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If you have three things in a bra ket what does it mean?

  1. Sep 24, 2014 #1
    say you have <ψ|x|ψ⟩ or <0|F|k⟩ where F is an operator, what does this actually mean? I understand C|ψ⟩ would be the operator C acting on PSI and <ψ1|ψ2⟩ is the inner product of two wavefunctions but what would a third term inbetween them mean?

    thanks for any help
     
  2. jcsd
  3. Sep 24, 2014 #2

    stevendaryl

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    Well, when an operator [itex]F[/itex] acts on a state [itex]|\psi\rangle[/itex], it produces a new state: [itex]|\psi'\rangle = F |\psi\rangle[/itex]. So the expression [itex]\langle \phi | F | \psi \rangle[/itex] is just the inner product of [itex]|\phi\rangle[/itex] and [itex]F |\psi\rangle[/itex].

    The nice thing about Hermitian operators is that the inner product of [itex]|\phi\rangle[/itex] and [itex]F |\psi\rangle[/itex] is the same as the inner product of [itex]F |\phi\rangle[/itex] and [itex]|\psi\rangle[/itex]. In [itex]\langle \phi | F | \psi \rangle[/itex], you can view [itex]F[/itex] as acting to the left on [itex]\phi[/itex] or to the right on [itex]\psi[/itex].
     
  4. Sep 24, 2014 #3
    ahh okay many thanks! its the bra notation for the wavefunction in front of the operator that confuses me, could you explain how it differs from the wavefunction in the ket?

    also if F can act left or right, is that like saying the two wavefunctions are overlapping (or bound) so it doesn't matter which the operator acts on first?

    thanks again for the reply!
     
  5. Sep 24, 2014 #4
    It doesn't matter which wavevector (the bra or the ket) the operator acts on, but that doesn't have anything to do with the wavefunctions overlapping or bound.
     
  6. Sep 24, 2014 #5

    dextercioby

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    The true mathematical justification of the use of bras and kets cannot be formulated in the absence of rigged Hilbert spaces. But this state of the art, which many people wouldn't understand. To be 'sandwich'ed between a ket and a bra an operator needs to be selfadjoint. One then has that

    $$\langle A\psi, \phi\rangle = \langle \psi, A\phi\rangle \equiv \langle \psi |A|\phi\rangle $$
     
  7. Sep 24, 2014 #6

    Avodyne

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    An operator need not be self-adjoint to be sandwiched, e.g.,
    ##\langle m|a^\dagger|n\rangle = \sqrt{m}\delta_{m,n+1}##
     
  8. Sep 24, 2014 #7
    I think the opposition to sandwiching operators which aren't self-adjoint is that it is ambiguous notation. You have to think about acting to the left with the adjoint of the operator that you see in the middle. The notation generally employed by mathematicians avoids such ambiguities.
     
  9. Sep 24, 2014 #8

    bhobba

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    Very true.

    But everyone into QM should have a smattering of knowledge of it:
    http://physics.lamar.edu/rafa/cinvestav/second.pdf [Broken]

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
  10. Sep 24, 2014 #9

    Avodyne

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    Nothing ambiguous about it. ##\langle m|a^\dagger|n\rangle = \langle m|(a^\dagger|n\rangle) = (\langle m|a^\dagger)|n\rangle##, and ##\langle m|a^\dagger =(a|m\rangle)^\dagger##.
     
  11. Sep 24, 2014 #10

    bhobba

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    The issue is in Rigged Hilbert spaces - care is required in exactly what one can apply an operator to.

    Thanks
    Bill
     
  12. Sep 25, 2014 #11
    Excellent! Thanks for all the replies!

    Very good link! Thank you!
     
    Last edited by a moderator: May 7, 2017
  13. Sep 25, 2014 #12

    vanhees71

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    More caveats can be found in the following very illuminating paper:

    Gieres, F.: Mathematical surprises and Dirac's formalism in quantum mechanics, Rep. Prog. Phys. 63, 1893, 2000
    http://arxiv.org/abs/quant-ph/9907069
     
  14. Sep 26, 2014 #13
    Last edited: Sep 26, 2014
  15. Sep 26, 2014 #14
    rather than start a new thread could I throw in a quick question about domains and ∈,

    What is the domain of an observable? Is it simple the place when it can be seen? and also for this domain:

    D(Q) = {f∈L^2|xf∈L^2}

    I understand Q is the position operator, this this is where its position will reside, HOWEVER, what does f∈L^2 actually mean? Wiki says f would be an element of the set L^2, whats a set in this instance?

    Thanks again
     
  16. Sep 26, 2014 #15

    dextercioby

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    This is really abstract and the chosen example is a little tricky. D(Q) is the (maximal) domain of the operator, the set of all states for which you can measure the position of a particle. As for <f(x) is an element of L^(2)>, well it suffices to say that |f|2 is Lebesgue integrable over a certain open domain of R3.
     
  17. Sep 26, 2014 #16

    Doug Huffman

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    Leonard Susskind at Stanford U.'s YouTube channel has marvellous introduction and reviews from basic mechanics to advanced QM and cosmology.
     
  18. Sep 26, 2014 #17
    ahh i think i see now, it is the limit of where the operator can operate.

    no idea what that means but it gives me something to look into, thanks for your reply!

    Yes, I have reached Lecture 5, I agree it's very good. Thanks.
     
  19. Sep 28, 2014 #18
    I would recommend Leonard Susskind theoretical minimum series he covers the basics of q bits and twospin system and derives all the basics equations such as heisenbergs uncertanty priciple and schrodinger time(in)dependent equation from then on it's just harmonic oscillators and so on...
     
  20. Oct 10, 2014 #19
    I have ordered this book. thank you!
     
  21. Oct 10, 2014 #20
    Good it's really great, so lots of fun,if you have any questions on the stuff just ask me.
     
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