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Negative variance of an observable quantity

  1. Nov 16, 2013 #1
    Quantum mechanics has a well-known procedure for evaluating the expectation value of an observable quantity in a given quantum state. First one must obtain the quantum operator O that is associated with the observable quantity. Then the rule for computing the expectation value is: Apply O to the wave function psi, multiply the result with psi*, and finally integrate over spatial variables.

    The procedure described above can be extended to evaluate the variance of an observable quantity. Now the rule is: apply the operator O twice in succession to the wave function psi, multiply with psi* and integrate over space. From this result, subtract the square of the expectation value.

    All this is in clear analogy with the standard rules of statistics. However, in QM things are less clear-cut. For example, it can be demonstrated that the variance may acquire a negative value!

    My question is: How should one interpret a negative value for the variance of an observable quantity?
  2. jcsd
  3. Nov 16, 2013 #2


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    No, this is false. In QM as well, the variance of an observable is nonnegative.

    <A2> = <ψ|A2|ψ> = <ψ|A(Aψ)> = <Aψ|Aψ> ≥ 0
  4. Nov 16, 2013 #3


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    Last edited: Nov 16, 2013
  5. Nov 16, 2013 #4
    Okay, thank you! I think I have a counter-example, but I could be wrong of course.

    Sorry, but I don't understand your third equality. Applying the operator A twice to the wave function on the right may well result in something that correlates negatively with the wave function on the left. However, if you are allowed to shift one operator to the left, then the result becomes indeed positive (or better non-negative) as you conclude.

    For example, let psi(x) = sin(x) and operator A is the first derivative with respect to x.
    Then your third term equals -sin(x)*sin(x) whereas your fourth term equals cos(x)*cos(x).
    Last edited: Nov 16, 2013
  6. Nov 16, 2013 #5
    All observables are Hermitian; the definition of a Hermitian operator is that it satisfies

    ##\langle \phi | A \psi \rangle = \langle A \phi | \psi \rangle##

    for any states ##| \phi \rangle## and ##| \psi \rangle##. As Bill_K showed, this immediately implies that ##A^2## is "positive semidefinite," i.e., that its expectation value is always nonnegative. This is intuitive if you know that Hermitian operators have only real eigenvalues. Squaring the operator squares the eigenvalues, so the squared operator has only real nonnegative eigenvalues.
  7. Nov 16, 2013 #6
    Thank you. I will have to check whether my example is in accordance with the Hermitian rules.
  8. Nov 16, 2013 #7


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    You might also want to check whether your example is indeed a normalizable wave function. :devil:

    If it's not, then it's not in the usual Hilbert space and everything becomes far more challenging...
  9. Nov 16, 2013 #8
    I spent a considerable amount of time checking my calculations. There are things I can't get right.

    Observable quantity: Kinetic energy
    Associated operator: constant * Laplace operator (L)
    Wave function: ground state of Hydrogen atom in spherical coordinates (psi)

    A simple example, everything is real. Now according to the Hermitian postulate (psi, LL psi) = (L psi, L psi). But this can not possibly be true! Here is just of many reasons: L takes first and second derivative with respect to r, LL third and fourth derivative. So if psi is a polynomial of degree 1 or 2, then the left hand side is zero, whereas the right hand is some positive function.
  10. Nov 17, 2013 #9


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    You're hitting well-known problems associated with unbounded operators and continuous spectra.

    If you want more explicit help, you'll have to show your work... :wink:

    Firstly, write down your example ##\psi## explicitly (here) as such a polynomial. Then calculate ##(\psi,\psi)## explicitly (here). I presume your inner product is just an integral? What do you get?
  11. Nov 17, 2013 #10
    Okay, thank you very much for your offer! Here is my calculation. For clarity, I will omit all physical parameters and constants such as Planck's constant, electron mass, Bohr radius and pi.

    psi(r) = exp(-r) ................................... [wave function; the n=1 orbital of the Hydrogen atom]
    L = (2/r)*(d/dr) + d^2/dr^2 .............. .. [radial part of the Laplace operator in spherical coordinates]
    LL = (4/r)*(d^3/dr^3) + d^4/dr^4 ........ [radial part of the Laplace operator applied twice in succession]

    I calculate the expectation value of the Kinetic energy squared in two ways, which should be identical [if the Hermitian criteria are met]. First I give the result of the projection of the wave functions (f and g); next the result of multiplying by (r^2)*dr and integrating over the spatial variables, yielding F and G. Note that a * means multiplication (not complex conjugate!).

    [1] f(r) = psi * (L(L psi)) = (-4/r + 1) * exp(-2*r)
    F = integral (0 to inf) f(r) * r^2 * dr = -3/4

    [2] g(r) = (L psi) * (L psi) = (4/r^2 - 4/r + 1) * exp(-2*r)
    G = integral (from 0 to inf) g(r) * r^2 * dr = +5/4

    I chose method [1], but this leads to a negative value. Why? Method [2] should be identical, but in fact leads to a different result (positive, and therefore perhaps the correct result?).
  12. Nov 17, 2013 #11


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    No, when you calculate LL, there's a couple of terms you forgot. (... d2/dr2 )(2/r d/dr + ...) when expanded produces d/dr and d2/dr2 terms in addition to the ones you have.
  13. Nov 17, 2013 #12
    Did I make a mistake? I believe I performed the calculation with considerable care. The result for LL is what I got after expanding all the terms; in the final result most of the terms (those with the lower order derivatives) cancel. I also performed the calculation by first computing L(psi) explicitly and only then applying the second operator L.
  14. Nov 17, 2013 #13


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    Sorry, your expression is right. :redface:
  15. Nov 17, 2013 #14
    Thank you! The mystery continues..... :smile:

    In my calculations I obtained the general expression for applying N consecutive (radial) Laplace operators:

    L^N = (2*N/r) * d^(2*N-1)/dr^(2*N-1) + d^(2*N)/dr^(2*N).
    Last edited: Nov 17, 2013
  16. Nov 17, 2013 #15


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    Well, congratulations! Your calculations of those integrals above look correct to me. :confused:
    So then it took me most of today to figure out WTF is going on here... :blushing:

    Here's an exercise to help you resolve the puzzle:

    Exercise: Determine (carefully!) whether ##L## is/isn't self-adjoint. I.e., complete the following:
    (\phi, L\psi) ~\equiv~ \int_0^\infty\!\! dr\; r^2 \phi^* (2r^{-1}\partial_r
    ~+~ \partial^2_r ) \psi ~=~ \cdots ~,
    $$where ##\phi,\psi## are functions of ##r##. (Hint: use integration by parts twice.) What conditions on the wavefunctions must be satisfied for ##L## to be self-adjoint? I.e., under what conditions is it true that:
    $$(\phi, L\psi) ~=~ (L\phi, \psi) ~~?$$
    And btw, PLEASE learn how to use Latex on this forum. :grumpy:
  17. Nov 18, 2013 #16
    Thank you very much for your time and assistance!

    I followed your advice and performed partial integration. The result I get is that the Laplace operator indeed satisfies the Hermitian criterium. However it is necessary that there is no contribution from the boundary terms. These terms can be combined into a single function f.

    Define: f(r) = (r^2) (phi* psi' - psi phi*').

    [* = complex conjugate; ' = derivative with respect to r]

    The condition is that f(inf) - f(0) must be zero.
  18. Nov 18, 2013 #17


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  19. Nov 18, 2013 #18
    I will continue the calculation started in my previous post. We are ready to apply the Hermitian rule derived in my previous post to the case discussed in this thread. So we set phi = exp(-r) and psi = L(phi) = (1 - 2/r)*exp(-r). It turns out that the
    r -> inf boundary term vanishes, but the r=0 term doesn't. It picks up a contribution equal to -2.

    Therefore, in this particular case, the Hermitian rule adjusts to:

    (phi, L psi) = (L phi, psi) - 2 or equivalently (phi , LL phi) = (L phi, L phi) - 2.

    This explains the difference between the two integrals, evaluated as -3/4 and +5/4 respectively!
    An elegant line of reasoning, leading to a surprising and thought provoking result!
    Last edited: Nov 18, 2013
  20. Nov 18, 2013 #19


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    Correct so far, afaict. For wavefunctions which vanish at infinity like ##e^{-r}##, i.e., faster than any power of ##r##, the condition is indeed:
    \lim_{r\to 0} \Big( r^2 (\phi^* \psi_r - \phi^*_r \psi) \Big) ~=~ 0 ~,~~~ \forall \phi, \psi ~,
    $$ (where my subscripts denote a partial derivatives). However, this must hold for arbitrary ##\phi,\psi##, so probably a stronger condition that each term vanish separately is justified.

    Indeed. Here's some more food for thought...

    Back in post #9, I made the cryptic remark:
    The problems that arise in this case can broadly be expressed in the Hellinger--Toeplitz theorem:

    Wiki's explanation is rather brief, so a Functional Analysis textbook is needed to delve deeper. Kreyszig's textbook on FA is one that's reasonably friendly to physcists and engineers, though still quite difficult. Typically, most QM textbooks don't explore such arcane results of Functional Analysis in much detail. The better texts might make a brief remark about how one must be careful about "domains of definition" and "domains of self-adjointness" when working with such operators.

    But this still doesn't explain thoroughly why your kinetic energy operator behaves in such a recalcitrant manner for this (supposedly basic) problem of the nonrelativistic H-atom. Naively, one might complain: "well, I solved the Schrodinger equation and normalized the resulting wavefunctions, so what's gone wrong??"

    If you'd like to explore further, try this:

    Exercise: Repeat the earlier self-adjointness exercise, but this time for:
    (i) the potential energy operator ##V## associated with this problem,
    and then,
    (ii) the full Hamiltonian ##H := L+V## .
    Last edited: Nov 18, 2013
  21. Nov 18, 2013 #20
    I think it is slightly off the topic Mandragonia eventually wanted, but you might find the Wigner function interesting. It is basically an attempt to provide a probability distribution in phase space, but in some regions associated with unphysical states, the function takes negative values.
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