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The reciprocal expectation value

  1. Dec 27, 2009 #1
    I am aware of the expectation value [tex]\left\langle\ r \right\rangle[/tex]. But I was wondering what is physically meant by the expectation value: [tex]\left\langle\frac{1}{r}\right\rangle[/tex]

    The reason I am asking is because calculating this (reciprocal) expectation value for the 1s state of hydrogen, one obtains [tex]1/a_{o}[/tex]
    This result happens to be the reciprocal of the most probable value of r for the 1s state of hydrogen. [but expectation values gives the mean value of the quantity working on, so there must be a coincidence occurring here]. I hope someone can enlighten me here.
     
  2. jcsd
  3. Dec 28, 2009 #2
    I would say it's coincidence off the top of my head. There was a recent discussion on the calculation for the expectation value of <r>, which can be found here:

    https://www.physicsforums.com/showthread.php?t=362988

    but the main result is that <r>=.5*a0[3n2-l(l+1)]

    The expectation value <1/r> is much easier to calculate than <r>, and it's [tex]\frac{1}{n^2a_0}[/tex].

    Since the expectation value of the radius <r> depends on the angular momentum, that means the probability does too. But the expression for <1/r>, oddly enough, doesn't depend on the angular momentum! So the coincidence stops.
     
  4. Dec 28, 2009 #3
    What probability do you mean?
     
  5. Dec 28, 2009 #4
    The probability of finding the electron at a distance 'r'. The expectation value of the radius, denoted <r>, in general, depends on the angular momentum. That means the probability value of the radius, depends on the angular momentum. However, the expectation value of the reciprocal, <1/r>, does not depend on the angular momentum. So there is no longer any equality.

    Maybe the ground state is special, but the relationship you found between the ground state expectation value of the reciprocal and the probability of finding it at a radius r, does not hold for higher states.
     
  6. Dec 29, 2009 #5

    dextercioby

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    It turns out that in quantum mechanics there is a version of the http://en.wikipedia.org/wiki/Virial_theorem" [Broken]. This leads to a computation of [itex] \langle\frac{1}{r}\rangle_{nlm} [/itex] without an integral, since there's a derivation of the hamiltonian eigenvectors without involving any integral of special functions (associated Laguerre functions).

    So to justify the lack of dependence of [itex] \langle\frac{1}{r}\rangle_{nlm} [/itex] on the angular momentum quantum numbers is the same as justifying why the hamiltonian eigenstates are independent of "l" or "m". The latter is purely group theory plus the symmetry of the Coulomb interaction.

    A reference on the quantum virial theorem can be found in Fluegge's exercise book "Practical QM", 2nd volume, problem 151, Merzbacher's QM text, section 8.9, Pauli's classic 1933 text (in which he quotes as the sourse the first ever treatise on quantum mechanics, namely Sommerfeld's), section 4.
     
    Last edited by a moderator: May 4, 2017
  7. Dec 29, 2009 #6
    That's fascinating. The virial theorem is easy to prove classically and says that for inverse potentials, 2<T>=-<V>, or that 2<T>+<V>=0. Of course <V> is negative (so that <T> can be positive), i.e., the potential is attractive, for the virial theorem to be valid.

    For the energy states of hydrogen, you now have:

    <T>+<V>=E
    2<T>+<V>=0

    So that:
    <T>=-E
    <V>=2E

    So the expectation value of [tex]<\frac{-e^2}{4\pi \epsilon_0 r}>[/tex] is 2E and from this you can already see that since E doesn't depend on angular momentum, that <1/r> does not either.

    But trying to derive the exact expression:

    [tex]<\frac{e^2}{4\pi \epsilon_0 r}>=2\frac{m e^4}{32 \pi^2 \hbar^2 \epsilon_{0}^2 n^2}[/tex]


    [tex]<\frac{1}{r}>=\frac{m e^4}{32 \pi^2 \hbar^2 \epsilon_{0}^2 n^2}=\frac{1}{a_0 n^2}[/tex]

    So the classical virial theorem works exactly for quantum mechanics!

    Calculating <1/r> is easy with Laguerre polynomials, since the resulting integral is just the L^2 product. Calculating <1> or <r> is much harder with Laguerre polynomials.

    I know you gave references, but in a nutshell, is the classical virial theorem always valid in quantum mechanics?
     
    Last edited by a moderator: May 4, 2017
  8. Dec 30, 2009 #7

    dextercioby

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    The quantum version of the virial theorem is derived under some rather general conditions, but a rigorous application of it can be found only for bound (as opposed to scattering) states of atomic systems. The H-atom is the perfect example.There's usually no problem with the definition of time derivatives and averages over time for operators, but the computing expectation values for scattering states is a tricky business.

    I think this theorem in QM goes along with Ehrenfest's equalities and if, let's say, one could give a derivation of its classical version in the Hamilton formulation (the widely known derivation posted on Wikipedia uses the Newton formalism of classical dynamics), then the passage to QM would be immediate and no proof of the quantum version would be necessary anymore.
     
  9. Dec 31, 2009 #8
    thanks for the discussion, was very helpful!
     
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