In wikipedia, it is said

The experiment is normally conducted using electrically neutral particles such as silver atoms. This avoids the large deflection in the path of a charged particle moving through a magnetic field and allows spin-dependent effects to dominate.[7][8]

But if I tried to add the Lorentz force from electrons and nuclei to the atom, it looks confusing. For example, a hydrogen atom $$ \mathbf{F}_e = q_e \mathbf{E} + q_e \mathbf{v}_e \times \mathbf{B} + \mathbf{F}_{p\,\, to \,\, e} $$ $$ \mathbf{F}_p = q_p \mathbf{E} + q_p \mathbf{v}_p \times \mathbf{B} + \mathbf{F}_{e\,\, to \,\, p} $$

$e$ and $p$ stand for electron and proton. It is known $q_e = - q_p$ and I can expect $\mathbf{F}_{p\,\, to \,\, e} = - \mathbf{F}_{e\,\, to \,\, p}$. But, it is unlikely $ \mathbf{v}_e = \mathbf{v}_p $.

I can think in terms of the Ehrenfest theorem to make the above argument somehow more quantum. But, this is also unlikely to reconcile this issue. So the question is, how to obtain the net Lorentz force for a neutral atom to be zero.

  • 2
    $\begingroup$ As you wrote, $q_e=-q_p$. If ${\bf v}_e={\bf v}_p$, that's enough to have a zero net Lorentz force. $\endgroup$ Jan 24, 2023 at 7:33
  • $\begingroup$ Yes, oops, I missed the sign in the velocities. $\endgroup$
    – RandomUser
    Jan 26, 2023 at 19:54

2 Answers 2


The argument is not that total Lorentz force on neutral atom is zero, but that by atom being neutral, this total Lorentz force is much smaller than it would be if the atom carried net charge, and second, that it is so small that it is negligible when compared to force due to magnetic field on intrinsic magnetic moment.

The first part is easy - the force is smaller than for a charged system for obvious reason - oppositely charged particles have all roughly the same laboratory "macroscopic" velocity $\mathbf V$ in the experiment, so those parts of all the Lorentz forces on all particles inside that are proportional to $\mathbf V$ cancel each other out.

However, there is possibly a non-zero force remainder due to the fact that particles' velocities can differ somewhat.

It then becomes the question of whether this force remainder is negligible when compared to force due to intrinsic magnetic moment (spin) in magnetic field gradient. This is much harder question to analyze exactly in quantum theory and it is not easy to answer off-hand, since expectation value of the Lorentz magnetic force is not, in general, the product of $\langle \Delta \mathbf v \rangle$ and $\langle \mathbf B(\mathbf r)\rangle$:

$$ e\bigg\langle (\mathbf v_p - \mathbf v_e) \times \mathbf B(\mathbf r) \bigg\rangle \neq e \langle \mathbf v_p - \mathbf v_e\rangle \times \langle\mathbf B(\mathbf r) \rangle. $$

However, we can make a classical estimate. Net Lorentz magnetic force due to differing velocities can be reformulated as force on orbital magnetic moment implied by those different velocities, in gradient of magnetic field, described by the well-known formula

$$ \mathbf F_{B~gradient~on~\mu_{orb}} = \boldsymbol{\mu}_{orb} \cdot \nabla \mathbf B. $$ The "spin force" is the same thing, but due to intrinsic magnetic moment $\boldsymbol{\mu}_{spin}$. So we can say, in classical theory, that non-spin forces affecting the atom trajectory (really due to Lorentz magnetic forces not cancelling each other perfectly) are certainly negligible when orbital magnetic moment is negligible to spin magnetic moment.

Borrowing facts from quantum theory, this seems to be possible in some cases, e.g. when the atom is in ground state with zero orbital angular momentum in direction of magnetic field gradient, but non-zero intrinsic angular momentum in that same direction. But this is not necessarily true for all atoms in all states. It seems that for the silver atoms in the SG experiment, it is always assumed that their angular momentum component in direction of the field gradient is zero, so the only important contribution to force that deflects the atoms is due to spin magnetic moment.

  • $\begingroup$ Thanks for your answer. For the orbital angular momentum, since silver has many electrons, it can form |j1j2;j1_z j2_z> <-> |j1j2;JJz>, uncoupled or coupled basis. In > it is always assumed that their angular momentum component in direction of the field gradient is zero, does it refer to uncoupled or coupled basis? and why? $\endgroup$
    – RandomUser
    Jan 27, 2023 at 3:52
  • $\begingroup$ I meant the component of net orbital angular momentum of the whole atom, the quantum number $m$ associated with orbital angular momentum operator $L_z = \sum_a L_{a,z}$, sum over all particles. If non-zero, I would expect strong gradient force on the atom in non-uniform field. But take this with a grain of salt, I don't really understand intricacies of angular momentum in quantum theory. $\endgroup$ Jan 27, 2023 at 12:12
  • $\begingroup$ If the Lorentz force for a neutral atom is not exactly zero, does it provide a way to detect if a particle is composite? Namely, if some particle is elementary and neutral, the Lorentz force will be zero; composite may be very small but not zero. Is there any experiment in this direction? $\endgroup$
    – RandomUser
    Jan 30, 2023 at 6:09
  • $\begingroup$ The Lorentz force is just the simplest model of EM force acting on charged particle and can explain force on composite particles (multiparticle systems) with electric or magnetic moment in nonuniform fields. But it does not work in reverse - experimental detection of such force does not on its own prove the particle is composite. There may be point particle with magnetic moment experiencing the above force in nonuniform magnetic field, without it being composite. $\endgroup$ Jan 30, 2023 at 9:48

The electrons in an atom do not have a velocity in the classical sense. They exist in a superposition of velocities and the expectation value of the velocity, i.e. their average velocity, is the same as the velocity of the atom. This also applies to the nucleus.

You say:

it is unlikely $\mathbf v_e = −\mathbf v_p$

but if we replace $\mathbf v_e$ and $\mathbf v_p$ by their expectation values they will be the same and hence the average Lorentz force on the protons and electrons will be equal and opposite.

  • $\begingroup$ But does it contradict the underlying picture of the Born-Oppenheimer approximation, that nuclei move slower than electrons, and the degrees of freedom are separated? $\endgroup$
    – RandomUser
    Jan 25, 2023 at 0:40
  • $\begingroup$ @RandomUser: In the Born-Oppenheimer approximation, $\vec{v}_p = 0$ and $\langle \vec{v}_e \rangle = 0$ for any stationary state. (And even for non-stationary states, the time-average of $\langle \vec{v}_e \rangle$ will still be zero.) $\endgroup$ Jan 25, 2023 at 18:59
  • $\begingroup$ @MichaelSeifert I think it is a bit tricky. BO approximation separates nuclear and electronic degrees of freedom. After that, the nuclear motion follows Boltzmann distributions, such that $v_p \neq 0$. If $v_e$ is derived as p/m ~ [H,x], for bound state, it would be zero, not scattering state (suppose an electron beam bends in a magnetic field); technically, <n| [H,x] |n> = <n| Hx - xH|n> = E_n (<x> - <x>), which <x> is not well defined in scattering state. $\endgroup$
    – RandomUser
    Jan 26, 2023 at 19:53
  • $\begingroup$ For John Rennie's answer, I guess from a phenomenological point of view, both electron and proton go in one direction together in the beam, so I can expect $\langle v_e \rangle = \langle v_p \rangle$, in the lab frame, thank you. (In center of mass frame, there will be $\langle v_e \rangle $ = 0 as in Michael Seifert's comment) $\endgroup$
    – RandomUser
    Jan 26, 2023 at 20:07
  • 1
    $\begingroup$ Unfortunately, in SG experiments the effect observed is due to magnetic field being non-uniform, and in such field, expectation value of the Lorentz magnetic force need not factorize exactly into product of exp. value of $\Delta \mathbf v$ and exp. value of $\mathbf B(\mathbf r)$. So $\langle \Delta \mathbf v\rangle = 0$ does not imply that exp. value of Lorentz magnetic force is zero. Contribution to force due to velocity difference (probably expressible as function of orbital magnetic moment) is not obviously negligible compared to contribution due to spin (intrinsic magnetic moment). $\endgroup$ Jan 27, 2023 at 0:01

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