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I am supremely confused when something has spin or when it does not. For example, atomic Hydrogen has 4 fermions, three quarks to make a proton, and 1 electron. There is an even number of fermions, and each fermion has a 1/2 spin. Since there are an even number of fermions, the total spin value is an integer. This spin number is the "intrinsic" spin number that cannot be changed, but its orientation "up" or "down" can be changed.

For atomic Hydrogen, it is a Boson because it has integer spin, however it also has a single electron. I read on physics forums, http://www.physicsforums.com/showthread.php?t=69992, that the spin of atom comes from the electrons and not its nucleus. I also read on here, How to find that a molecule has zero spin?, that the spin of atomic Hydrogen is 1/2! The answer says atomic Hydrogen has spin 1/2 because it ignores the nuclear spin.

This is one thing that is confusing me. Shouldn't atomic Hydrogen have an integer spin because of the nuclear component? So does atomic Hydrogen have spin and is affected by a magnetic field? Nuclear spins are affected by magnetic fields, but they aren't as affected as electrons according to the discussion on physics forums.

Why do we ignore nuclear spin sometimes? Also, can someone help me out here with all the possibilities?

Is there a Boson with an half integer spin value? (Surely, there must not be) However atomic Hydrogen is one of those cases! (It seems...) (Why don't we cancel out the nuclear spin with the electron spin?)

Say we have another atom that is a Boson, It has unpaired electrons in different orbitals, so what determines whether or not electrons fill in orbitals as spin up or down? Does spin down nuclear spin cancel out a electron up spin?

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this is not a bad summary answers.yahoo.com/question/index?qid=20060821140105AApFczb . This site also has usually simple explanations hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nspin.html –  anna v Apr 2 '13 at 6:04
    
and here for a discussion of the addition of angular momentum in quantum mechanics (or pick up any QM textbook). Short answer: yes, the nuclear spin does count, but it doesn't have a big effect on the energy levels so you don't hear about it much in atomic physics. –  Michael Brown Apr 2 '13 at 6:27
    

4 Answers 4

In contrast with the previous incorrect answers that I hadn't noticed, there isn't any ambiguity or confusion about the Bose-Einstein or Fermi-Dirac statistics for composite systems such as atoms.

A particle – elementary or composite particles – that contains an even number of elementary (or other) fermions is a boson; if it contains an odd number, it is a fermion. Bosons always have an integer spin; fermions always have a half-integer spin. By the spin-statistics theorem, the wave function of two bosons is invariant under their exchange but it is antisymmetric under the exchange of two identical fermions. These two rules are theorems for elementary particles and assuming this theorem, it's also trivial to prove these statements for composite particles.

In particular, for a neutral atom, the numbers of protons and electrons are equal so their total parity is even. That's why only neutrons matter. An isotope with an even number of neutrons is a boson (the whole atom: e.g. helium-4); an isotope with an odd number of neutrons is a fermion (the whole atom: e.g. helium-3).

That doesn't mean that composite bosons exhibit all the same physical phenomena like superfluidity that we may observe with some bosons.

The magnetic moment of a charged "elementary enough" particle scales like $1/m$ where $m$ is the mass of the particle. That's why the magnetic moment of the protons, neutrons, and nuclei are about 1,000-2,000 times smaller than the magnetic moments of the electrons. That's why the nuclear spins are largely negligible for the behavior of the atom in a magnetic field.

This is no contradiction because the whole atoms have a much larger magnetic moments than the nuclei separately – because of the neutrons: atoms are not "elementary enough" in this definition. Both the electrons' spins and their orbital angular momentum contribute to an atom's magnetic moment. Also, there exist a higher number of states because an atom is a typical example of the "addition of several angular momenta". The tensor product Hilbert space may be decomposed as a direct sum of Hilbert spaces with fixed values of the total angular momentum. The degeneracy and the magnetic moment of these components depend on the total angular momentum i.e. on the relative orientation of the nucleus-based and electron-related angular momenta.

In effect, the spectral lines of the whole atom exhibit the so-called hyperfine structure. Up to some approximation, the nuclear spin may be totally ignored. But when the considerations from the previous paragraph are properly account for, each spectral line is actually split to several nearby (3 or so orders of magnitude closer to each other) finer spectral lines, each of which corresponds to a different value of the total angular momentum of the whole atom (or, equivalently, a different value of $\vec J_{\rm electrons}\cdot \vec J_{\rm nucleus}$).

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Nope, it's not that simple. See P. Ehrenfest and J. R. Oppenheimer, "Note on the Statistics of Nuclei," Phys. Rev. 37 (1931) 333, link.aps.org/doi/10.1103/PhysRev.37.333 , DOI: 10.1103/PhysRev.37.333 . The abstract makes it clear that the application of the spin-statistics theorem to composite systems is an approximation. –  Ben Crowell Aug 27 '13 at 4:55
    
Hi Lumo. I like this, but it confuses me further. Obviously it's true when the particles are far apart, but when they are close enough that their size matters I'm not so sure... Would you mind having a quick look at my answer. It's possible I made an elementary mistake, but I don't think so. The trouble comes in precisely at short distances. I think what the calculation is showing is that at short distances it's really best to give up on the notion of atoms/nucleons/whatever all together and just work with the constituents. Would like some clarification on this. :) –  Michael Brown Aug 27 '13 at 4:56
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@BenCrowell That article sounds interesting. It will have to wait until I'm behind the paywall though... That Oppenheimer and Ehrenfest are talking about it is slightly reassuring to me. :) –  Michael Brown Aug 27 '13 at 4:58
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Dear @Ben, the only point at which the paper says "it's more complicated" is the part of the wave function for which the positions of the two identical composite particles are too close - because they overlap or strongly interact. Indeed, when they do, they disrupt each other and the whole effective description in terms of the composite particles becomes inapplicable. But whenever it's applicable - when the particles are far enough from each other - the (anti)symmetry of the wave function under the permutation is totally exact. The 1930-1 paper has an outdated discussion of effective theories. –  Luboš Motl Aug 28 '13 at 8:32
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Dear @MichaelBrown, I totally agree - and wrote the comment above this one before I read yours. When the distance of the composite particles is too short so that they overlap, or due to strong interactions etc., the whole description - an effective description - in terms of the composite particles (not just the statistics) becomes inapplicable and one must switch to a finer theory of the constituents. But whenever it makes sense to talk about the composite particles at all - when the effective theory based on them works - their statistics follows the simple rule linked to the spin. –  Luboš Motl Aug 28 '13 at 8:35

Actually, there are two different and inequivalent definitions of bosons. On the one hand, they are often defined as particles with integer spin, on the other hand, sometimes they are defined as particles for which only symmetric states exist in nature (see, e.g., Dirac's "Principles of Quantum Mechanics"). Composite particles, such as hydrogen atoms, can be bosons under the first definition, but not under the second one.The commutation relations for operators of creation/annihilation of hydrogen atoms are derived from the anticommutation relations for the operators of creation/annihilation of protons and electrons that are parts of hydrogen atoms, and those commutation relations approximately coincide with the commutation relations for boson creation/annihilation operators in the limit of low density. See the details (for the example of deuterons) in the book by Lipkin called "Quantum Mechanics. New Approaches to Selected Topics"

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Haven't read the Lipkin book (though it sounds good), but would it be fair to say that when the density is so high that bose statistics are no longer a good approximation it is no longer useful to think about the system as a system of Hydrogen atoms, but rather consider the electrons and protons directly? Or is it still ok? I.e., does the effective theory of hydrogen atoms break down at or before the compositeness scale? –  Michael Brown Apr 2 '13 at 8:57
    
I am reluctant to proclaim something useful or useless - it depends on what you need and what precision you require, but behavior of composite objects can be more complex at high density than that of bosons. For example, Lipkin considers, among other things, behavior of Cooper pairs: they behave as bosons at low density, but display new important features, such as superconductivity, at high density. –  akhmeteli Apr 2 '13 at 9:28
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Although my field theory is weak and I haven't read the Lipkin book, I think our answers show, using different approaches, the same fact: that it is only an approximation to think of an atom as a boson or fermion. –  Ben Crowell Aug 26 '13 at 20:44

This is confusing enough to make me want to scribble down a few equations. This is a creation operator for deuterium:

$$D_{\rho}^{\dagger}\left(k\right)=\sum_{\alpha\beta\gamma}\int\mathrm{d}^{3}l\mathrm{d}^{3}p\mathrm{d}^{3}q\delta^{\left(3\right)}\left(l+p+q-k\right)\ \psi_{\alpha\beta\gamma}\left(l,p,q;\rho\right)e_{\alpha}^{\dagger}\left(l\right)p_{\beta}^{\dagger}\left(p\right)n_{\gamma}^{\dagger}\left(q\right), $$

where $e^\dagger,p^\dagger,n^\dagger$ are creation operators for electrons, protons and neutrons with a hopefully obvious notation for spin and momentum indices. $\psi_{\alpha,\beta,\gamma}(l,p,q;\rho)$ is a wavefunction whose detailed form I don't care about. $\rho$ is an internal configurational index which includes spin, internal energy level etc.

By counting fermions this should be a fermion. Indeed all the Clebsch-Gordon coefficients (which I've hidden away in $\psi$) will vanish whenever the total angular momentum is not a half integer, and the spin-statistics theorem still applies. So let's compute the anticommutator:

$$ \begin{array}{ll} \left\{ D_{\mu}\left(r\right),D_{\rho}^{\dagger}\left(k\right)\right\} &= \sum_{\alpha'\beta'\gamma'}\int\mathrm{d}^{3}l'\mathrm{d}^{3}p'\mathrm{d}^{3}q'\delta^{\left(3\right)}\left(l'+p'+q'-r\right)\ \psi_{\alpha'\beta'\gamma'}^{\star}\left(l',p',q';\mu\right) \\ &\times\sum_{\alpha\beta\gamma}\int\mathrm{d}^{3}l\mathrm{d}^{3}p\mathrm{d}^{3}q\delta^{\left(3\right)}\left(l+p+q-k\right)\ \psi_{\alpha\beta\gamma}\left(l,p,q;\rho\right) \\ &\times\left\{ e_{\alpha'}\left(l'\right)p_{\beta'}\left(p'\right)n_{\gamma'}\left(q'\right),e_{\alpha}^{\dagger}\left(l\right)p_{\beta}^{\dagger}\left(p\right)n_{\gamma}^{\dagger}\left(q\right)\right\} \end{array}. $$

The meat is the anticommutator of the field operators. Working that out I find:

$$\begin{array}{ll} \left\{ \cdots\right\} & =-\delta_{\alpha\alpha'}\delta_{ll'}\delta_{\beta\beta'}\delta_{pp'}\delta_{\gamma\gamma'}\delta_{qq'}\\ & +\delta_{\alpha\alpha'}\delta_{ll'}\delta_{\beta\beta'}\delta_{pp'}n_{\gamma'}\left(q'\right)n_{\gamma}^{\dagger}\left(q\right)\\ & +\delta_{\alpha\alpha'}\delta_{ll'}p_{\beta'}\left(p'\right)p_{\beta}^{\dagger}\left(p\right)\delta_{\gamma\gamma'}\delta_{qq'}\\ & -\delta_{\alpha\alpha'}\delta_{ll'}p_{\beta'}\left(p'\right)p_{\beta}^{\dagger}\left(p\right)n_{\gamma'}\left(q'\right)n_{\gamma}^{\dagger}\left(q\right)\\ & +e_{\alpha'}\left(l'\right)e_{\alpha}^{\dagger}\left(l\right)\delta_{\beta\beta'}\delta_{pp'}\delta_{\gamma\gamma'}\delta_{qq'}\\ & -e_{\alpha'}\left(l'\right)e_{\alpha}^{\dagger}\left(l\right)\delta_{\beta\beta'}\delta_{pp'}n_{\gamma'}\left(q'\right)n_{\gamma}^{\dagger}\left(q\right)\\ & -e_{\alpha'}\left(l'\right)e_{\alpha}^{\dagger}\left(l\right)p_{\beta'}\left(p'\right)p_{\beta}^{\dagger}\left(p\right)\delta_{\gamma\gamma'}\delta_{qq'} \end{array}, $$

which starts out with a promising delta function (obvious abuse of notation here: Kronecker = Dirac delta), but quickly turns into something awful. The first term clearly gives the $\delta_{\mu\rho}\delta_{rk}$ we want (up to a sign which I'm not sure what to do with). The other terms give one and two particle reduced density matrices. In general these will have nontrivial off-diagonal components measuring the correlation between two particles induced by the third particle.

For example focus on the one electron term (fifth term of the anticommutator). Doing the $p',q'$ integrals, sums and rearranging delta functions gives

$$ \left\{ D_{\mu}\left(r\right),D_{\rho}^{\dagger}\left(k\right)\right\} =\cdots+\sum_{\alpha'}\sum_{\alpha\beta\gamma}\int\mathrm{d}^{3}l\mathrm{d}^{3}p\mathrm{d}^{3}q\delta^{\left(3\right)}\left(l+p+q-k\right)\int\mathrm{d}^{3}l'\delta^{\left(3\right)}\left(l'-l+k-r\right) \times\psi_{\alpha\beta\gamma}\left(l,p,q;\rho\right)\psi_{\alpha'\beta\gamma}^{\star}\left(l',p,q;\mu\right)e_{\alpha'}\left(l'\right)e_{\alpha}^{\dagger}\left(l\right) $$

Looking at the second delta function we see that this term is generically spread out in $k-r$ by as much as the wavefunction is spread out in $l$. This is just saying that the electron is correlated with the other particles, but it has the effect of spreading the anticommutator out in momentum space. Going back to position space we will find that the anticommutation relationship is modified at short distances, comparable to the size of the composite state.

It's possible for a miracle cancellation to happen here and give a nice local anticommutator, but it's not obvious to me and it seems rather unlikely. So from this calculation I conclude that either:

  • It is implicit in one of the assumptions of the spin-statistics theorem that one is talking about fundamental rather than composite objects, or
  • I've made a mistake, or
  • the spin-statistics theorem is wrong. (I include this possibility only to rule it out immediately. ;))

EDIT: The other anticommutators do vanish: $\left\{D,D\right\}=\left\{D^\dagger,D^\dagger\right\}=0$, but the failure of the anticommutation relation above is enough that one cannot build the usual Fock space. Perhaps a field definition would be able to fix up the algebra, it's not obvious to me...

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There are two different issues here. (1) Is it a good approximation to describe a particular composite system as a boson or a fermion? (2) If so, which is it?

Question #2 is the easy one. The spin-statistics theorem tells us that iff the spin is an integer, the object is a boson. For an atom or ion, this is determined by whether the total number of electrons and quarks is even.

Question #1 is more complicated. Akhmeteli's answer has explained, based on general ideas about quantum mechanics, why the answer may be no. For an atom, I think this issue comes down to the energy/temperature regime you're dealing with and the strength of the interaction between the nucleus and the electrons (hyperfine interaction). If the nucleus didn't interact at all with the electrons, then it would be meaningless to consider them as a composite system. They do interact, but the interaction is very weak; it amounts to $\sim 10^{-4}$ eV, or about 1 K in terms of temperature.

At room temperature, we're dealing with temperatures hundreds of times higher than this scale, so any hyperfine effects are too delicate to matter. This is why, for example, the properties of 3He and 4He gases differ only due to their differing masses.

At temperatures below about 1 K, hyperfine effects start to matter. At these temperatures, 3He and 4He liquids differ qualitatively, because 4He is a boson, and it forms a superfluid due to effects analogous to Bose-Einstein condensation.

To make this more clear, it may be helpful to observe that there are two independent temperature scales here. The first one is the one described in the first paragraph above, the temperature corresponding to the strength of the hyperfine interaction. Let's call this $T_{hf}$. The second one is the temperature at which the de Broglie wavelength of the atom is equal to the typical spacing $n^{-1/3}$ between the atoms, where $n$ is the number density. Below this temperature, we expect the substance to be highly quantum-mechanical, so let's call it $T_q$. This temperature is given by $T_q=\hbar^2 n^{2/3}/2mk$. For superfluid 4He, this comes out to be about 0.4 K. For helium, these two temperatures happen to be about the same, but in principle they are completely separate. If we want to see any effect from quantum statistics, we need a temperature $\lesssim T_q$. It happens that for helium, this also guarantees that we're at temperatures low enough so that the nucleus couples to the system and affects the statistics, but that is an accident of nuclear physics, which happens to give magnetic dipole moments on a certain order of magnitude.

In general, it is not always correct to try to treat composite systems as elementary. It may or may not be a good approximation to do so. For example, in nuclear physics we can try to treat nucleons as elementary particles, and talk about two-body interactions between them, but this is messy and involves approximations, because really whe a nucleon interacts with another nucleon, it's just six quarks interacting. Similarly, it doesn't always make sense to attribute Bose or Fermi statistics to a composite system. At temperatures $\lesssim T_{hf}$, it makes sense approximately to treat an atom as a composite system whose statistics are defined by its total spin (nuclear coupled to electronic).

In a case like 4He where the nucleus has zero spin, there are two separate reasons why the nucleus doesn't affect the statistics. One is that the nucleus has integer spin, and adding an integer to another number doesn't affect whether it's an integer or a half-integer. The other reason is that a system with zero spin can't have a magnetic moment, so there's no way to couple it to the electrons magnetically, and therefore $T_{hf}$ is effectively zero.

[EDIT] Prompted by Lubos Motl's skepticism, I looked around for some more general treatment of the basic issue of whether, when, or to what approximation spin-statistics applies to composite systems. It turns out that the classic paper on this topic is Ehrenfest 1931. Unfortunately this scientific paper paid for by my grandparents' income taxes is behind a paywall, but here is the abstract:

From Pauli's exclusion principle we derive the rule for the symmetry of the wave functions in the coordinates of the center of gravity of two similar stable clusters of electrons and protons, and justify the assumption that the clusters satisfy the Einstein-Bose or Fermi-Dirac statistics according to whether the number of particles in each cluster is even or odd. The rule is shown to become invalid only when the interaction between the clusters is large enough to disturb their internal motion.

This makes it clear that the application of the spin-statistics theorem to composite systems is only an approximation. I can't be sure, because I haven't yet found a more complete presentation of the argument online, but the abstract quoted above does seem to be consistent with my analysis above of liquid 3He and 4He. At temperatures above $T_{hf}$, the interaction is "large enough to disturb" the "internal motion," i.e., the delicate hyperfine coupling of the nuclear spin to the electrons.

P. Ehrenfest and J. R. Oppenheimer, "Note on the Statistics of Nuclei," Phys. Rev. 37 (1931) 333, link.aps.org/doi/10.1103/PhysRev.37.333 , DOI: 10.1103/PhysRev.37.333

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Sorry, Ben, but there is no "confusion" about the statistics of a composite system. Someone's being a boson or a fermion doesn't necessarily include the assumption that it undergoes Bose-Einstein condensation in the same way as some bosons we know etc.; that's a stronger statement. Something's being a boson or a fermion is about the sign flip or the absence of it when two particles of the same kind are exchanged and this may be done both with elementary and composite particles and the behavior is always linked to the integrality/half-integrality of the spin i.e. to the #(neutrons) for atoms. –  Luboš Motl Aug 27 '13 at 0:58
    
Ben - you need to distinguish the question "how does the wavefunction of two similar clusters of fermions behave under swaps of these clusters" from the question "can a cluster be described by its centre of mass coordinate". Ehrenfest and Oppenheimer seem to consider the combination of both questions. - Furthermore, claiming that the interaction between the nucleus and electrons is weak is obviously ignoring the Coulomb interaction. Bottom line is that also at room temperature a hydrogen atom is a boson, and a deuterium atom a fermion. –  Johannes Aug 27 '13 at 12:32
    
@Johannes: claiming that the interaction between the nucleus and electrons is weak is obviously ignoring the Coulomb interaction The Coulomb interaction of course strongly couples the momenta. But the magnetic interaction only weakly couples the spins. Bottom line is that also at room temperature a hydrogen atom is a boson, and a deuterium atom a fermion. I don't think this statement makes sense, because at room temperature the classical limit applies and the statistics are undetectable. –  Ben Crowell Aug 27 '13 at 14:20

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