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Force to change momentum?

  1. Jan 18, 2010 #1
    consider an isolated electron present at (0,0,0). therefore it is smeared in the vicinity of the point(0,0,0). now its momentum and position changes every instant of time. But force is responsible for the change in momentum.what is the force acting ?

    the probability of this content to contain irony is closer to one.
  2. jcsd
  3. Jan 18, 2010 #2
    The uncertainty principle:
    Δp Δt ≥ h-bar/2.
    F = dp/dt.
    The force is on average zero, but indeterminate for times less than Δt.
    Bob S
  4. Jan 18, 2010 #3
    what is agent which causes this force?
  5. Jan 18, 2010 #4
    or to which part of the universe will receive the impact in response to which the momentum changes?
  6. Jan 18, 2010 #5


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    We don't know that. We don't know what the electron is "really doing" before we measure its position or momentum. The mathematics of QM simply does not address this question. This is the subject of various interpretations of QM which give different answers.
  7. Jan 18, 2010 #6


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    There is no such smeared electrons in nature. Electrons are point particles, they are not smeared. What is smeared is the probability wave function.

    Momentum doesn't "change" from time to time. The momentum is a well defined number once we measure it. Before we measure it, there is uncertainty as to the momentum (of course, measuring it won't yield ONE single number, but a range of values depending on the measurement errors). The Heisenberg uncertainty principle never says that momentum of particles change for no reason.

    There is a difference between change and uncertainty.
  8. Jan 18, 2010 #7
    For example, see the Shroedinger equation (of the constant energy (E)).

    [tex]\frac{P^2}{2m}\psi -\frac{ke^2}{r} \psi = E \psi[/tex]

    So, the change of each value is,

    [tex]\Delta(\frac{P^2}{2m}) - \Delta(\frac{ke^2}{r})=\Delta(\frac{P^2}{2m}) + \frac{ke^2}{r^2}\Delta(r) = \Delta(E) = 0[/tex]

    We arrive at the following relation,

    [tex]\Delta(\frac{P^2}{2m}) = - \frac{ke^2}{r^2} \Delta(r)[/tex]

    This means that the energy of the motion is influenced by the Coulomb force (X the moving distance).
    Actually if we use the reduced mass of the electron considering the motion around the center of mass, the better results is obtained.
  9. Jan 18, 2010 #8


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    Of course, ytuab is talking about a particle in a potential. In QM we deal with potentials and not forces. On the QM scale, all forces are conservative, so we can do this (Since F=-del U for conservative forces). So, in those cases, the potential provides the "force". We just decided not to talk about forces, and just talk about potentials.

    I believe Peeyush's question involved a free particle. In that situation, there is no force or potential. What dictates the motion of the particle is the initial conditions. The average (expected) momentum does not change in such a situation. The fluctuations around this average value is uncertainty and not changes.

    It may interest Peeyush to know Ehrenfest's theorem. Which is that the expectation values of observables follow traditional classical mechanics. Which means d<p>/dt = -del U (which is F). Since -del U is 0 in terms of a free particle, d<p>/dt=0 for a free particle. The expectation value of momentum does not change, just as in classical mechanics.
  10. Jan 18, 2010 #9


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    I think the part in bold is mistaken. If you have no force acting on the electron (isolated) then the potential is not a function of space and so the Lagrangian is symmetric under spatial shifts. Therefore, per Noether's theorem momentum is conserved, so it doesn't just randomly vary. It is true that you cannot measure both the position and the momentum to infinite precision, but that is not the same as the momentum not being conserved.
  11. Jan 20, 2010 #10
    There is no force acting. Force is a classical concept. If the electron has a known position (0,0,0) then it has an infinite uncertainty in its momentum. When there is an uncertainty in momentum, the probability of the electron being deflected (without any classical forces acting on it) is non-zero.

    It is all explained in arxiv.org/pdf/quant-ph/0703126.
  12. Jan 20, 2010 #11
    But actually, in QM we use the classical concept force such as the Coulomb force(f).
    See the reduced mass.

    When in the two-body system, there is a force(f) between the two particles(for example nucleus(1) and the electron(2)),

    [tex]m_{1}\frac{d^2r_{1}}{dt^{2}} = -f, \quad m_{2}\frac{d^2r_{2}}{dt^{2}} = f[/tex]

    Here we define r as [tex]r = r_{2}-r_{1}[/tex]. we arrive at the following relation.

    [tex]\mu = \frac{m_{1}m_{2}}{m_{1}+m_{2}}, \quad \mu\frac{d^2r}{dt^2}=f[/tex]

    If we use this [tex]\mu[/tex] instead of the electron mass([tex]m_{2})[/tex], the energy levels of the Schroedinger equation becomes more correct.

    And the virial theorem is valid also in the quantum mechanics. (Wiki)
    In the virial theorem, using <T> (the average kinetic energy) of a stable system and the force(F),

    [tex]2<T> = - \sum_{k=1}^{N} <F_{k}\cdot r_{k}> [/tex]

    When the force between the two particles of the system is the Coulomb force,

  13. Jan 20, 2010 #12

    The electron is a group of waves otherwise known as a wave-particle. I thought we did away with thinking of electrons as point particles 100 years ago.

    Sure when you collapse the wave function it is measured as a point particle, that doesn't mean its not a wave before we touch it.

    If an electron was a point particle there would be angular momentum in the lowest energy state of a hydrogen atom.
  14. Jan 20, 2010 #13
    That's a very tricky question. How things are when they are not observed? Is there Moon when nobody is looking? Is there sound when a tree is falling and nobody's around to hear it?

    Physics is a science about observable things. Every statement made by a physical theory should be (at least in principle) possible to confirm/reject by experiment. If a statement cannot be verified (because we are not allowed to look) then it doesn't belong to physics. It belongs to philosophy. The whole argument about waves and particles is of that philosophical kind.

  15. Jan 21, 2010 #14


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    Actually, I was just reading a paper online (http://www.scribd.com/doc/4462521/Derivation-of-Schrodinger-Equation-and-Klein-Gordon-Equation-From-Lagrangian-Dynamics-and-Discreteness" [Broken]) that seems to address the OP's question. The author claims to derive the Schrodinger and Klein-Gordon equations from classical Lagrangians, assuming only quantization of energy and momentum, along with a background "zero-point field". The author argues that it is the action of instantaneous fluctuations of the background field that give rise to the uncertainty in position and momentum of a (free) particle.
    Last edited by a moderator: May 4, 2017
  16. Jan 21, 2010 #15
    And so I agree if it can't be measured it can't interact with anything and is not science.

    However measurements have shown that before a particle is measured it could not possibly be a particle in the billiard ball sense and is in fact a collection of waves with different frequency.
  17. Jan 21, 2010 #16


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    Hmmm ... I am not aware of any experiments that support your statement as phrased. Experiments on the quantum mechanical properties of a system by their very nature cannot address the characteristics of said system previous to the measurement taking place.

    Perhaps what you mean is something closer to, "An experiment designed to measure the particle properties of an electron will observe it as a particle, and an experiment designed to measure the wave properties of an electron observes it as a wave. A mathematical model that is consistent with both these experimental observations is a quantum wave-packet, where the electron is represented as a superposition of quantum states with different momenta."?
    Last edited: Jan 21, 2010
  18. Jan 21, 2010 #17

    We do not use the classical equations you have written in quantum mechanics (QM) and I know of no way to obtain the stationary states of Hydrogen, for example, from them alone. We must remember that observables in QM are operators and the mathematical formalism is radically different from the way we do classical mechanics.

    You may be thinking of the Bohr Hydrogen atom (1913) where Bohr used classical equations, plus quantization of angular momentum, to obtain the correct energy levels. But this ad hoc approach fails elsewhere in spite of massive efforts to save it before the invention of quantum mechanics (1927).

    However, Erhenfest’s theorem is often used to demonstrate a classical limit of the quantum approach. It tells us how to obtain classical-like equations in terms of expectation values. For example, classically Newton’s second law can be written as dp/dt =-dV/dx. In QM, p and V are operators. But, for any quantum state, we can calculate the average values <p> and <V> (called expectation values, as you may know), which obey the equation d<p>/dt=-d<V>/dx that does looks like its classical counterpart. Maybe you were thinking of this?
    Best wishes.
  19. Jan 21, 2010 #18
    My apologies to matterwave. I had not read his response and mention of Ehrenfest's theorem
    before posting my reply to ytuab.
  20. Jan 21, 2010 #19
    I see what you mean, eaglelake.
    But I'm sorry to say you probably misunderstand what I said in #11.
    Because the reduced mass in #11 is used also in the hydrogen solution of the Schroedinger equation (this is the same as the Bohr model.)

    See this site. Of course this site is about the Shroedinger equation of the hydrogen atom
    Here, [tex]\mu=m_{e}m_{p}/(m_{e}+m_{p})[/tex] is the reduced mass, which takes into account the fact that the electron and the proton both rotate about a common centre, which is equivalent to a particle of mass [tex]\mu[/tex] rotating about a fixed point.

    As you know, if you don't use this form of the reduced mass, the correct energy levels of the Schroedinger equation would not be obtained. This is a well-known fact. But if you want to arrive at this equation of the reduced mass, you have to think about the classical rotaion of the electron. Do you know some other ways of getting this [tex]\mu=m_{e}m_{p}/(m_{e}+m_{p})[/tex] using only quantum mechanics?

    If you know more about the Bohr model, see this thread.
    I think the historical parts (in 1920's) are not explained in detail in the ordinary QM textbooks.

    For example, the solution of Bohr-Sommerfeld model completely coincides with that of the Dirac equation. (See this Wiki.)
    Why does the Bohr-Sommerfeld model which has no electron spin coincide with the Dirac equation which includes the spin-orbital interaction?
    Because the Dirac hydorogen model includes "many accidental coincidences" as follows (See my post #20 in the Bohr thread as I said above.)

    In page 167 Atomic physics by Max Born
    The case of hydrogen is peculiar in one respect. Experiment gives distinctly fewer terms than are specified in the term scheme of fig 9; for n=2 only two terms are found, for n=3 only three, and so on.
    The theoretical calculation shows that here (by a mathematical coincidense, so to speak) two terms sometimes coincide, the reason beeing that the relativity and spin corrections partly compensate each other. It is found that terms with the same inner quantum number j but different azimuthal quantum numbers l always strictly coincide.

    Of cource, the Bohr-Sommerfeld model doesn't have the electron spin, so it doesn't contain these accidental coincidences.
    Generally speaking, the probability that many coincidences occur is very small.
    Last edited: Jan 21, 2010
  21. Jan 21, 2010 #20
    An example would be quantum tunneling. In order for an entire particle to tunnel it must be made up of waves.
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