I am currently reading up on the use of group theory in physics using Peter Woit's book draft (available on his homepage). I do understand the mathematical concepts but have a bit of a problem making the connection to the physical notation I am used to.

Specifically, if I have U(1) symmetry, the only irreducible representation is one dimensional meaning that the Hilbert space is also one dimensional or a direct sum of one dimensional Hilbert spaces. A state in this Hilbert space is then represented as a vector with complex numbers.

Now, how does this relate to a state represented in the occupation number formalism? When I see U(1) explained in physical texts, it is always represented as acting on the wave function or creation and annihilation operators with a global phase, but I do not see how to translate this to the one dimensional or multidimensional sum Hilbert space dictated by U(1).

Edit: By translate I mean the following: In the SU(2) case, I know that it admits a representation in terms of Pauli matrices which act on two dimensional vectors with complex entries and these complex entries represent my physical state. So it is very clear how I construct a representation of my group and how the space it acts on represents physics. In the U(1) case, I have simple exponentials labelled by the charge and they act on complex numbers. This complex number is then taken to be the phase of the wave function. To me, that is very different from the SU(2) case because the phase is not the wave function but might be a very different object such as an element of Fock space when we say that a many-body state is U(1) invariant.

I'd be grateful for any answer helping me to clear up this problem!

  • $\begingroup$ I don't understand the question - what do you mean by "translate"? Also, note that there are countably many different irreducible representations of $\mathrm{U}(1)$, which are all one dimensional. $\endgroup$
    – ACuriousMind
    Jun 4, 2015 at 14:34
  • $\begingroup$ Thank you for your reply! Specifically, I am thrown off by the statement that the Hilbert space has to be one dimensional which is not the case for a Fock space, for example. $\endgroup$
    – derpeter
    Jun 4, 2015 at 14:59
  • $\begingroup$ Noone requires the Fock space to be an irreducible representation of $\mathrm{U}(1)$. $\endgroup$
    – ACuriousMind
    Jun 4, 2015 at 15:03
  • $\begingroup$ Then how do you decompose an element of the Fock space into a reducible representation of U(1)? Elements of U(1) are complex numbers but a state in Fock space is given by the integer occupation numbers. That is what I mean by translation. $\endgroup$
    – derpeter
    Jun 4, 2015 at 15:31

1 Answer 1


I have worked recently on something related to this and I have understood that if you have a time evolution that respects a certain symmetry group, then the state space breaks down into its algebra multiplets. For instance, if you have 2 bosonic modes, you can recover the angular momentum algebra su(2) by labelling them by the total photon number and by building the ladder operators as $a^\dagger b$ and $a b^\dagger$. From there, since the ladder operators for the angular momenta are $J_x + iJ_y$ and $J_x -iJ_y$, you can keep playing the game and find all the equivalent operators of the rotation operators. So a Hamiltonian containing those operators will be equivalent to the Hamiltonian of a system under rotations, as long as you phrase the analogy correctly. For three modes you find the multiplets of su(3) and so on. So for one single mode it's a bit trivial because a Hamiltonian will contain only terms in $a$ and $a^\dagger$. So your state space does indeed break down into 1-dimensional subspaces.


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