##\hat x <x|\psi>## to ##<x|\hat x|\psi>##?
     
  5. Mar 11, 2017 #4

    hilbert2

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    If you're talking about the abstract ##\mathbf{x}## operator, it can only act on a vector ##|\psi >##, not on the wavefunction ##<x|\psi >##. But if you mean the position space representation of the operator ##\mathbf{x}##, then it is something that acts on the wavefunction, not an abstract state.
     
  6. Mar 11, 2017 #5
    Ah, ok.
    What is the motivation for defining such a operator? In Quantum Mechanics one can define a time-evolution operator which gives you a state vector at a later time, one can define a rotation operator which gives you a rotated state vector, etc... What about the position operator?
     
  7. Mar 11, 2017 #6

    hilbert2

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    The matrix element ##<\psi |\mathbf{x}|\psi >## gives the expectation value of the position of a particle that is in state ##|\psi >##. That's one motivation. The wavefunctions describing eigenstates of ##\mathbf{x}## are not proper functions like those of a momentum operator or hamiltonian operators, which makes them a bit difficult to handle.
     
  8. Mar 11, 2017 #7
    I see
    In what does the position operator differs from the translation operator?
     
  9. Mar 11, 2017 #8

    PeroK

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    Where and how are you learning QM? Your questions are seemingly random.
     
  10. Mar 11, 2017 #9
    Most from Sakurai's book and McIntyre (on introductory QM)
    It's just that one question leads me to another
     
  11. Mar 11, 2017 #10

    PeroK

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    How much linear algebra do you know? Sakurai, in my opinion, assumes a good grasp of the relevant undergraduate maths - especially linear algebra.
     
  12. Mar 11, 2017 #11

    PeroK

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    I have the "revised" edition of Sakurai. Page 42 has the relevant section on "Position Eigenkets and Position Measurements". Page 44 describes "Translation" and the Translation operator.
     
  13. Mar 11, 2017 #12

    hilbert2

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    There's not much in common, because the translation operator is not hermitian. The momentum operator is the generator of translation, and I guess that the position operator is a generator of a galilean frame change, which is a translation in momentum space.
     
  14. Mar 11, 2017 #13
    Well, I attended Linear Algebra classes last semester at university. So I would say I know the basics of it.
    I think this is not avaiable in older editions of the book.

    How could we see this? Can you give me an example where we can see that the position operator is the generator of a galilean frame change?
     
  15. Mar 11, 2017 #14

    hilbert2

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    If you have a momentum eigenstate ##\psi (x)=e^{ipx/\hbar}## and multiply it with ##e^{i(\Delta p) x/\hbar}##, you get a state with a momentum that's increased by ##\Delta p##. Just like operation with ##e^{i(\Delta x) \mathbf{p}/\hbar}## where ##\mathbf{p}## is the position space momentum operator is equivalent to space translation by ##\Delta x##.
     
  16. Mar 11, 2017 #15
    Thank you
     
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