I have been studying symmetries in quantum mechanics and I have come across two types:

  1. Given some transformation on a Hilbert space of states $\mathcal{H}$, an operator $U: \mathcal{H} \rightarrow \mathcal{H}$ is a symmetry operator if $$ U^\dagger H U = H$$ where $H$ is the Hamiltonian of the system.
  2. A symmetry transformation is either a unitary or anti-unitary operator on $\mathcal{H}$ (Wigner's Theorem).

Are these two definitions of symmetry equivalent?


2 Answers 2


There is some confusion between symmetries and dynamical symmetries. A Wigner symmetry is a bijective map $S$ associating a pure state, i.e. a normalized vector of the Hilbert space $\cal H$ up to phases, namely a set $[\psi]= \{e^{ia} \psi | a \in \mathbb C\}$, to a normalized vector up to phases $[\psi']$ (another pure state) preserving the probability transition: $$|\langle \psi| \phi \rangle |^2 = |\langle \psi'| \phi' \rangle |^2\:. \tag{1}$$ Wigner theorem proves that for every such $S$ there is an operator $U_S : {\cal H}\to {\cal H}$ which is unitary or antiunitary depending on $S$ and it is determined up to phases by $S$ itself, such that $$[U_S\psi] = S[\psi]\:.$$ It is evident that every unitary or antiunitary operator induces a map satisfying (1). This is the reason why one says that symmetries are all of unitary or antiunitary operators over ${\cal H}$.

Since $\langle U\psi| A U\psi\rangle = \langle \psi| (U^*A U)\psi\rangle$, acting on pure states with symmetries $\psi \to U\psi $ is equivalent to acting on observables $A \to U^*AU$. Hence, one can define a "dual" action on symmetries on observables: $$A \to U^*AU$$

All that makes sense if no superselection rules occur so that every unit vector represent a pure state (up to phases), otherwise a different notion of symmetry due to Kadison is more appropriate (equivalent to Wigner's one in the standard case).

A dynamical symmetry is a (unitary) symmetry $U$ with the following property with respect to the temporal evolutor of the system $V_t = e^{-itH}$: if I modify the initial state $\psi$ with $U$, the evolution $\psi_t = V_t\psi$ of $\psi$ changes with the same rule $V_t U\psi = U\psi_t$ (this is by no means obvious for arbitrary symmetries and evolutors, in general $U\psi_t$ is not the evolution of any initial state). In other words $$V_tU = UV_t \quad \mbox{for every $t \in \mathbb R$}\tag{2}\:.$$ (2) is equivalent to $$U^*V_tU = V_t \quad \mbox{for every $t \in \mathbb R$}\tag{2'}\:.$$ Taking the (strong) derivatives of both sides at $t=0$ we get $$U^*HU=H\:.$$ It is possible to prove that this identity is equivalent to (2') (there are subtleties with domains).

  • $\begingroup$ Is it correct to think of Wigner's definition of a symmetry as a passive transformation of a state due to a change of frame of reference? This is something which Weinberg suggests i.e. the ray $[\psi]$ is what one observer would say the state of the system is, whilst $[ \psi']$ is what the other observer would say it's in. It is at this point I am a bit confused about whether $\psi$ is the abstract state vector that is basis independent or $\psi$ is in fact the representation w.r.t. a particular basis. I.e. an abstract vector $X$ versus the column vector $\mathbf{x}$. $\endgroup$ Sep 30, 2018 at 18:53
  • $\begingroup$ My idea is that symmetries are active: they represent active actions on the state. Changes of reference frame indirectly define similar transformations since they are (passive) isometries...but this is an involved (for the reason you remark) and reductive point of view in my opinion. $\endgroup$ Sep 30, 2018 at 19:21
  • $\begingroup$ Everyrhing is basis independent. $\endgroup$ Sep 30, 2018 at 19:23
  • $\begingroup$ Imagine a system with a Hamiltonian $H=P^2/2m+V(x)$ that doesn't have translation invariance (due to the potential). The translation operator $T=e^{-iP}$ is a symmetry in the Wigner sense, just due to being unitary, but it's clearly not a symmetry in the usual sense, since it doesn't commute with the Hamiltonian. Shouldn't we be calling unitary operators "physical transformations" instead of symmetries? They are clearly not necessarily symmetries, as the translation operator for systems without translation symmetry shows. $\endgroup$ Nov 14 at 17:52

In the first definition you have not made $U$ preserve the $\mid \langle \psi \mid \phi \rangle \mid$. As an example, suppose $H=0$ (a trivial system), then you would get all operators $U$ as symmetries even if they aren't even unitary or anti-unitary. Those shouldn't have been called symmetries. You want both conditions not either.

The implications go the other way

Symmetry => Unitary or Anti-unitary

Symmetry => Preserves Hamiltonian under $U^\dagger H U$

You need both conditions to say something might be called a symmetry.


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