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Physical Meaning of Bra-kets

  1. Oct 16, 2011 #1
    Hi, I some basic questions on the physical meaning of bra-ket notation. I have looked at a lot of material, and I have seen descriptions of the algebraic properties of bra-kets, and I have seen hints at it having meaning regarding the probability of events happening/state changes, but I can't seem to find an exact depiction of this relationship. Hopefully this will be an easy one. If anyone could just let me know which, if any, of my statements are true, it would be most appreciated. Thanks, and feel free to elaborate.
     

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  2. jcsd
  3. Oct 16, 2011 #2
    There is a little book 'The Quantum World' by Polkinghorn. Perhaps you find it a good introduction.
     
  4. Oct 16, 2011 #3

    tom.stoer

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    A ket is an abtract notation for a quantum state. You can project it on a bra-basis to get (e.g.) a wave function in position or momentum space

    [itex]\psi(x) = \langle x|\psi\rangle[/itex]

    There are some spezial kets (special basis) with a physical interpretation; |x> is a position-eigenket, |p> is a momentum-eigenket, |E> is an energy-eigenket of a Hamiltonian H

    For the wave functions you get

    [itex]\psi_y(x) = \langle x|y \rangle = \delta(x-y)[/itex]

    [itex]\psi_p(x) = \langle x|p \rangle = e^{ipx}[/itex]

    i.e.a delta-function as a position-eigenfunction in position space and a plane-wave as a momentum-eigenfunction in position space, respectively.
     
  5. Oct 16, 2011 #4
    Here is an analogy:

    A vector v can be represented by its components in some coordinate basis.

    e.g. v = ivx + jvy +kvz

    Similarly a quantum state ψ can be represented by a ket |ψ> which can be expanded in terms of its components in some basis |k>.

    i.e. |ψ> = Σ|k><k|ψ>
     
  6. Oct 16, 2011 #5
    Okay, I did a little more research and found something, which I hope to be true. Lemme revise my earlier question according to the attached.
     

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  7. Oct 17, 2011 #6

    Fredrik

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    Regarding your questions about time:

    No, there's isn't one Hilbert space for each value of t, or one inner product for each value of t. There's one Hilbert space that contains all the state vectors, and it's equipped with an inner product. I suggest that you think of it this way: Define [itex]\psi_t(\vec x)=\psi(\vec x,t)[/itex]. For each [itex]\vec x[/itex], [itex]\psi_t:\mathbb R^3\rightarrow\mathbb C[/itex] is a state vector. The function [itex]t\mapsto\psi_t[/itex] is a curve in the Hilbert space of state vectors. If we define [itex]\psi[/itex] such that each [itex]\psi_t[/itex] is normalized, it's actually a curve on the unit sphere in that Hilbert space. The Schrödinger equation tells you how to find that curve, given a single point on the curve.

    There's a minor inaccuracy in what I just said. The "inner product" on the set of square-integrable functions is actually a "semi-inner product". See posts #4 and #6 here if you care about those details.
     
  8. Oct 17, 2011 #7
    Okay, cool, this is great.
     

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  9. Oct 19, 2011 #8
    Oh, sorry I didn't mention it, but the pic on my last reply contains further questions if anyone can answer them. Especially the one at the end. Thanks.
     
  10. Feb 2, 2012 #9
    Re: Reply #3 <x|y> and <x|p> are interpreted as inner products correct? Thus, I assume these can expanded in terms of integrals involving delta functions. <x|p> is easy to figure out but I don't understand how to get δ(x-y) for <x|y>.
     
  11. Feb 2, 2012 #10

    bhobba

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    You have hit upon something actually quite difficult here - the use of Delta Functions in QM. Basically they do not belong to a Hilbert Space but to something called a Rigged Hilbert Space and coming to grips with that is a whole other story - you need to be very adept at analysis and delve into tombs like Gelfand and Vilenkin - Generalized Functions - tough going even for guys like me into that sort of stuff - but it must be said not impossible if you persevere.

    At the beginner level however its best to simply view Delta Functions as an ordinary function that for all practical purposes behave like a delta function - but really isnt. Under that view <x|y> = integral delta (x-x') delta (y-x') dx' = delta (x-y).

    Basically the idea of a Rigged Hilbert Space is this. In a Hilbert space H bras and kets can be put into one to one correspondence ie its vectors and the linear functions defined on those vectors - called its dual -can be put in one to one correspondence. But if instead of a Hilbert space (which consists of all square integrable functions) you consider the space of all test functions (which is a subset of square integrable functions) then its dual T* (ie the space of all linear functions defined on that space) is much larger than H and includes stuff like the delta function. So what you have is T subset of H subset of T* - which is called a Gelfland Triple and the basis of Rigged Hilbert Spaces.

    For a bit of an introduction to Rigged Hilbert Spaces check out (which also explains what a test function is if you do not know it):
    http://en.wikipedia.org/wiki/Distribution_(mathematics [Broken])
    http://www.abhidg.net/RHSclassreport.pdf [Broken]

    As to its physical meaning that is contained in a theorem you wont find mentioned that often in textbooks but IMHO is very fundamental - Gleason's Theorem which basically says Born's rule is the only way you can define probabilities on a Hilbert Space.
    http://en.wikipedia.org/wiki/Gleason's_theorem

    In fact from the assumption that QM is theory whose pure states are complex vectors all of QM basically follows. Check out:
    http://arxiv.org/pdf/quant-ph/0111068v1.pdf

    Its slightly different to the way I mentioned (from Gleasons Theorem) but it gives you the gist.

    Thanks
    Bill
     
    Last edited by a moderator: May 5, 2017
  12. Feb 3, 2012 #11
    Thank you very much for your response. I had done a little research prior to my post and concluded that this was indeed a complicated issue. I think I will stick with the beginner view for my purposes, given that I am re-learning a lot of this stuff 25 years after leaving college! Thanks again.
     
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