I study particle physics and am finally tired of pushing through QFT with annoying doubts which seem to be both very simple and fundamentally important, and to which several professors of mine couldn't give proper (and coherent with each other) answers. So, forgive if the question is stupid and know that I'm aware that this has already been answered probably dozens of times, for example here.

So, suppose we are considering some symmetry (Lie) group $G$, whose abstract elements $g$ act on our Hilbert space through a representation $U(g)$. If we want the transformations to conserve probabilities, the representation better be (projectively) (anti)unitary. Now, how this transformations act on our fields and states? The question sound very silly considering that the representation itself is defined to do what my question regards in a 'what does it do' sense, but I am confused with three situations:

(i) The transformations acts just as in the classical scenario, but with fields evolved to operators, that is, the (operator) fields transform as: $\phi \to U(g)\phi$. Then it appears that we use the result that this must have to be possibly represented by a unitary similarity transformation, and write

\begin{equation} \phi \to U(g)\phi=U'^{\dagger}(g)\phi U'(g). \end{equation}

Now, a representation (basis) $\{T_a\}_a$ of the algebra $\mathfrak{g}$ defines through the exponential map a representation of the group, and if we have the representation we want, we write

\begin{equation} \phi \to e^{i\alpha_aT_a}\phi=e^{-i\alpha_aT_a'}\phi e^{i\alpha_aT_a'}, \end{equation}

where the set $\alpha_a$ is one-to-one related to $g\in G$. I know this description is very sketchy, but hopefully understandable. Now, what is the relation between $T_a$ and $T'_a$ -- if there is one and this is not just something that works in some specific cases, for $U(1)$, for instance -- and why can the primed transformations $U$ also be written as exponentials? And is the reasoning I made for the equality in the first equation correct or is the cause and effect non-existing?

(ii) The transformations act as

\begin{cases} \left |\psi \right> \to U \left |\psi \right>, \\ \phi \to U\phi U^{-1} \end{cases}

preserving expected values. This is, of course, just the old linear algebra change of basis. I'm pretty sure that this is not how internal symmetry transformations act and there is not really a specific doubt here.

(iii) The transformation acts straight as it would be expected in the Heisenberg picture (in which we are working), leaving the states alone and changing the fields by

\begin{equation} \phi \to U^{\dagger}(g)\phi U(g)=e^{-i\alpha_aT_a}\phi e^{i\alpha_aT_a}. \end{equation}

This is the one I find most likely to be the correct affirmation, yet I ask:

If it is the case, what is wrong with the presented in (i)? And I would be immensely grateful if a summary of the formal aspects of the 3 'cases' could be made.


1 Answer 1


It looks like there are a few independent confusions in this question, so maybe doing a full example will help. Let's consider a complex scalar field in a theory with the $U(1)$ symmetry $$\phi \to e^{i \theta} \phi$$ where I suppress the $x$ coordinate. Indeed, the symmetry in the quantum case is simply $$\hat{\phi} \to e^{i \theta} \hat{\phi}.$$ However, we might want to know how the symmetry operators act on states, and to do that we'll have to work harder. By Noether's theorem, the conserved current is $$j^\mu = i \, (\phi \partial^\mu \phi^* - \phi^* \partial^\mu \phi).$$ The conserved charge is $$Q = i \int d\mathbf{x} \, (\phi \dot{\phi}^* - \phi^* \dot{\phi}).$$ When we pass to quantum field theory, the $\phi$'s here simply become field operators.

Now, the conserved charge always generates the symmetry, which is to say that $$U(e^{i \theta}) = e^{i \theta Q}.$$ This is not something foreign or unexpected; it occurs even in Hamiltonian mechanics, where conserved charges are the generating functions for the corresponding symmetries. The symmetry acts on states directly and on fields by conjugation, $$\hat{\phi} \to e^{i \theta Q} \hat{\phi} e^{-i\theta Q}, \quad |v \rangle \to e^{i \theta Q} |v\rangle.$$ Of course in a practical situation, you wouldn't do both of these transformations, you would do one or the other depending on whether you're thinking in Schrodinger or Heisenberg picture. As you said in point (ii), doing both is a no-op. The whole point of a symmetry is to relate different expectation values to each other, not rewrite a single one in a fancy way. If you're getting confused on this point, think back to a simpler case, such as the rotational symmetry of the hydrogen atom. We're not doing anything fundamentally differently here.

In the case of an infinitesimal symmetry, we have $$\hat{\phi} \to \hat{\phi} + i \theta Q \hat{\phi} - i \theta \hat{\phi} Q$$ to first order in $\theta$, which means $$\delta \hat{\phi} = i \theta \, [Q, \hat{\phi}].$$ Thus you'll sometimes hear the statement "[symmetry] operators act on operators by commutators". Again, this is not unfamiliar; you already saw this in Hamiltonian mechanics but with a Poisson bracket instead of a commutator.

Earlier I said $\hat{\phi} \to e^{i \theta} \hat{\phi}$, which implies $$\delta \hat{\phi} = i \theta \, \hat{\phi}.$$ Are these two statements consistent? Yes. All you have to do is evaluate the commutator explicitly, using the canonical commutation relations. These are valid even for interacting field theories, since we work in interaction picture. Since the commutation relations only hold at equal times, you must evaluate $Q$ at the same time you evaluate $\hat{\phi}$. (This is not an issue because, by the definition of a symmetry, $[Q, H] = 0$ so $Q$ is time independent.)

Your case (i) is bad, because you should distinguish between the mappings $\phi \to f(\phi, g)$ associated with classical symmetries and the unitary operators $U(g)$ that act on the Hilbert space. The mapping $f(\phi, g)$ does not have to be unitary, or even a linear transformation.

In the example above, you don't act on the states by just multiplying them by $e^{i \theta}$, which would be trivial. You instead act with $U(g)$. You can check that what it actually does is rotate with a factor of $e^{i \theta}$ for every matter particle and $e^{-i\theta}$ for every antimatter particle in the state.

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    $\begingroup$ Thanks so much for the answer, it! There is something I still don't quite understand: in the beginning you said that 'indeed, the symmetry in the quantum case is simply...' and then, after, showed the correct symmetry transformation (by conjugation). I understand, of course, that the transformations are generated by the Noether charges, and what I cannot exactly follow is the difference between the two statements reside. Is it that the first one is merely the realization of what symmetry the theory possess and the second is about what corresponding transformation leaves the physics invariant? $\endgroup$
    – GaloisFan
    Jul 22, 2018 at 23:27
  • $\begingroup$ @GaloisFan I'm not really sure what you mean here. Symmetries are transformations that leave the physics invariant. (I'm not sure what else you would mean by a "symmetry".) Given a classical symmetry, I've shown you two ways to write down how a symmetry acts on the fields: either copy what it did in the classical case, or more formally conjugate by an operator. Both are ways to write down precisely the same underlying symmetry. Please tell me if this doesn't resolve your confusion! $\endgroup$
    – knzhou
    Jul 23, 2018 at 9:45
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    $\begingroup$ @GaloisFan The reasoning is identical except that indices appear in most of the equations. The equation $U(e^{i\theta}) = e^{i \theta Q}$ simply becomes $U(e^{i T_a \theta^a}) = e^{i Q_a \theta^a}$. (This is consistent with the abelian case, where the generator $T$ is one.) $\endgroup$
    – knzhou
    Jul 24, 2018 at 14:20
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    $\begingroup$ @GaloisFan The quantity $T_a$ is not a conserved charge in any sense, neither classically nor in the quantum theory. I mean, look at the abelian case -- there $T_a$ is literally just the number $1$. $\endgroup$
    – knzhou
    Jul 24, 2018 at 14:22
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    $\begingroup$ @GaloisFan Ah, that's probably where you were confused. Both the $T_a$ and the $Q_a$ are generators for a Lie group of symmetries, classical and quantum ones respectively. That means that they represent infinitesimal symmetries, and you exponentiate them to get finite symmetries. But $T_a$ is not a conserved charge is, $Q_a$ is the corresponding conserved charge. $\endgroup$
    – knzhou
    Jul 24, 2018 at 15:19

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