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I Representation between State Vector & Wave Function

  1. Mar 2, 2016 #1
    By the Principle of Superposition, a state vector can be defined as pic.01

    also, the state vector can represent a wave function in a continuous case as pic.02

    My (1) question is, in pic.03, why the state vector can be pulled out from the integral???

    I have an idea but I think it should be wrong:
    ------------
    As shown in pic.02, the state vector is a set of values that each element is the value of the wave function, e.g. phi(0), phi(0.002), etc.

    Thus, it can be considered as a constant since every elements will never change

    Therefore, it can be pulled out from the integral
    -------------

    However, this idea is fail to explain in the observable case.

    Considering, in pic.04, question (2), Q is an operator which contain a momentum component, a differential operator. If the state vector is constant, it should give zero value when a momentum operator act on it and the Dirac representation will be failed.

    So, if it is not a constant, how can it be pulled out from the integral????????
    ========================

    And my (2) question is that, in pic.04,

    step 4 to step 5,

    why the operator Q can skip the bra(x) vector and act on the state vector???????
    =========================

    Thank you so much as I'm new in Quantum Mechanics!
     

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  2. jcsd
  3. Mar 2, 2016 #2

    DrClaude

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    Staff: Mentor

    For a continuous basis such as ##| x \rangle##, it doesn't make sense to express it as a column vector.

    ##| \psi \rangle## is a state vector, corresponding to a ray in Hilbert space. In a sense, it is an abstract construct that has no particular dependence on ##x## or any other variable (except maybe time), so you can take it in and out of the integral as you wish.


    Saying that ##\hat{P}## is a differential operator is incorrect. Only in position representation is it a differential operator, i.e., when you have already projected the state onto the position basis ##\langle x | \psi \rangle##. You could as well express the state vector as
    $$
    | \psi \rangle = \sum_i c_i | p_i \rangle
    $$
    with ##\hat{P} | p_i \rangle = p_i | p_i \rangle##. In the momentum basis, ##\hat{P}## is definitely not a differential operator.

    In summary, you have to be careful to distinguish ##| \psi \rangle## from its various representations.
     
  4. Mar 2, 2016 #3

    blue_leaf77

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    Science Advisor
    Homework Helper

    The correct transition between the position representation and the braket notation in the first line of the integral is actually
    $$
    Q_x\langle x|\psi\rangle = \langle x|\hat{Q}|\psi\rangle
    $$
    where ##Q_x## is the operator ##\hat{Q}## in position representation.
     
  5. Mar 2, 2016 #4

    bhobba

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    Science Advisor
    Gold Member

    Your question is why is the following valid - ∫<x|u><u|y> du = <x|∫|u><u|du |y>

    The rigerous answer lies in an advanced area of mathematics called Rigged Hilbert Spaces:
    http://artssciences.lamar.edu/_files/documents/physics/webdis.pdf

    It is NOT recommended for the beginner.

    Non rigorously its justified by sloppy use of limits
    ∫<x|u><u|y> du = limit Δu → 0 Σ <x|ui><ui|y> Δu = limit Δu → 0 <x|Σui><ui| Δu |y> = <x|limit Δu → 0Σ|ui><ui Δu |y> = <x|∫|u><u|du |y>

    Justifying and giving meaning to each of those steps is what's hard.

    As a build up to understanding this I recommend you become acquainted with distribution theory:
    https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

    Thanks
    Bill
     
    Last edited: Mar 2, 2016
  6. Mar 2, 2016 #5

    vanhees71

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    Science Advisor
    2016 Award

    Your pics are a bit hard to read. It doesn't hurt to learn LaTeX in posting here!

    I answer the question concerning pic 1. Your confusion is due to mixing notations in a quite confusing way. You have to distinguish between abstract vectors, which you write in the Dirac bra-ket notation like ##|\psi \rangle##, and components with respect to a complete orthonormal basis. The mapping is one-to-one, i.e., you can use the completeness relation of the basis,
    $$\sum_n |\epsilon_n \rangle \langle \epsilon_n|=1.$$
    Defining
    $$c_n=\langle \epsilon_n|\psi \rangle$$
    you get
    $$|\psi \rangle = \sum_n |\epsilon_n \rangle \langle \epsilon_n|\psi \rangle=\sum_n c_n |\epsilon_n \rangle.$$
    The scalar product of the state vector with itself, i.e., its norm squared reads
    $$\langle \psi|\psi \rangle = \sum_n \langle \psi|\epsilon_n \rangle \langle \epsilon_n|\psi \rangle=\sum_n c_n^* c_n=\sum_n |c_n|^2.$$
    Now you can use as well the generalized basis of the position vector, which leads to a representation in terms of wave functions,
    $$\psi(x)=\langle x|\psi \rangle.$$
    Then the completeness relation reads
    $$\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|=1,$$
    and the norm squared is thus
    $$\int_{\mathbb{R}} \mathrm{d} x \langle \psi|x \rangle \langle x |\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \psi^*(x) \psi(x)=\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2.$$
    Now, if you have given the wave function you can as well calculate the components ##c_n## with respect to the complete orthonormal basis ##|\epsilon_n##:
    $$c_n=\langle \epsilon_n|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle \epsilon_n|x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x u_n^*(x) \psi(x),$$
    where
    $$u_n(x)=\langle x|\epsilon_n \rangle,$$
    are the wave functions related to the basis vectors.

    The abstract vectors and the components with respect to a basis or the wave functions, which are "components" with respect to a generalized basis are all equivalent descriptions of the same physical situation, encoded in the state vector, which describes a (pure) state of the quantum system, but it's not the same. Using a complete basis with normalizable basis vectors only, you have a description of the Hilbert space of state vectors in terms of square summable complex series, ##(c_n)##, and with the wave functions a representations in terms of square integrable functions. These Hilbert spaces are called ##\ell^2## and ##L^2## respectively. They are equivalent, if they have a countable complete basis, and that's the case for these to Hilbert spaces. The abstract Hilbert space is in some sense an equivalence class of all such concrete realizations, and you should cleanly distinguish these various realizations. Then you'll have much less confusion in dealing with them and switching from one description to the other in concrete problems. That's the strength of using the abstract Dirac formalism: Switching from one to the other description is just a clever insertion of completeness relations at the right places, as in the examples above.
     
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