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A Integral ##\int_{-1}^{1} [P_{l}^{m}]^2 ln [P_{l}^{m}]^2 dx##

  1. Nov 1, 2017 #1
    Hi, todos:

    Do you know how to calculate the definte integral for Integral for ##\int_{-1}^{1} [P_{l}^{m}]^2 \ln [P_{l}^{m}]^2 dx##, where ##P_{l}^{m} (x)## is associated Legendre functions. Thanks for your time and help.
     
    Last edited by a moderator: Nov 1, 2017
  2. jcsd
  3. Nov 1, 2017 #2

    PeterDonis

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    Is there a particular reason why you posted this in the Quantum Physics forum instead of the math forum?
     
  4. Nov 2, 2017 #3

    Demystifier

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    The integral seems to be a kind of entropy in a specific quantum state. But note that it is not von Neumann entropy. Let me explain. Let ##\psi(x)## be a wave function. The integral above can be interpreted in terms of entropy defined as
    $$S=-\int dx \rho(x){\rm ln}\rho(x)$$
    where
    $$\rho(x)=\psi^*(x)\psi(x)$$
    The von Neumann entropy is
    $$S_{vN}=-\int dx \int dx' \rho(x,x')({\rm ln}\rho)(x',x)$$
    where
    $$\rho(x,x')=\psi^*(x')\psi(x)$$
    Note that ##({\rm ln}\rho)(x',x)\neq {\rm ln}(\rho(x',x))##. Clearly, the two entropies are different. It can be shown that ##S_{vN}=0## (since wave function represents a pure state), while ##S>0## (unless ##\psi(x)## is something like a ##\delta##-function).
     
    Last edited: Nov 2, 2017
  5. Nov 2, 2017 #4
    This function is used widely in physics. Physicist is familiar with this topic not the mathematician.
     
  6. Nov 2, 2017 #5
    Thanks. The problem is how to calculate it? Or there is no analytical result?
     
  7. Nov 3, 2017 #6

    Demystifier

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    I don't know how to calculate it analytically. Perhaps a mathematician might know. But am I right that your physical motivation is to compute a kind of entropy?
     
  8. Nov 4, 2017 #7
    right.
     
  9. Nov 4, 2017 #8
    Dear Demystifier,

    Byproduct, I have another question. The wave function is position space is normalized. However, after Fourier transform using mathematica, why some of wave function in momentum is normalized, but some of them are not normalized. We have to renormalize them once again. I use
    \psi(p)=\frac{1}{(2\pi)^{3/2}}\int\psi(r) \exp{- i p\dot r} d^3 r.
    Thanks.

    I doubt there is some problem in mathematica 10.
     
  10. Nov 6, 2017 #9

    vanhees71

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    If you mean
    $$\tilde{\psi}(\vec{p})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{1}{(2 \pi)^{3/2}} \exp(-\mathrm{i} \vec{x} \cdot \vec{p}) \psi(\vec{x}),$$
    then
    $$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} |\psi(\vec{p})|^2=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(\vec{x})|^2,$$
    because with the chosen prefactor the Fourier transformation is a unitary mapping ##\mathrm{L}^2(\mathbb{R}^3) \rightarrow \mathrm{L}^2(\mathbb{R}^3)##.
     
  11. Nov 7, 2017 #10

    DrDu

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    Maybe you can express one of the outer P's in terms of its derivative and perform a partial integration to get rid of the ln?
     
  12. Nov 7, 2017 #11

    MathematicalPhysicist

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    Did you mean ##\ln \rho(x',x) \ne \ln \rho(x)##, cause I don't understand mathematically why what you wrote is not equal, it doesn't matter if you evaluate ##\ln \rho## at ##(x,x')## inside the log or outside the log.

    I don't understand this inequality you wrote.
     
  13. Nov 8, 2017 #12

    Demystifier

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    You must think of ##\rho(x',x)## as matrix elements ##\rho_{x'x}## of the matrix ##\rho##. Similarly, ##\rho(x)=\rho(x,x)## are diagonal matrix elements ##\rho_{xx}##. From that point of view we have, for instance,
    $$(\rho^2)_{x'x}=\sum_{x''}\rho_{x'x''}\rho_{x''x}\neq (\rho_{x'x})^2$$
    Similarly,
    $$({\rm ln}\rho)_{x'x}\neq {\rm ln}(\rho_{x'x})$$
    To define logarithm of a matrix you must expand the logarithm into a Taylor series, which reduces the logarithm of a matrix to the well defined multiplication and summation of matrices.
     
  14. Nov 8, 2017 #13

    MathematicalPhysicist

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    Ah , OK agreed.
     
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