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Very stupid question about Wigner's Theorem

  1. Aug 30, 2010 #1
    Wigner's representation theorem says that any invertible transformation between rays of a Hilbert space that preserves transition probabilities can be implemented by a transformation on the Hilbert space itself, which is either unitary or antiunitary, depending on the particular transformation considered, right?

    For example, you can find in any QM book that almost all symmetries are represented by linear operators, the only significant exception being time inversion, right?

    I present here the most trivial example I can imagine: a one-dimensional Hilbert space. The two complex function of complex variable

    [tex]f(z)=z\qquad\textrm{and}\qquad g(z)=z^*[/tex]

    are, respectively, unitary and antiunitary, and they both induce the same transformation between rays of the Hilbert space [tex]H=\mathbb{C}[/tex], the identity transformation.

    I know the solution of this apparent paradox should be easy, but I really can't see it! I know the proof of the theorem, and it doesn't help!

    Any hint woukd be appreciated.
    Last edited: Aug 31, 2010
  2. jcsd
  3. Aug 30, 2010 #2


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    I don't understand your example. How is the operation of complex conjugation
    equivalent to the identity transformation? If you're saying the z (a complex number) is
    a complex multiple of z*, then effectively there's only one state (ray), and all (invertible)
    transformations are the identity transformation. The transformations are different if
    the Hilbert space is considered as a vector space, but coincide if we revert to a projective
    space of rays.

    Wigner's theorem doesn't say (iirc) that the unitary and antiunitary transformations
    must be distinct for every possible type of Hilbert space.

    Maybe try a slightly less trivial example of a 2D Hilbert space?
  4. Aug 31, 2010 #3
    Yes, using the definition of rays, the two transformations coincide (you have to consider the Hilbert space as a space on complex numbers, but that's what you usually do in QM). In a one-dimensional Hilbert space there are actually two rays, because (but this probably depends on the definition you adopt) z=0 is a ray all alone. Not only the invertible transformations induce the identity on rays, it is sufficient, for example, that f(0)=0 and that f(z) is different from zero for z different from zero (you can find horrible functions that belong to this class!!) and the induced map will be the identity anyway!

    Well, if a transformation is unitary, then it can't be antiunitary and vice versa...so I suppose they must be different!
  5. Aug 31, 2010 #4
    No. Should be: "of a Hilbert space of dimension at least two."
    Last edited: Aug 31, 2010
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