# (Coming from Wigner's Theorem): What is a Symmetry in QFT?

In classical mechanics, classical field theory and QM, I was introduced to the concept of "Symmetry" as some kind of active transformation of either spacetime / time or configuration space (or of the phase space in hamiltonian-formalism) that leaves the "laws" untouched, that is, the transformed positions / fields / whatever timely evolve exactly as the untransformed would do.

Now I read about Wigner's Theorem, which somehow relates the notion of a symmetry to the scalar-product of two states in a Hilbert space - Why is this scalar-product for any two states connected to a symmetry, and can you describe what exactly constitutes a symmetry in this formalism?

EDIT: Some of the answers (and a comment) suggest that the "transition probabilities" don't change, if the condition that $$| \langle \Psi | \Phi \rangle | = | \langle T \Psi |T \Phi \rangle |$$ holds. This is probably at the heart of my question, as I'm not used to the skalar-product being linked to a "transition probability". In what formalism, / which quantum picture is this transition probability set? What transition does happen? Or is it about the quantum collapse in the moment of a measurement, when the state $$|\Psi\rangle$$ is projected $$|\Phi\rangle$$ with exactly that probabillity? In any case, does that mean "symmetry" in this context means that the transition probabilities don't change (whatever this transition is)?

• This extract from Weinberg's Lectures in QM explains it nicely: "A symmetry principle is a statement that, when we change our point of view in certain ways, the laws of nature do not change. For instance, moving or rotating our laboratory should not change the laws of nature observed in the laboratory.Such special ways of changing our point of view are called symmetry transformations. (continued in the next comment)
– user137661
Nov 7 '19 at 15:55
• This definition does not mean that a symmetry transformation does not change physical states, but only that the new states after a symmetry transformation will be observed to satisfy the same laws of nature as the old states.In particular, symmetry transformations must not change transition probabilities.". As you probably know, transition probabilities are given by (the modulus square of) scalar products. That is why Wigner's theorem involves them.
– user137661
Nov 7 '19 at 15:57
• @SV "As you probably know, transition probabilities are given .... ": This is precisely the thing I don't know (I edited my question to point this out): In what formalism is a transition probability given by a mere modulus square of scalar products? What transition? Valter Moretti's answer even suggests that this notion of "transition" doesn't deal with the dynamics of the system at all. Nov 11 '19 at 9:38
• Yes your last two sentences are precisely what's happening. See my edit.
– lcv
Nov 11 '19 at 11:22
• @Quantumwhisp although the answer by Icv is comprehensive there is one more thing I would add that can be more intuitive to think about. In QM we associate basis vectors in Hilbert space to the eigenvectors of observables (ideally we want a "complete set of commuting observables" CSCO) . Any state vector can be expressed as a superposition (i.e., linear combination) of this basis, and thr coefficients of the superposition are the scalar products themselves (if the basis is orthonormal) $\vert\Psi\rangle=\sum\limits_{i}\langle\Phi_i\vert\Psi\rangle\vert\Phi_i\rangle$.
– user137661
Nov 11 '19 at 17:53

There are two distinct notions of symmetry in quantum theory which does include QFT. The former regards the structure of the space of the states (or also the algebra of observables). The latter uses the former notion but concerns the dynamical evolution.

Wigner theorem regards the first type of symmetry.

A symmetry (of first type), is a map $$S$$ from the set of the states to the set of the states which is bijective (so that, in particular, it is physically reversible) and preserves some structure which properly defines which notion of symmetry one choose (always unrelated with the dynamical evolution). Two possibilities are Wigner symmetries and Kadison symmetries.

(a) Wigner symmetry: the space of the states is the space $${\cal S}_p$$ of pure states and the bijective map $$S : {\cal S}_p \to {\cal S}_p$$ preserves the transition probability(see the final remark) of a pair of states.

As is well known, in the absence of superselection rules, pure states are unit vectors $$\psi$$ up to phases of the Hilbert space of the system. So $$\psi$$ and $$e^{i\alpha} \psi$$ represent the same pure state $$[\psi] := \{e^{i\alpha} \psi\:|\: \alpha \in \mathbb R\}\:.$$ The transition probability of two states respectively represented by unit vectors $$\psi$$ and $$\phi$$ is the non-negative number $$P([\psi],[\phi]):=|\langle\psi|\phi\rangle|^2$$ These numbers can be experimentally measured. A Wigner symmetry is a bijective map $$S : {\cal S}_p \ni [\psi] \mapsto [\psi']\in {\cal S}_p$$ such that $$P([\psi],[\phi]) = P([\psi'],[\phi'])$$ for all choices of $$[\psi]$$ and $$[\phi]$$.

(b) Kadison symmetry the space of the states is now the whole space $${\cal S}$$ of generally mixed states and the bijective map $$S : {\cal S} \to {\cal S}$$ preserves the convex structure of $${\cal S}$$.

In the absence of superselection rules, generally mixed states are trace-class unit-trace positive operators $$\rho : H \to H$$ also known as statistical operators. If $$\rho, \pi \in {\cal S}$$ and $$p,q \in [0,1]$$ with $$p+q=1$$, the convex combination $$p\rho+q\pi$$ is still a state ($$p$$ and $$q$$ are the weights of the mixture).

A Kadison symmetry is a bijective map $$S : {\cal S} \ni \rho \mapsto\rho \in {\cal S}$$ such that $$(p \rho+q\pi)' = p\rho' + q \pi'$$ for all choices of $$\rho, \pi \in \cal S$$ and $$p,q \in [0,1]$$ with $$p+q=1$$.

I stress that there are something like 6 or 7 notions of symmetry in quantum theory (some deals with the space of observables instead of that of states as I said above, for instance isomorphisms of the orthomodular lattice of elementary observables, or isomorphisms of the von Neumann algebra of observables). All these notions are however finally proved to be identical and related with the unitary or anti unitary operators in the Hilbert space of the system.

The Wigner/Kadison theorems characterize the corresponding symmetries as follows.

If $$S$$ is such a symmetry, there is a unitary or antiunitary (depending on the nature of $$S$$) linear operator $$U : H \to H$$ which implements $$S$$, in the sense that, respectively, $$\psi' = U\psi\quad \mbox{or}\quad \rho' = U\rho U^{-1}\:.$$ Furthermore $$U$$ is determined by $$S$$ up to a phase: $$U$$ and $$e^{i\alpha}U$$ define the same symmetry and there are no other possibilities to represent $$S$$ (unless $$\dim H=1$$).

Notice that, in particular, Wigner symmetries and Kadison symmetries coincide though constructed referring to different aspects of the theory.

The notion of symmetry is therefore quite abstract, but it is always related with some concrete idea. For instance, if we perform quantum experiments staying in an inertial reference frame, we expect that all our experiments on isolated systems are translationally invariant. I meant that if I perform the experiment here or there on an isolated quantum system then, intrinsic properties, like transition probabilities of pairs of states (prepared here and there with the same procedure) must be identical. This idea automatically implies that

spatial translations must act in terms of Wigner symmetries on the quantum system

and, in turn, Wigner's theorem entails that

there must be a unitary operator (antiunitary is impossible here) which represent a given spatial translation.

This notion of symmetry is strictly entangled with that of dinamical evolution first and that of dynamical symmetry next.

Suppose that $$S_t : {\mathbb R} \in t \mapsto [\psi_t]$$ defines the evolution law of pure states of a given quantum system.

It is possible to prove from basic principles and fair mathematical assumptions that $$S_t$$ is a Wigner symmetry for every fixed $$t$$, so that it is represented by a class of unitary operators $$U_t$$ defined up to phases. In view of a theorem due to Bargmann, it is possible to prove that phases can be rearranged in order to satisfy $$U_{t+t'}= U_tU_{t'}$$ and $$U_0=I$$ so that $$\psi_t = U_t \psi_{0}$$ Furthermore, a natural continuity requirement (already used actually in applying Bargmann's theorem) imply $$U_t\psi \to \psi$$ for $$t\to 0$$. In view of Stone's theorem, we can write $$U_t = e^{-itH}$$ for some selfadjoint operator $$H$$ (defined up to additive constants changing the arbitrary phase of $$U_t$$) called the Hamiltonian of the system.

Here the notion of dynamical symmetry comes out. Suppose that we are studying a Wigner symmetry $$S$$ represented by the (anti)unitary operator $$V$$. And consider $${\mathbb R} \ni t \mapsto V\psi_t$$ In other words we change the state at every time with that symmetry. The natural question is whether or not $$V\psi_t$$ is still the evolution of the system with another initial state: $$VU_t\psi_0 = U_t \phi_0 \tag{1}$$ (There is a subtlety regarding the arbitrary phase in passing from the real state $$[\psi_t]$$ to a representative vector $$\psi_t$$, but I will not enter into the details here.) Taking $$t=0$$ in (1), we have that $$V\psi_0=\phi_0$$. Since $$\psi_0$$ is arbitrary, the requirement above is $$V U_t = U_t V \quad \mbox{for all t\in \mathbb R}\tag{2}$$ I.e., the action of the symmetry $$S$$ commutes with the time evolution. Requirement (2) is the definition of a dynamical symmetry.

If $$S$$ is in turn a continuous symmetry, a quantum implementation of Noether theorem arises, but it is far form the initial question so I stop here.

QFT in Hilbert space (there are other more abstract formulations) is nothing but a quantum theory in a specific Hilbert space (usually taken as a Fock space) and where the algebra of observables is of a specific kind: generated by elementary observables called field operators.

In the case of a real scalar field, it is a map associating smooth real-valued functions on the spacetime $$f: M \to \mathbb R$$ to selfadjoint (essentially selfadjoint in general) operators $$f \mapsto \Phi(f) = \int_{M} \Phi(x) f(x) d^4x$$ The notion of symmetry simply specialises to this context, so that it is still given in terms of unitary or antiunitary operators, but also it is often defined by fixing its action on the said elementary observables.

For instance if $$T_y$$ is the action of spacetime translations on smearing functions $$(T_y f)(x) = f(x-y)$$ we can assume that there is a corresponding symmetry (unitary operator) $$U_y :H \to H$$ whose action is compatible with $$U_y \Phi(f) U_y^{-1} = \Phi(T_y f)\tag{3}$$ Usually there is a preferred unit vector $$\Omega \in H$$ (the vacuum) such that the linear combinations of vecotors $$\Phi(f_1) \cdots \Phi(f_n)\Omega$$ is dense in $$H$$. If (3) is accompanied by $$U_y \Omega = \Omega \tag{4}$$ it is not difficult to prove that there exists, in fact, a unique unitary map satisfying both (3) and (4).

Many symmetries in QFT are constructed in that way. There are more complicated symmetries arising for instance from the fact that there is an internal Hilbert space for the fields, describing charges, or the field is not a simple scalar but a section $$\Phi_a$$ of some vector bundle over the spacetime so that geometric symmetries $$T$$ act into a more complicated way in the right-hand side of (3) (I pass to a pointwise representation of fields) $$U_T \Phi_a(x) U_T^{-1} = \sum_b M^{(T)}_{ab}(x) \Phi_b(T^{-1}x)$$

Remark. The notion of probability transition is pervasive in Quantum Theory. If we start with a pure state state represented by the unit vector $$\psi$$ and we simultaneously measure a set of observables $$A_1,\ldots, A_n$$ which are maximally compatible, the post-measurment state is a unique state (uniqueness corresponds to "maximality") represented by the unit vector $$\psi_{a_1,\ldots, a_n}$$ (defined up to a phase), where $$a_1,\ldots, a_n$$ are the values of the outcomes the mesurments of the observables $$A_1,\ldots, A_n$$.

We have such a vector $$\psi_{a_1,\ldots, a_n}$$ for every choice of the numbers $$(a_1,\ldots, a_n)$$.

The probability to obtain a certain set of outcomes $$(a_1,\ldots, a_n)$$ when the state is represented by $$\psi$$ is therefore, according to the postulates of QT, $$P(a_1,\ldots, a_n, \psi) = |\langle \psi_{a_1,\ldots, a_n} | \psi\rangle |^2\:.$$ Since, according to the axioms of QT, the state immediately after the measurement is $$\psi_{a_1,\ldots, a_n}$$, it is natural to call $$P(a_1,\ldots, a_n, \psi)$$ the transition probability from $$\psi$$ to $$\psi_{a_1,\ldots, a_n}$$, even if that probability concerns a set of outcomes of simultaneous measurments of observables.

It is also important to stress that, when assuming that every selfadjoint operator of the Hilbert space represents an observable, as is postulated in QT in the absence of superselection rules and gauge groups, it turns out that

every state vector $$\phi$$ can be written in the form $$\phi = \psi_{a_1,\ldots, a_n}$$ for a suitable (generally non-unique) choice of the compatible observables $$A_1,\ldots, A_n$$.

The transition probability $$|\langle \phi| \psi\rangle|^2$$ can be experimentally measured in terms of statistical frequencies (after having fixed a set of observables $$A_1,\ldots, A_n$$ for $$\phi$$ as above). This is the reason why Wigner decided to use it a a building block to formulate his notion of symmetry.

• Thank you for your long and elaborate answer. It seems that there is sadly one detail missing which makes me still wonder (I edited the original question to explicitly point this out): Why are the squared absolute values of the scalar-product associated with a "transition probability"? What is transitioning here? I'm especially wondering why this "transition" seems to be an intrinsic property of the hilbert space, and no property of the systems dynamics (as you suggest in your answer). Is the term misleading me here? It would be nice if you could add this "missing bit" in your answer. Nov 11 '19 at 9:30
• @Quantumwhisp I now added a final remark on that issue. Nov 11 '19 at 10:41

This answer is about symmetry in quantum systems in general in the context of Wigner's theorem; however, it is applicable to the case of quantum field theory. I'll consider, as a motivation, the case of classical systems which has analogous definitions of symmetry.

A classical Hamiltonian system can be defined by the principle of least action, which in the Hamiltonian picture has the following form (in the case of systems with translational degrees of freedom): $$I = \int p dq – H(p, q) dt$$ For more general phase spaces , which may include spin for example, the action can be written as: $$I = \int \alpha(x) – H(x) dt$$ Where $$\alpha$$ is a symplectic potential (one-form)and $$x$$ is a coordinate in phase space.

The classical system has two properties:

1. The phase space, entering the least action principle through the symplectic potential $$\alpha$$ (or $$p dq$$ in the special case).

2. The Hamiltonian $$H$$ which controls the dynamics of the system.

The phase space (together with its symplectic form) is the primary characteristic of the system, for example, an electron has three translational degrees of motion and one spin degree of motion. This defines its phase space. This electron can be subjected to various Hamiltonians, for example, coupled to a uniform electric field as in a capacitor, or a central force electric field as in atoms, or a magnetic field, etc.. In each case, it will be an electron, but in each case, the Hamiltonian is different.

Considering a single electron, the phase space is the space of states that a classical electron can assume. Its state at any time is determined by the three coordinates of its position, its three momentum components and its spin direction.

A transformation of the phase space: $$x\rightarrow x'$$ Or in the special case $$q\rightarrow q'$$ $$p\rightarrow p'$$ Which preserves the symplectic potential $$\alpha$$ ( $$p dq$$ in the special case) up to boundary terms, but not necessarily preserves the Hamiltonian, is a canonical transformation. This transformation is a symmetry of the phase space, despite the fact that it doesn't Poisson commute with the Hamiltonian. If in addition, the symmetry Poisson commutes with the Hamiltonian, it is called a symmetry of the dynamics (This distinction is emphasized by Freed and Moore (text before equation (9.4)), although for the quantum case which I'll discuss soon). Of course, the group of dynamical symmetries is a subgroup of the group of canonical transformations.

So, the intuition about the symmetries which are canonical transformations, (which will be generalized to the quantum case) is that they do not change the primary character of the systems: electrons remain electrons.

Before going to the quantum case, let me remark that a phase space is not the most general description of a classical state space. A phase space is rather the state space of a classical pure state space. Classical mechanics can describe more general (statistical) systems whose states are defined by probability distributions over the phase space. These are just the classical analogs of mixed states. If this is our case, then a symmetry should be such that these by probability distributions should be invariant. I am mentioning this point, because, in the main reference that I'll follow in the answer, namely, Landsman, the author defines a symmetry on various descriptions of the quantum state space, pure, mixed, etc. and furthermore defines the equivalence between these definitions.

In order to "quantize" our definition of symmetry above, we need first to think what the corresponding quantum state space is. The system's Hilbert space doesn't qualify because the overall magnitude and phase of a state vector are immaterial to a quantum system state. It is known, for example, that a projective Hilbert space (in which collinear vectors are identified) is a right description of a quantum state space, however, only in the case of pure states.

Therefore, restricting our attention for a moment to pure states, a symmetry should be primarily defined as a transformation of the projective Hilbert space. But which transformations qualify; remember that in the classical case, there was a restriction that the transformation should preserve the symplectic form up to boundary terms. Interestingly, projective Hilbert spaces are also symplectic manifolds and the restriction in the quantum case remains the same, i.e., a (quantum) symmetry of the projective Hilbert space should preserve its symplectic structure. Moreover, the projective Hilbert space possesses a metric compatible with its symplectic structure, thus the same symmetry should be an isometry of this metric. This is the first definition of symmetry that Landsman gives in equations (5.7), (5.8) in his book. This condition can be also expressed by means of the preservation of the inner product as in the question above.

What Wigner's theorem specifies is how these symmetries of the quantum state space look in the underlying Hilbert space (which, as argued above, is not the state space); and the answer is unitary or anti-unitary transformations.

Please, observe that in the above discussion the Hamiltonian was not mentioned, because we are dealing with symmetries of the state space, and not with symmetries of the dynamics.

However, Landsman, is not satisfied with the single definition of symmetry above valid only for pure states (which he calls Wigner symmetry). He defines quantum state spaces of ascending degrees of generality, equations (5.1)-(5.6). For example, the space in (5.2) is the space of density matrices which covers also mixed states. The analogous definition of symmetry in this case is that it preserves convex combinations of density matrices.

The spaces (5.3)-(5.6) are spaces of operator algebras which can be used to characterize a quantum system. Landsman gives the analogous definitions of symmetry for his various description of the quantum state spaces in equations (5.7)-(5.17) and proves the equivalence of all definitions.

I would say a way to understand why symmetries and scalar product are related is the following.

Physical quantum (pure) states are not really vector in a Hilbert space, rather they are rays, that is equivalence classes of vectors which differ by a complex number (or normalized vectors which differ by a phase).

Now the natural distance between quantum states, understandably, is not the standard euclidean distance such as $$\parallel x - y \parallel$$, because this function changes if we multiply $$x, y$$ by phases.

Instead a meaningful distance for quantum states is the following Bures distance:

$$D(\phi,\psi) :=\sqrt{2(1- | \langle \phi | \psi \rangle | )} \ \quad \quad (1)$$

where $$\phi,\psi$$ are pure quantum states, that is, rays (equivalence classes of...).

Now it is clear that Eq. (1) is invariant if the modulus of the scalar product is invariant, i.e., a transformation $$T$$ leaves the distance between quantum states invariant if

$$| \langle T \phi | T \psi \rangle | = | \langle \phi | \psi \rangle |$$

for all quantum states $$\phi, \psi$$. This is precisely the hypothesis of Wigner's theorem. The answer is that $$T$$ must be unitary or anti-unitary.

$$| \langle \phi | \psi \rangle |^2. \ \ \quad (1)$$
Where the system is in state $$|\psi\rangle$$ and the operator we are measuring has an eigenvector $$|\phi\rangle$$. This is the probability that after the measurement the system will be in state $$|\phi\rangle$$. That is why it's called transition probability. All this is the content of Born rule.