Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How does spin work in the position representation?

  1. Apr 14, 2012 #1
    When you have an ordinary (read: spin zero) particle with a quantum state |ψ>, the corresponding position-space wave function <x|ψ> is an ordinary scalar function of position. How do things work when the particle has spin? In that case its quantum state will live in the tensor product of two Hilbert spaces: an infinite-dimensional (rigged) Hilbert space for position, and presumably a finite dimensional Hilbert space for its spin states. So its state will be of the form |ψ>=|ρ>|χ>, where |ρ> is the positional state and |χ> is the spin state, and the amplitude to find it in a particular position and spin is given by <x|<s||ρ>|χ>. So then the position space wave function should be written as ψ(x)=ρ(x)χ(s), where ρ(x)=<x|ρ> and χ(s)=<s|χ> are scalar-valued wave functions representing the positional and spin states.

    But instead I think I've seen people write ψ(x)=ρ(x)χ(x), where χ(x)=<x|χ> is called a "spinor-valued wavefunction" representing the spin state. What's going on here? Is this correct, and what does this mean? I don't know too much about spinors, other than the fact that they can be represented in terms of matrices. Are spinors supposed to be the irreducible representations of the group SU(2) in the positional Hilbert space?

    Any help would be greatly appreciated.

    Thank You in Advance.
     
  2. jcsd
  3. Apr 15, 2012 #2
    The way to deal with this is to extend the list of parameters to the wavefunction. It's no longer enough to ask "what is the probability that the particle is at position x?". Instead, you have to ask "what is the probability that the particle is at position x and with spin state s?". To use your notation, if you have a state [itex]|\psi\rangle = |\rho\rangle | \chi\rangle[/itex], then converting to the position representation will result in a wavefunction with an extra spin state parameter: [itex]\psi(x, s) = (\langle x| \langle s|)(|\rho\rangle | \chi\rangle)[/itex].

    Another equivalent way to write this is to say that we'll still just have a function of position [itex]\psi(x)[/itex], but that instead of being a simple complex number at each point in space, it's a pair of complex numbers, one for the probability that it's spin-up, and the other for the probability that it's spin-down. This method ends up being useful for a number of reasons as one gets further into the mathematics of spin, so it's very common. These two-component objects are spinors.
     
  4. Apr 15, 2012 #3
    But is it valid to factor [itex]\psi(x, s)[/itex] into a scalar-valued function of position and a "spinor-valued" function of position?
    So which is the spinor, the ordered pair of complex numbers or the wavefunction which outputs such pairs? Also, can this notion be generalized to arbitrary spin, not just for 1/2, by using ordered n-tuples of complex numbers. And do the complex numbers represent probabilities or just amplitudes?
     
  5. Apr 15, 2012 #4

    tom.stoer

    User Avatar
    Science Advisor

  6. Apr 15, 2012 #5

    martinbn

    User Avatar
    Science Advisor

    You can use some of those isomorphisms as [tex]V^*\otimes W \cong Hom(V,W)[/tex] and if you identify $V^*$ and $V$, you can represent the states as elements in the tensor product or as vector valued functions. For example for a single particle you can think of [tex]L^2(\mathbb R^3)\otimes \mathbb C^n[/tex] as [tex]Hom(L^2(\mathbb R^3), \mathbb C^n)[/tex].
     
  7. Apr 15, 2012 #6
    Well, I guess it depends on what you mean by "valid". Mathematically it's a thing you can do, but probably not a useful thing. You usually either write the wavefunction as a spinor-valued function of position, or a complex-valued function of position and spin.

    The ordered pair. The wavefunction which outputs such pairs is a spinor-valued function.

    Sorry, I meant amplitude. You have to square them just like usual to get a probability.

    Sure. For spin 1, it's a 3-component object, aka a vector. In general, for spin [itex]n[/itex], it's a [itex]2n+1[/itex] component object, so you can go as high as you want.

    The important thing about these multi-component objects, though, is that there's a meaningful concept of "rotation" on them. When I say meaningful, I mean that if you apply two rotation operations to one of them, the resulting object is consistent with what you'd get if you'd applied the single combined rotation to it--the rotation operations compose just like real rotations do. For the vector (spin 1) case, we already know what operations will do this--it's just the set of 3-dimensional rotation matrices. What's novel is that in the spinor case, we can find a set of 2-dimensional rotation matrices which do the same thing. For instance, a rotation of [itex]\theta[/itex] degrees about the Z axis is represented by [itex]e^{i \theta\sigma_z}[/itex], where [itex]\sigma_z[/itex] is the Z Pauli spin matrix.

    To use the fancy terminology, these matrices form a two-dimensional representation of SU(2). It's necessary to use objects which have a representation of the rotation group so that you can meaningfully talk about taking a system and rotating it in space--if there were no such representation, then it would be impossible to solve the Schrodinger equation and then change to a rotated basis, because there would be no way to transform the spinors.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How does spin work in the position representation?
Loading...