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Does spin have a spatial direction

  1. Mar 29, 2014 #1
    Spin is an operator with components (Sx,Sy,Sz) but do these represent "spatial" components. The question arose because the spin of an electron sets up an angular momentum
    L = -e/m * S (S and L are vectors)
    So inverting the spatial coordinate system and requiring that the physical result must be the same, the spin has to be reversed right? So what is spin up and what is spin down seems to be fixed by our choice of coordinate system. Is that correct or am I misunderstanding something? It just seems a bit weird to me, since spin is not really a true angular momentum but something intrisic and non spatial if I remember correct.
     
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  3. Mar 29, 2014 #2

    edguy99

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  4. Mar 29, 2014 #3
    Yes but does that mean they are spatial?
     
  5. Mar 29, 2014 #4

    edguy99

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    The idea of spin components being Sx,Sy,Sz in the traditional thinking has a problem. If say along the Z-axis you define the spin as a tangent vector in the x-y plane, you will not have properly described the spin of the particle. Just like an orbit when thinking of angular momentum, a particle can "precess" around its axis. To properly describe the particle, you not only have to define its spin at a particular point, but you must specify its precession. In this way you can get intrinsic spin within the particle. Ie. You could define quantum state up/down as if the particle spin is the same direction as the precession, its up. If the particle spin is opposite to the precession, its down.

    This style of definition using 3 components of spin s1,s2,s3 means the particle spin is independent of the coordinate system. Ie, take a 'up' particle and send it through a stern-gerlach device and it goes up. Turn the particle upside down and it will go through the same device and go down. You are absolutely correct in saying the spin of the particle is independent of the choice of the coordinate system.
     
  6. Mar 30, 2014 #5
    Well the problem is that the particle has a magnetic moment directly proportional to the spin, which is dependent on the choice of coordinates. I.e. if we turn the z-axis upside down we should obviously have that the magnetic moment is inverted. But if the spin is the same how does that happen?
     
  7. Mar 30, 2014 #6
    The expectation values of the three spin operators form a vector. If you picked a different coordinate system and took the expectation values of those three operators, you would get a vector which pointed in the same direction. This vector couples to a magnetic field in a way which is similar to the way that classical angular momentum does.

    So while the spin itself is a quantum object, its expectation value acts like a classical angular momentum vector, which is why it was originally interpreted as a physical spin.
     
  8. Mar 31, 2014 #7
    For quantum systems, the best approach is to abandon the concept of spin, as you understand it.
    Any quantum particle has a property called spin, but it is not spin in a macroscopic sense. Perhaps you should think of it as an operator... This property is intrinsic to the particle and is NOT dependent on the coordinate system you choose. However, when you make a measurement on the particle's spin, you have explicitly selected a particular coordinate system. Hence the result of your measurement WILL be dependent on that coordinate system. When the (charged) particle is placed in a non-isotropic environement (ie placed in an electric or magnetic field) the spherical symmetry is broken and certain directions are lowest energy. Meaning that direction becomes part of the system, rather than just an arbitrary (random) choice. Point being that you need to LOGICALLY separate choice of YOUR coordinates from the directions inherent in the system...your coordinates are NOT (logically) part of the system. In Euclidean 3-space, any coordinate system can be rotated by two orthogonal angles, and the properties of the particle will NOT be affected by this (arbitrary) change of coordinates. You are confusing the representation of a property with the property. Physics, since the 1800's, has been coordinate independent. A vector exists independent of the coordinate choice used to represent it.
    When dealing with QM systems, superposition is real. But you clearly are confusing your selection of coordinate axes, with something "real".Flipping your axis does NOTHING to the system, flipping the field DOES.
     
    Last edited: Mar 31, 2014
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