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I Understanding the wavefunction for a free particle

  1. Sep 6, 2016 #1
    Hi everybody,
    I was reading about the free particle in a textbook and I got confused by the line:
    "If we adopt the convention that k and k are real, then the only oscillating exponentials are the eigenfuntions with positive energy" [Also see the attached picture with the paragraph.]
    What is the rationale for that line? Furthermore, does any body have a link where I can read about those exponentials? I couldn't find them online and I am curious to know how and when they are used to solve the Schrodinger Equation!
     

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  3. Sep 6, 2016 #2

    BvU

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    Insert the ##\psi## in the (time independent) SE and see they satisfy the equation !
     
  4. Sep 6, 2016 #3
    I don't doubt that they satisfy the equation. My question is about the line,
    If we adopt the convention that k and k are real, then the only oscillating exponentials are the eigenfuntions with positive energy"
    Why is there a need for both k and kappa be real? And what are those exponentials e^(+-ikx) and e^(+-kx) called and where can I read more about them?
     
  5. Sep 6, 2016 #4

    DrClaude

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    Because if they are not real, then it is difficult to separate the two cases. For instance, an imaginary κ would correspond to a real k.

    I don't know of any particular name for these exponentials. They come from the mathematics of differential equations.
     
  6. Sep 6, 2016 #5

    PeroK

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    If you let ##k## be complex, then you get all the solutions with ##k^2 = -\frac{2mE}{\hbar^2}## hence:

    ##k = \pm i \frac{\sqrt{2mE}}{\hbar}##

    And:

    ##\psi(x) = Ae^{\pm i \frac{\sqrt{2mE}}{\hbar}x}##

    So, it amounts to the same thing. Taking ##k## as real initially is just a bit of a shortcut, based to some extent on knowing the solution in advance!
     
  7. Sep 6, 2016 #6

    vanhees71

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    What's the source? It's very inaccurate.

    The rationale is the following: In wave mechanics the vectors of the quantum mechanical Hilbert space are represented by square integrable functions ##\psi:\mathbb{R}^3 \rightarrow \mathbb{C}##. The position vector ##\vec{x}## is represented by the multiplication of the wave function with ##\vec{x}##, and momentum by ##\hat{\vec{p}}=-\hbar \vec{\nabla}##.

    These operators are defined on a dense subspace of these Hilbert space of square-integrable functions. Now you can ask for the generalized eigenfunctions of momentum. They obey the equation
    $$\hat{\vec{p}} u_{\vec{p}}(\vec{x})=\vec{p} \psi(\vec{x}).$$
    The operators have to be understood as self-adjoint operators, and their eigenvalues are thus real. For real ##\vec{p}## the solution of the eigenvalue problem reads
    $$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi \hbar)^{3/2}} \exp\left (\frac{\mathrm{i} \vec{x} \cdot \vec{p}}{\hbar} \right).$$
    I've normalized this generalized functions "to the ##\delta## distibution", i.e.,
    $$\langle u_{\vec{p}'}|u_{\vec{p}} \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}'}^*(\vec{x}) u_{\vec{p}}(\vec{x})=\delta^{(3)}(\vec{p}-\vec{p}').$$
    These generalized eigenfunctions do NOT represent states since they are not square integrable, but you can describe all wave functions in terms of generalized momentum eigenstates:
    $$\psi(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}} (\vec{x}) \tilde{\psi}(\vec{p}),$$
    where ##\tilde{\psi}(\vec{p})## is a square integrable function. Then also ##\psi(\vec{x})## is a square integrable function and represents a state of the particle. For the given wave function ##\psi(\vec{x})## you find the ##\tilde{\psi}(\vec{x})## by the inverse transformation (it's a Fourier transformation!):
    $$\tilde{\psi}(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u^*(\vec{p})(\vec{x}) \psi(\vec{x}).$$
     
  8. Sep 6, 2016 #7

    BvU

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    I agree the wording is odd. The time-independent SE leads to a time part of the solution of the time-dependent SE of the form ##\exp({-iEt/ \hbar})##. The continuity equation forces ##(E-E^*) = 0## so E is real.

    Then: with the most general form, a complex ##e^{{\bf \alpha} x} ## where ##{\bf \alpha} = \kappa + ik \ \ ## (##\kappa## and ##k## real), you get from the time-independent SE: ##\ \ -{\hbar^2\over 2m}{\bf\alpha}^2 = E\ ## . Positive, real-valued E requires ##\kappa = 0##.

    As vanHees points out, these are solutions, eigenfunctions of the Hamiltonian. They are not physically realizable states (not normalizable). But that is (I expect) a subject further on in your curriculum.
     
  9. Sep 7, 2016 #8

    vanhees71

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    That the plane waves are not representing states, is a very important point. A lot of confusion arises from not making this very clear in the very first encounter of these generalized solutions of the Schrödinger equation!
     
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