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I Is dψ/dx zero when x is infinite in QM?

  1. Sep 25, 2016 #1
    In QM, we all know that the wavefunction ψ is zero when x is infinite. However, Is dψ/dx also zero when x is infinite? And the d2ψ/dx2?
     
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  3. Sep 25, 2016 #2
    the wave function of a QM system vanishes at the infinite distances but its a well behaved function and while asymptotically going to zero it must go smoothly means its slope i/e. the first differential should tend towards zero ...when we plot it it goes very nearly parallel to x axis and slowly turning closer to x-axis.
    my hunch is that the first and 2nd order differential must be going smoothly to zero to avoid any singularity or kinks in the wave functiion.
     
  4. Sep 25, 2016 #3

    Vanadium 50

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    If y is a constant (and zero is a constant) for large x. what does that say about dy/dx?
     
  5. Sep 25, 2016 #4

    PeroK

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    ##f(x) = \frac{1}{x} sin(x^2)##

    is an example of a function where ##f(x) \rightarrow 0## as ##x \rightarrow \infty## but ##f'(x)## does not tend to 0 as ##x \rightarrow \infty##.

    In fact, there is a monotonic function ##f## with this property. See:

    https://www.physicsforums.com/threa...rexample-challenge.869304/page-3#post-5459479

    So, functions like these have to be excluded from what can be considered a valid QM wavefunction.
     
  6. Sep 25, 2016 #5

    Vanadium 50

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    You got me there - but functions if that form will not be normalizable.
     
  7. Sep 25, 2016 #6

    rubi

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    That is a common misconception. In QM, a wave functions can be any square-integrable function. However, there are square-integrable functions that don't vanish as ##x\rightarrow\infty##. An example would be ##f(x)=\sum_{n=0}^\infty \chi_{[n,n+2^{-n}]}##, where ##\chi_B(x)=0## for ##x\neq B## and ##\chi_B(x)=1## for ##x\in B##. One finds that ##\int_{\mathbb R} |f(x)|^2\mathrm d x = \sum_{n=0}^\infty 2^{-n} = 2##. You can also find counterexamples for ##f'##. In rigorous proofs in QM, one must work a little harder to work around such issues.
     
  8. Sep 25, 2016 #7
    Is that function a possible solution of the Schroedinger equation?
     
  9. Sep 25, 2016 #8

    PeroK

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    That function needs to be patched up for small ##x## but otherwise it's square integrable as it looks like ##1/x## for large ##x##.
     
  10. Sep 25, 2016 #9

    rubi

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    Wave functions are never solutions to the Schrödinger equations. Instead, solutions to the Schrödinger equation assign a wave function to every time ##t##. If you have any admissible wave function ##\psi(\vec x)## (i.e. a square integrable function), you can construct a solution to the Schrödinger equation by setting ##\Psi(\vec x, t) := \left(e^{-\frac{\mathrm i}{\hbar}t\hat H} \psi\right)(\vec x)##. So Perok's wave function (if regularized properly at ##x=0##) generates an admissible solution to the Schrödinger equation.
     
  11. Sep 25, 2016 #10

    Nugatory

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  12. Sep 25, 2016 #11

    dextercioby

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    While it is undoubtedly true that there are L2 (R) functions not going to 0 as the argument goes to infinity, thus these could be called wavefunctions, they are hardly of any use. The only really useful wavefunctions are those in the domain of the operators (coordinate, momentum, Hamiltonian, angular momentum). The Born-Jordan commutation relation in L2 (R) pretty much forces the fact that the only admissible wavefunctions are the Schwartz test functions. On this space the 2 operators are essentially self-adjoint and its rigging of L2 (R) solves the spectral problem for the operators.
     
  13. Sep 25, 2016 #12
    So it's a starting point? By the time it gets to infinity will it still not be a solution of it?
     
  14. Sep 25, 2016 #13

    rubi

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    The function ##\Psi(\vec x,t)## is always a solution, no matter what ##\psi(\vec x)## is. And ##\Psi(\vec x, 0) = \psi(\vec x)##, so there exists a solution to the Schrödinger equation, which is equal to Perok's function at ##t=0##.
     
  15. Sep 25, 2016 #14
    Oh yes, I see.
     
  16. Sep 25, 2016 #15

    dextercioby

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    But is that "wavefunction" really in the domain of a certain necessarily (essentially) self-adjoint Hamiltonian ? If not, your whole argumentation is moot.
     
  17. Sep 25, 2016 #16

    rubi

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    It doesn't need to, since ##e^{-\frac{\mathrm i}{\hbar} t\hat H}## is a bounded operator and thus has an extension to the whole Hilbert space by the BLT theorem.

    (Of course, such solutions are still physically not important in practice. :wink:)
     
  18. Sep 25, 2016 #17

    PeroK

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    One problem with the function I gave is that, if it is the solution at ##t = 0##, then

    ##m \frac{d \langle x \rangle}{dt} \ne \langle p \rangle##

    And, something like:

    ##f(x) = \frac{1}{x}sin(x^3)##

    would be worse, as the modulus of the wavefunction does not remain at a constant value of ##1## after ##t = 0##.

    So, the basic theorems of QM break down for such wavefunctions (assuming they are a solution to the Schroedinger equation at ##t=0##).
     
  19. Sep 25, 2016 #18

    Vanadium 50

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    OP, just so you don't get confused by the advanced-level arguing. The answer is "yes".
     
  20. Sep 25, 2016 #19

    dextercioby

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    While it's again beyond any doubt that the unitary evolution applies in principle to any wavefunction, because any unitary operator is bounded, Stone's theorem which advocates the existence of a self-adjoint Hamiltonian will not render this Hamiltonian necessarily bounded, thus as per Hellinger-Toeplitz theorem, the Hamiltonian's domain will be smaller than the domain of its exponential. One more time: e^(itH) is unitary, if H is self-adjoint. I cannot find any self-adjoint Hamiltonian which has that wavefunction in its domain. Unitary time evolution in Quantum Mechanics makes sense only for wavefunctions in the self-adjointness domain of the Hamiltonian.
     
  21. Sep 25, 2016 #20

    dextercioby

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    How did you compute the averages (expectation values) and why do you expect the Ehrenfest's theorem to hold in the first place?
     
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