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I What is the origin of Van der Waals force?

  1. May 25, 2017 #1

    goodphy

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    Hello.

    What I know about the Van der Waals force is that it comes from an instantaneous electronic cloud flucutation of the netural atom. This instantaneous electric dipole of the atom attracts electrons in nearby neutral atoms so other electric dipoles are induced on those atoms. As a result, the atoms are attracting each other. This is what we call the Van der Waals force.

    My question is what is the origin of the instantanous electron cloud fluctuation of the atom? Is it from the uncertainty principle, saying the position and momentum of electrons are not simultaneously measured?
     
  2. jcsd
  3. May 25, 2017 #2
    It is due to the varying electron density that the force arises of.
    With movement of the electrons the density changes which leads to the electric polarization.
     
  4. May 25, 2017 #3

    blue_leaf77

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    As a matter of convention, actually Van der Waals force is traditionally defined as the force that binds a molecule with another and is not used for addressing the force between atoms forming a molecule, nevertheless the two kind of forces are similar in nature in that they are a manifestation of Coulomb forces between nuclei and electrons.
    Some sources usually mention the origin of Van der Waals force as the formation of time-dependent/momentary dipole moment formed in the participating atoms or molecules however this is an incomplete explanation without mentioning the reason why a time-dependent dipole moment arises. When an atom or molecule is isolated in space, they should be in the ground state which is a stationary state. Any physical quantity measured under stationary state should be time-independent.
    However when a perturbation comes about such as external electric field, magnetic field, or an approaching atom and molecule then the time dependent effect can take place during which the state of the atom or molecule changes to that which is not a stationary one. In this moment, a time-dependent dipole moment arise in the two approaching atoms or molecules and depending on the properties of the two atoms or molecules they can bind together to form a bigger quantum system.
    In short, the origin of the instantaneous electron cloud fluctuation is the presence of another atom or molecule.
     
  5. May 25, 2017 #4

    goodphy

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    Hello. Thank you for giving me the explanation.

    I thought the electron cloud fluctuation occurs even at completely isolated ground state neutral atom. However, you said such a fluctuation occurs only when there is external interaction with it. So, for the isolated atom in the ground state, the electron cloud is stationary?
     
  6. May 26, 2017 #5

    blue_leaf77

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    Yes. This concept is related with the Hamiltonian (energy) operator being the symmetry operator related to time (for now neglecting QED effect). If initially the atom is in the ground state and the atom is alone in the space so that that atom can be treated as an entire system, then the energy should be conserved with time, i.e. it will stay in the ground state forever.
     
  7. May 26, 2017 #6

    DrDu

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    Yes, these fluctuations occur also in an isolated atom in its ground state. The point is, that while in an eigenstate the expectation value of the dipole moment is independent of time i.e. ## \langle d (t)\rangle =\langle d(0) \rangle=\bar{d} ##, there are nevertheless nonvanishing fluctuations, i.e. ##\langle (d(t)-\bar {d})^2\rangle > 0## and also the correlation function ## \langle(d(t)-d(0))^2\rangle =f(t)## is not constant.
     
  8. May 26, 2017 #7

    blue_leaf77

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    Is ##d(t) = U^\dagger (t) d(0) U(t)##, i.e. the dipole moment operator in Heisenberg picture? If yes, I can't see how ##\langle(d(t)-d(0))^2\rangle ## depends on time in a stationary state.

    EDIT:
    Yes the correlation function above has time-dependent expectation value which is caused by the fact that it's actually an explicitly time-dependent operator. Previously, I think I was too careless in having been able to show that this correlation function can be factorized as ##U^\dagger(t) O U(t)## where O is a time-independent operator (did the calculation in my head), which led me think that its expectation value was under no circumstances time-dependent in a stationary state. However I made a mistake as one can easily show by expanding the power.
     
    Last edited: May 26, 2017
  9. May 26, 2017 #8

    Demystifier

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    In quantum mechanics, one should not think of "fluctuations" as changes in time. Fluctuations are just uncertainties. An observable ##A## fluctuates if
    $$\langle A^2\rangle - \langle A \rangle^2 \neq 0.$$
    In general, it does not need to be time dependent.
     
    Last edited: May 26, 2017
  10. May 26, 2017 #9

    Demystifier

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    Consider, for instance, the vacuum ##|0\rangle## of the harmonic oscillator and consider the (rescaled) position observable
    $$x(t)=ae^{-i\omega t}+a^{\dagger}e^{i\omega t}$$
    Using ##a|0\rangle=0## and ##\langle 0|aa^{\dagger}|0\rangle=1##, it is not difficult to obtain
    $$\langle 0|[x(t)-x(0)]^2|0\rangle=2(1-\cos\omega t)$$
    It is stationary in the sense that it is periodic in time.
     
  11. May 26, 2017 #10

    blue_leaf77

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    Actually, this is the first time I see the expression for position operator as given above applied for a harmonic oscillator potential. However, it does remind me of the electric field operator in the topic of photon quantization, do you possibly refer to this?
     
  12. May 26, 2017 #11

    Demystifier

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    In a sense yes, electric field (or any other free field) can be thought of as a collection of many harmonic oscillators.

    For a simple harmonic oscillator analogue of Casimir force (which is just a manifestation of van der Waals force) you might want to look at
    https://arxiv.org/abs/1702.03291
     
    Last edited: May 26, 2017
  13. May 26, 2017 #12

    blue_leaf77

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    I see, so in post #9 the operator ##x(t)## is an operator which inherently depends on time just like the classical EM wave when brought in to the semi-classical theory of light matter interaction. However as far as I know, the dipole moment operator is proportional to the ordinary position operator ##\mathbf r## (not a function of time), this is why some molecules e.g. symmetric molecules do not have permanent dipole moment (if this operator were defined to be inherently dependent on time then this statement would be false).

    Note: I think somewhere in my previous posts I said that the expectation value of all observables are independent of time when measured in stationary state. I admit this was an overgeneralization for I forgot that operators which genuinely time-dependent will still have time-dependent expectation value.
     
  14. May 26, 2017 #13

    Demystifier

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    It is a function of time in the Heisenberg picture. In the Schrodinger picture it is not a function of time, but then the state is time dependent because even a stationary state depends on time as ##e^{-i\omega t}##. In each picture, the final result is that an expectation value in a stationary state can depend on time.

    Note also that this time dependence has nothing to do with quantum fluctuations. This time dependence is deterministic and predictable, while quantum fluctuations are random and unpredictable.
     
  15. May 26, 2017 #14

    Demystifier

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    This is a false reasoning. My operator ##x(t)## above has an dependence on time, yet you can easily check that
    $$\langle x(t) \rangle =\langle 0|x(t)|0\rangle=0$$
    does not depend on time. Likewise, a symmetric molecule in a stationary state has zero (and hence time-independent) expectation of dipole moment, yet the dipole-moment operator in the Heisenbeg picture is non-zero and depends on time.

    The moral is, when you say "dipole moment", you need to specify do you mean the expectation value of dipole moment, or dipole moment operator, or dipole moment eigen-value. (In addition, in hidden-variable interpretations such as Bohmian mechanics, there is also a fourth possible meaning, namely the hidden actual value of dipole moment.) And when you say "dipole moment operator", you need to specify do you mean the operator in the Heisenberg picture, or operator in the Schrodinger picture, or operator in the Dirac interaction picture. It is a big mistake to mix those things up.
     
    Last edited: May 26, 2017
  16. May 26, 2017 #15

    blue_leaf77

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    Yes the operator in Heisenberg picture is time-dependent but its expectation value vanishes, the expectation value evaluated in both pictures should give the same result. I think I have been emphasizing on the expectation value instead of the operator itself.
    The use of the more exact expression is actually crucial here, which is ##e^{-iE_nt/\hbar}## for a stationary state, since this is what makes the two pictures have the same expectation values.
    The expectation value can depend on time if the operator depends on time explicitly, such as a time-sinusoidal potential. But for operator such as ##\mathbf r## in either pictures the expectation values are independent on time for stationary states.
     
  17. May 26, 2017 #16

    Demystifier

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    Yes, it's true for operator such as ##\mathbf r (t)##, but not for operator such as ##[\mathbf r (t)- \mathbf r(0)]^2##. For the latter operator, the expectation value depends on time (in any picture).

    I have a tricky test question for you. If ##\mathbf r(0)## is an operator in the Heisenberg picture, what is the corresponding operator in the Schrodinger picture?
     
  18. May 26, 2017 #17

    blue_leaf77

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    Let's make it clear first, what is ##r(t)## here? If it is ##U^\dagger (t) \mathbf r(0) U(t)##, then I can't think of any situation in which you have an operator with such an expression. By definition ##\mathbf r(t) = U^\dagger (t) \mathbf r(0) U(t)## is the position operator in Heisenberg picture and ##\mathbf r(0)## (or the better notation, just ##\mathbf r##) is the one in Schroedinger picture. If you put them in one expression to form a single operator, things looks strange now.
    If ##U(t) = \exp (-iHt/\hbar)## then ##U(0) = I## and ##I O I## for any operator ##O## is trivial.
     
  19. May 26, 2017 #18

    Demystifier

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    Well, we need more general definitions to make sense of it. In general, even an operator in the Schrodinger picture can be time dependent. When this is the case, we say that the operator has an explicit time dependence. So if ##A_S(t)## is an operator in the Schrodinger picture, then the corresponding operator in the Heisenberg picture is
    $$A_H(t)=U^\dagger (t) A_S(t) U(t)$$

    Now let us consider the case ##A_H(t)=r(0)##, i.e. a case in which an operator in the Heisenberg picture does not depend on time. From the expression above, one can show (I leave it as an exercise) that
    $$A_S(t)=r(-t)$$
    In other words, when the operator does not depend on time in the Heisenberg picture, it is because it has an inverted explicit dependence on time in the Schrodinger picture.

    With all this in mind, the expression ##r(t)-r(0)## makes perfect sense because it is really ##A_H(t)-B_H(t)##, with ##A_H(t)\equiv r(t)##, ##B_H(t)\equiv r(0)##. In the Schrodinger picture, the corresponding quantity is
    $$A_S(t)-B_S(t)=r(0)-r(-t)$$
     
    Last edited: May 26, 2017
  20. May 26, 2017 #19

    blue_leaf77

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    If you put it that way that makes sense now.
    Yes I am aware of that that, in fact that's my point in the 2nd paragraph in post #12. I called this "an operator which is genuinely time-dependent".

    Still, back to the original problem, I am still not convinced that what gives rise to the Van der Waals force is that a fluctuation of dipole moment, namely the fact that
    ##\langle(d(t)-d(0))^2\rangle =f(t)##. The gradual change (rather than fluctuation which completely random in nature) in time of the electron cloud in one atom due to another approaching atom looks much more natural to me since after some distance the cloud in one atom will begin to 'feel' the field from the nucleus and electron of the other atom and hence an induced dipole moment in the two atoms can form.
     
  21. May 26, 2017 #20

    TeethWhitener

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    Fluctuations are necessary for the vdw force because the expectation value of the electric field from a neutral atom is zero.
     
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