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Trying to understand particle polarization

  1. Apr 30, 2014 #1
    Hi,

    I'm trying to understand particle polarization. To my understanding if a particle is in a pure quantum state with non-zero spin (i.e. spin up) then it is fully polarized.

    If it is in a superposition of different spin states then it is weakly polarized and if its properties are uniform in all directions then it is completely un-polarized.

    I feel like I'm missing something here because it seems to me that a particle with spin would be completely un-polarized and then become fully polarized upon measurement. How would you get a value in between?

    Sorry if this is a stupid question. Thank you for your time.
     
  2. jcsd
  3. Apr 30, 2014 #2

    DrChinese

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    One of the issues is your terminology. Any particle whose spin or polarization is known (measured) along one axis has spin/polarization completely unknown along one or more other axes.

    A particle whose spin or polarization is in a superposition of states cannot exactly be said to have that property, either uniform or not. You could say that the likelihood of a measurement outcome is uniform across some range.

    When any quantum measurement occurs, it can be strong or weak in relative terms. I would call a strong (or stronger) measurement one that yields a certain value if repeated (say electron spin up on its x axis). All non-commuting observables (say electron spin axis y or z) will be completely unknown. A weak measurement is one in which the observable (for the sake of discussion) is partially collapsed or known. Keep in mind that the Heisenberg Uncertainty relations apply and it is the product of 2 observables' standard deviations that cannot be less than a constant value. Each individually can theoretically be made to have a range of values and some of those would correspond to ones we could call "weak" or "strong".
     
  4. Apr 30, 2014 #3
    Ah thank you DrChinese that clarifies some things.

    Doesn't this still mean that a particle that isn't observed at all still has no polarisation though? My problem is that in collider experiments they talk about the production polarisation of certain particles and I was wondering where these particles obtain their partial polarisations.

    I suppose particles in the collider are measured but only things like their track momentum and lifetime, as far as I know there is no measurement of any of the spin properties.

    Thank you for the reply though :)
     
  5. Apr 30, 2014 #4

    DrChinese

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    Although I do not know much about collider experiments, there are a number of others here that do. Perhaps if you can give just a little more context for your questions, one of those will speak up.
     
  6. Apr 30, 2014 #5
    You equate "full polarization" to a known spin (state), "weak polarization" with a superposition of spin states, and "unpolarized" with "uniform (isotropic) properties". "Properties"? What happened to only limiting discussion to spin? Any (quantum) state can be equivalently described as a mixture of states. A good way to look at it, is as a vector. Any vector has components which depend on the basis vectors. That is, if one coordinate system describes a vector as (1,0), then you can choose to create a different coordinate frame of reference/system/basis by a simple rotation about the origin. In the new coordinates, the same identical vector will have x and y components different from 100% x and 0% y. (Say a 45° rotation of the coordinates, NOT of the vector, will result in the vector being (√2,√2). This doesn't change the vector at all.)
    Also, apparently you are assuming your quantum particle can have ONLY two 'pure' polarizations - but I'm not sure about this. Usually, we attribute non-zero spins to each particle but this depends on the context. I'm all for +1 and -1 spins. The math is equivalent if you choose +1 and 0, instead. BUT if you allow +1,0 and -1 as possible pure states, then that completely changes the problem (and the system). Obviously, the simplest system is two state (well, the simplest is 1 state, but with one state there is nothing interesting that can happen, is there?). Spin angular momentum is quantized for real particles. I encourage you to read the Wikipedia article on Spin (quantum spin). There it discusses spins as having half integer values, so my convention above is different from its. It also states that particles spin is inherent. This is confusing if you are considering spin as being dependent on (say) field strength. You need to keep several factors in mind. One is that usually its the MAGNITUDE of the spin, not its direction, that they're talking about. The other is the fact that a "particle" can be elementary or composite. A composite particle can be viewed as a system of particles each with their own spin. In the simplest of (non-trivial) quantum systems all you can do is to flip the spin from +1 to -1 or vice versa (or equivalently from +½ to -½). There's NOTHING in between. BUT, we can think about the probability of the particle being in the +1 state - that clearly has to be between 100% and 0%. If we know nothing about the particle's spin, then a two-state particle (oops! we know that about it at least) has 50% chance of being in either state. Quantum Mechanics deals with what can be observed about a system. The math is expressed so that it provides the context for those observations. I don't think "full" or "weak" are very useful terms when discussing this type of system. For instance, lets say we have an electron (two spin states) in a device that allows us to measure spin in ANY direction. So, if we were to measure the spin along one axis and find it is +1 and then measure it along another axis, the classically trained physicist would assume the spin (if it hadn't changed - which is a whole 'nother can of worms) would be zero along that new axis. But wait! QM requires that spin be either +1 or -1. It doesn't ALLOW a spin of 0. Guess what? You won't measure a spin of 0, either! This is NOT intuitive! It is the strangeness of QM at work. (incidentially, it isn't actually possible to measure the spin and not mess with a single electron, so a more realistic experiment would be to have a device generating a stream of electrons all with the same state and using that stream in your measuring device. The logic is the same, you would never be able to find a direction where the spin was EVER anything but ±1.)
    Note also that "spin" can mean different things - angular momentum, magnetic spin, orbital spin, you need to be very careful about using a single term without being clear about context. Which gets us back to your terminology again. Full circle.
     
  7. May 1, 2014 #6
    Thanks for the reply abitslow.

    Yeah I didn't think about that, I guess spin up can be written as a superposition of different states in a different basis so my definition of full polarisation is wrong.

    When you say using terms like full or weak to describe the situation are not useful terms to describe the system I would agree but that's exactly what they do in particle physics.

    I guess I should give full context to my question since maybe the definition is slightly different in particle physics. I read the following paper on the measurement of the ##\Lambda_b## production polarization:

    http://arxiv.org/pdf/1302.5578v2.pdf

    apparently the Lambda_b inherits most of its polarisation from the b quark (it says this on the first sentence of the introduction page). I just want a good definition of production polarisation since clearly my understanding of it is flawed. I want to understand how the b-quark which is not observed can have a definite polarisation which can then be inherited by the ##\Lambda_b##.

    Sorry for not being more clear before.
     
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