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

A Why do space translations satisfy the Wigner's Theorem?

  1. Dec 28, 2017 #1
    Some books argue that typical coordinate transformations such as space translations and rotations are represented in quantum mechanics by unitary operators because the Wigner's theorem. However I do not find any clear proof of this. For instance, suppose 1D for the sake of simplicity, by definition spatial translations change the position operator as
    where "a" is a constant.

    I would want to prove that this transformation between operators can be implemented by a unitary operator. To this end, I try to apply the Wigner's theorem showing that the change X→X+a on the position operator induce a change in the state vectors |ψ>→|ψ'> which is one-to-one and preserves transition amplitudes, i.e. |<ψ|φ>|=|<ψ'|φ'>|.

    Any idea?
  2. jcsd
  3. Dec 28, 2017 #2
    Apply relativity. Your choice of basis is entirely arbitrary.
  4. Dec 28, 2017 #3

    Urs Schreiber

    User Avatar
    Science Advisor
    Gold Member

    Maybe noteworthy that there are two different but related statements by Wigner:

    On the one hand there is Wigner's theorem which says that if we already know/assume that a transformation is a symmetry (any kind of symmetry) in that it (is surjective and) preserves the norm of states, then it must be representable by a unitary or anti-unitary operator.

    On the other hand there is Wigner's classification which assuming that Poincare transformations (translations, rotations and boost) act as symmetries (as it does in relativistic field theories, by definition) classifies the irreducible representations of the corresponding group of unitary operators given by the first statement.
  5. Dec 28, 2017 #4
    I do not see how relativity allows you to prove that the map is a one-to-one correspondence between unit rays.
  6. Dec 28, 2017 #5
    Yes, it seems that space translations are always assumed to be a symmetry transformation. However, I wonder if there is a way to obtain this result from equation X→X+a.
  7. Dec 29, 2017 #6

    Urs Schreiber

    User Avatar
    Science Advisor
    Gold Member

    No, that's the thing: In general a QFT need not be invariant under translations. All the fundamental QFT that one considers are, so it's easy to forget that this is an assumption, but it is evident that one may consider Lagrangians which explicitly depend on spacetime.
  8. Dec 30, 2017 #7

    A. Neumaier

    User Avatar
    Science Advisor

    The unitary operator you ask for is ##e^{\pm iz\cdot p/\hbar}## with the appropriate sign, where ##p## is the momentum operator. In general, you exponentiate the corresponding infinitesimal generator.
  9. Dec 30, 2017 #8


    User Avatar
    Science Advisor

    Define [tex]T_{a} | x \rangle \equiv |x + a \rangle \ .[/tex] Then calculate the action of operator [itex]T_{a}[/itex] on the wavefunction as follows [tex]\begin{align*}\Psi^{\prime} (y) \equiv \left( T_{a} \Psi \right) (y) & = \langle y |T_{a} \Psi \rangle \\

    & = \int dx \ \langle y |T_{a}|x \rangle \langle x | \Psi \rangle \\

    & = \int dx \ \langle y | x + a \rangle \Psi (x) \\

    & = \int dx \ \delta (y - x - a ) \ \Psi (x) \\

    & = \Psi (y - a) \\

    & = e^{\left(- \frac{i}{\hbar} a \right) \cdot \left( - i \hbar \partial \right)} \ \Psi (y) \\

    & = e^{- \frac{i}{\hbar} a \cdot \hat{P}} \ \Psi (y) .

    \end{align*} [/tex]

    Thus [tex]\Psi^{\prime} = U(a) \Psi \ , \ \ U(a) = T_{a} = e^{- \frac{i}{\hbar} a \cdot \hat{P}} \ .[/tex] And [tex]\hat{X}(a) = U^{\dagger}(a) \hat{X}(0) U(a) = \hat{X}(0) + a I \ .[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted