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Doubt regarding rotations in spinor space

  1. Aug 25, 2013 #1
    Dear Sir,

    I am currently doing an advanced course in Quantum Mechanics. This current doubt of mine, I am unable to clarify it properly. It follows as:

    Spin 1/2 particles reside in 2dim-Hilbert space( Spinor Space)...However, we talk about rotations of states in this space where the angle of rotation is measured w.r.t Euclidean Space and we also build a Rotational Operator in this 2dim-space. My doubt is that why should spin states which reside in their own Hilbert Space respond to rotations that are carried in 3-dim Eucliean Space.? I was struck with this . Can you please help me to solve this ?
     
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  3. Aug 25, 2013 #2

    Bill_K

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    Well to begin with, be clear on one thing.. a Hilbert space is infinite dimensional. Spin space is 2-dimensional, i.e. not a Hilbert space.

    To say a spin-1/2 particle resides in spin space is a bit of an exaggeration. Its wavefunction is first and foremost, like all wavefunctions, a function of x and t. A spinor wavefunction has two components, whereas the vector wavefunction for a spin-1 particle has three. An important difference between the two is that the three components of the vector can be directly associated with the x, y and z axes, whereas the two components of a spinor cannot, and in this sense "lie in a different space."

    However, thanks to group theory (the two-to-one mapping of the SU(2) rotations in spin space to the SO(3) rotations in position space) a three-D rotation acts in a well-defined way on the two components of the spinor.
     
  4. Aug 25, 2013 #3

    dextercioby

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    Well, Bill, spin space is a 2-dimensional Hilbert space, it's actually [itex] \mathbb{C}^2 [/itex] endowed with a hermitean scalar product and complete wrt the metric topology induced by it via the norm.
     
  5. Aug 25, 2013 #4

    Bill_K

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    I love terminology debates! :smile:

    The Bible I was raised on, "Functional Analysis" by Angus E Taylor, says, "A complete, infinite-dimensional inner product space is called a Hilbert space."

    Wikipedia admits any number of dimensions, but then adds, "Hilbert spaces arise naturally and frequently in mathematics, physics, and engineering, typically as infinite-dimensional function spaces."

    What about Hilbert himself? Courant and Hilbert introduce the term when discussing the function space of solutions of a hyperbolic PDE. I don't imagine David Hilbert would call a two-dimensional space a "Hilbert space"! :wink:
     
    Last edited: Aug 25, 2013
  6. Aug 25, 2013 #5
    There is some ambiguity in the precise definition of "Hilbert Space", at least in some of the older literature. Some include only the infinite-dimensional case and others include finite-dimensional spaces too. What I think all definitions have in common is that a Hilbert space always has the L2 norm. For example, the classic "Principles of Mathematical Analysis" by Walter Rudin has this in its closing sentence:
    But in my own terminology, spin spaces are Hilbert spaces because they have an L2 norm.
     
    Last edited: Aug 25, 2013
  7. Aug 25, 2013 #6

    WannabeNewton

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    Sorry to jump in but a Hilbert space is simply a real or complex complete inner product space. For example ##l_2## is a Hilbert space under the usual inner product ##(x,y) = \sum \bar{x}_n y_n##. Another standard example is gotten by equipping ##\mathbb{R}^{n}## with a Borel measure ##\mu## and taking the set of all square-integrable complex-valued ##\mu##-measurable functions on ##\mathbb{R}^{n}## with the inner product ##(f,g) = \int \bar{f}(x)g(x)d\mu##. This is of course just ##L^2(\mathbb{R}^{n},d\mu)##.

    There is no restriction on the dimensionality of the space in full generality but that's just terminology :)
     
    Last edited: Aug 25, 2013
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