Physics Van 3-site Navigational Menu

Physics Van Navigational Menu

Q & A: spin

Learn more physics!

Browse our 5921 answers by or search term

Related Questions

Still Curious?
Explore Q&As in related categories

Quantum Mechanics
Q:
How is Spin (intrinsic angular momentum)calculated? When fundamental particles interact, the total spin doesn’t seem to be conserved. For example, when an electron and a proton reacte to form a neutron, instead of an integral spin (and accompanying bosonic properties) it retains a 1/2 spin and is fermionic. Why is that?
- Luke (age 16)
St. Matthews, Florida
A:

 The spin of a particle can be measured in various ways. For particles with magnetic moments, the spin determines the number of different beams that come out when the particles travel through a spatially varying magnetic field. (The number is 2s+1) (There are some case, such as electrons, for which experimental complications get in the way of the simple version of this experiment.) The spin of a nucleus determines the number of magnetic resonance lines of an electron near it, by the same formula and for essentially the same reason. The number of different spin states shows up when you count states in various thermodynamic problems, affecting equilibrium concentrations. I've undoubtedly left off numerous other ways of measuring spin.

Total spin is definitely not conserved under various reactions. For example, two free radical molecules, each s=1/2, can combine to form an s=0 molecule. The average of the component of angular momentum, of which spin is one piece, along any direction is conserved, but quantum mechanics doesn't require that states have sharply defined values for those components. The spread around the average is not conserved, and it's part of what determines the total spin.

In the particular case you cite, you've noticed that something seems fishy. You're right- something else is needed to make it all balance out. The something else is a neutrino, also spin 1/2. So you have an odd number of fermions going to an odd number of fermions. That obeys another conservation rule (parity), which would be violated if an even number of fermions formed a fermion.

Mike W.

There's a theorem you can prove about the possible allowed values of spin (and all angular momentum for that matter) of a particle has to be a half-integer multiple of Planck's constant divided by 2pi, but the axioms you start with in order to prove this are called the commutation relations for the angular momentum operators. The reasoning is somewhat circular -- the angular momentum commutation relations are a very compact statement of angular momentum physics from which other results can be derived, but the reason we believe them is because experiment agrees with their predictions. It is also true that the angular momentum operators you can build out of the regular momentum operator in quantum mechanics for a single particle obey the commutation relations, so we have a certain confidence in them.

We say electrons have "spin 1/2" and we mean by that their spin angular momentum has magnitude (1/2)(h/2pi). Planck's constant can be measured in other ways and compared with the spin of the electron.

Some newer models predict "anyons", which are neither Fermions nor Bosons, but have spins which are not integer multiples of (h/4pi). These can exist in reduced dimensional spaces (2 space dimensions + 1 time dimension) which are approximately true for some condensed matter systems. These are "quasiparticles" which are really combinations of more normal particles (like electrons, or holes) and possibly excitations of the surrounding lattice. Sometimes it is a lot easier to think of an interacting system in terms of quasiparticles than in terms of each of the real particles that's there. But all real particles we know about are either Fermions or Bosons, which reduces the problem of measuring their spins to choosing among a few small integers times (h/4pi).

Tom

(published on 10/22/2007)

Follow-Up #1: uncertainty for spin

Q:
You said (under http://van.physics.illinois.edu/qa/listing.php?id=21535), "The velocity of the wave function is not separate from how the wave is distributed about in space. In fact, the momentum is a function of how rapidly the quantum wave changes from place to place." "The more squashed in the cloud gets, the more spread-out the range of momenta has to get. That's called Heisenberg's uncertainty principle." So, is "the uncertainty principle(UP)" also applicable to the "spin" of electrons? Assuming it is applicable, how is "the angular momentum" of an electron "spread out" in space? (it cannot localize into a point but must spread out if UP is applicable, correct?)? What shape does it(=spread of the angular momentum) have? Does it look like a sphere/doughnut or something else? What does experiment/observation or theory says about its shape/distribution?
- Anonymous
A:
There is an uncertainty effect for spin, but it isn't connected with spatial spread. For certain types of particles, with spin-zero, there's no uncertainty, the spin on any axis is just plain zero. For common spin-1/2 particles (electrons, protons, neutrons...)if the spin is fully aligned on one axis (say z) then the spins on the other axes (say x and y) are completely uncertain, equally likely to give either possible sign, if measured.  There are similar uncertainty relations for all spin values except zero.

As it turns out, the square of the spin of a spin-s particle is s(s+1)=s2+s. Since by definition the maximum spin in any direction is s, that leaves an extra s in the square, coming form the uncertainty in the spin at right angles to the aligned part. The uncertainty can't be 0 unless s=0.

We have no information whatever about the internal structure of electrons. If any exists, it's likely to be on the Planck scale, or at any rate far smaller than anything that's been measured.

Mike W.

(published on 02/23/13)

Follow-Up #2: spin uncertainty

Q:
Thank you for your answer. You said "For common spin-1/2 particles (electrons, protons, neutrons...)if the spin is fully aligned on one axis (say z) then the spins on the other axes (say x and y) are completely uncertain, equally likely to give either possible sign, if measured." This is puzzling. If, the spin is fully aligned on one axis (say z) then there is zero probability of the spin pointing to x or y!? is this not the case? The conservation of angular momentum should preserve the direction of the spin(if no external influence is imposed on the electron), right? "There is an uncertainty effect for spin, but it isn't connected with spatial spread...": So, you are saying that uncertainty exists in "which direction (x, y, z)" spin is pointing to but not in the amount of spin angular momentum or its spread in space? Does this mean that we can measure the scalar amount of spin angular momentum and its localization in space at an arbitrary accuracy? Does this not violate uncertainty principle? First of all, could there such a phenomenon/entity that carry angular momentum but has no spacial spread? Intuitively, it looks like any entity with (angular) momentum should have some spacial spread. Or, there is no such a requirement/rule for momentum to exist? (is there a counterexample?)
- Anonymous
A:
You raise several separate questions.

Certainly you're correct that the location of the spin is just as uncertain as the location of any other property (mass, charge....) of an electron or other object. My point was that that uncertainty isn't anything special connected to spin, which has its own uncertainty relations in addition to that.

"If, the spin is fully aligned on one axis (say z) then there is zero probability of the spin pointing to x or y!? is this not the case? "
It is not the case. That's what we mean by spin uncertainty. Spin is not a classical quantity.

"Does this mean that we can measure the scalar amount of spin angular momentum ... at an arbitrary accuracy?"
Yes, the square of the spin operator, like the square of other vectors, is a scalar. For a spin s object, the square has definite value s(s+1). The biggest squared amount along any axis is just s2, so the extra is what gives the uncertainty along the other axes.

Mike W.


(published on 03/11/13)

Follow-up on this answer.