What is the origin of Van der Waals force?

  • #1
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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?
 

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  • #2
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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.
 
  • #3
blue_leaf77
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My question is what is the origin of the instantanous electron cloud fluctuation of the atom?
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.
 
  • #4
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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.

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?
 
  • #5
blue_leaf77
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So, for the isolated atom in the ground state, the electron cloud is stationary?
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.
 
  • #6
DrDu
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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?
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.
 
  • #7
blue_leaf77
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⟨(d(t)−d(0))2⟩=f(t)⟨(d(t)−d(0))2⟩=f(t) \langle(d(t)-d(0))^2\rangle =f(t) is not constant.
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.
 
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  • #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.
 
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  • #9
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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.
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.
 
  • #10
blue_leaf77
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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.
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?
 
  • #11
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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?
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
 
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  • #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.
 
  • #13
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the ordinary position operator ##\mathbf r## (not a function of time),
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.
 
  • #14
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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).
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.
 
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  • #15
blue_leaf77
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yet the dipole-moment operator in the Heisenbeg picture is non-zero and depends on time.
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.
then the state is time dependent because even a stationary state depends on time as e−iωt
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.
In each picture, the final result is that an expectation value in a stationary state can depend on time.
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.
 
  • #16
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But for operator such as ##\mathbf r## in either pictures the expectation values are independent on time for stationary states.
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?
 
  • #17
blue_leaf77
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ut not for operator such as [r(t)−r(0)]2[\mathbf r (t)- \mathbf r(0)]^2. For the latter operator, the expectation value depends on time (in any picture).
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 r(0)\mathbf r(0) is an operator in the Heisenberg picture, what is the corresponding operator in the Schrodinger picture?
If ##U(t) = \exp (-iHt/\hbar)## then ##U(0) = I## and ##I O I## for any operator ##O## is trivial.
 
  • #18
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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)## is the one in Schroedinger picture. If you put them in one expression to form a single operator, things looks strange now.
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
$$A_S(t)-B_S(t)=r(0)-r(-t)$$
 
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  • #19
blue_leaf77
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If you put it that way that makes sense now.
When this is the case, we say that the operator has an explicit time dependence.
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.
 
  • #20
TeethWhitener
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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.
Fluctuations are necessary for the vdw force because the expectation value of the electric field from a neutral atom is zero.
 
  • #21
blue_leaf77
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he expectation value of the electric field from a neutral atom is zero.
Let's an easy example, a hydrogen atom in the ground state and the proton fixed at the origin. Calculating the expectation value of the electric field at a point ##\mathbf R## in space is equivalent to calculating the electric field due to a continuous charge distribution ##-e|\psi_{100}(r)|^2## and a point charge ##+e## at the origin. We can use Gauss law to calculate the field since the density is spherically symmetric. It can easily be seen that the electric field does not vanish unless ##|\mathbf R| \to \infty##. Anyway, a molecule or atom being neutral is not always a good approximation, when you go closer and closer and eventually arrive in a location between the orbitals, you can no longer say that the atom is neutral. The same case as in a molecule interacting with another one to possibly form a bonding, as one of them gets closer to the other it will feel a net force due to the other molecule's electrons and nuclei and it is this "residual" force that is responsible for the bonding formation. Classically speaking, depending on whether this net force is repulsive or attractive, a bond may form.
 
  • #22
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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".
Well, my ##x(t)## was only time-dependent in the Heisenberg picture, so it was not really an explicit time-dependence.

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 quantum fluctuations of dipole moment are essential for van der Waals force, but their time dependence is not essential. After all, the average force does not depend on time.

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.
Maybe it looks natural because it looks classical in spirit, but, as you certainly know, atoms are not well described by classical physics. Attempts to describe intermolecular forces with such classical models are not in good agreement with observations.
 
  • #23
TeethWhitener
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Let's an easy example, a hydrogen atom in the ground state and the proton fixed at the origin. Calculating the expectation value of the electric field at a point ##\mathbf R## in space is equivalent to calculating the electric field due to a continuous charge distribution ##-e|\psi_{100}(r)|^2## and a point charge ##+e## at the origin. We can use Gauss law to calculate the field since the density is spherically symmetric. It can easily be seen that the electric field does not vanish unless ##|\mathbf R| \to \infty##. Anyway, a molecule or atom being neutral is not always a good approximation, when you go closer and closer and eventually arrive in a location between the orbitals, you can no longer say that the atom is neutral. The same case as in a molecule interacting with another one to possibly form a bonding, as one of them gets closer to the other it will feel a net force due to the other molecule's electrons and nuclei and it is this "residual" force that is responsible for the bonding formation. Classically speaking, depending on whether this net force is repulsive or attractive, a bond may form.
I'm pretty sure you can't use Gauss' Law here. (Though I'm happy to be proven wrong) For the expectation value, you have to integrate over all space. Once you perform the integration, assuming properly normalized orbitals, you get a contribution of ##-e/\mathbf{r}## from the electron and ##e/\mathbf{r}## from the proton. They cancel to give zero. Using Gauss' Law means that you're unphysically constraining the electron to be within a finite sphere.
 
  • #24
blue_leaf77
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Using Gauss' Law means that you're unphysically constraining the electron to be within a finite sphere.
No, that's not true. Remember when you calculated the electric field inside a dielectric with a spherically symmetric charge density. What matters is only the charges inside the spherical surface containing the observation point. Of course, however Gauss law is not the only tool to calculate the E field. One can use the basic equation
$$
E(\mathbf R) \propto \int_0^\infty \int_0^{2\pi}\int_0^{\pi} \frac{\rho(r)}{|\mathbf r - \mathbf R|^3} (\mathbf r - \mathbf R) d^3\mathbf r
$$
Now the expectation value of electric field operator in the ground state hydrogen atom should be
$$
\langle E(\mathbf R)\rangle \propto \int_0^\infty \int_0^{2\pi}\int_0^{\pi} \frac{|\psi(r)|^2}{|\mathbf r - \mathbf R|^3} (\mathbf r - \mathbf R) d^3\mathbf r
$$
You see that the two equations are different only by interchanging ##|\psi(r)|^2## and ##\rho(r)##. So, I don't see why the second equation cannot be alternatively calculated using Gauss law.
 
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  • #25
TeethWhitener
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I thought about it a little more and you're probably right.
 

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