[Weinberg QFT] 2.2 Symmetries

This article is one of the posts in the Textbook Commentary Project.

Symmetry means the indistinguishability of information.

When symmetry is imposed on a theory, many degrees of freedom are removed and the range of possible expressions is reduced. 

Let’s express this process in a formula.

Symmtery change 'rays'

A symmetry transformation is a change in our point of view that does not change the results of possible experiments. If an observer $O$ sees a sysyem in a state represented by a ray $\mathcal{R}$ or $\mathcal{R}_1$ or $\mathcal{R}_2\cdots$, then an equivalent observer $O'$ who looks at the same system will observe it in a different state, represented by a ray $\mathcal{R}'$ or $\mathcal{R}'_2$ or $\mathcal{R}'_2\cdots$, respectively, but the two observers must find the same probabilities $$\begin{align}P(\mathcal{R}\rightarrow\mathcal{R}_n) = P(\mathcal{R}'\rightarrow\mathcal{R}_n')\end{align}$$

Two Types of Symmetry on Hilbert Space

An important theorem proved by Wigner in the early 1930s tells us that for any such transforamtion $\mathcal{R}\rightarrow\mathcal{R}'$ of rays we may define an operator $U$ on Hilbert space, such that if $\Psi$ is in ray $\mathcal{R}$ then $U\Psi$ is in the ray $\mathcal{R}'$, with $U$ either unitary and linear $$\begin{align}\left\langle U\Phi | U\Psi \right\rangle &= \left\langle \Phi | \Psi \right\rangle \\ U(\xi\Phi + \eta\Psi) &= \xi U\Phi + \eta U \Psi\end{align}$$ or else antiunitary and antilinear $$\begin{align}\left\langle U\Phi | U\Psi \right\rangle &= \left\langle \Phi | \Psi \right\rangle^* \\ U(\xi\Phi + \eta\Psi) &= \xi^* U\Phi + \eta^* U \Psi \end{align}$$ Complete proof is given at the end of this chapter in Appendix A.

As already mentioned, the adjoint of a linear operator $L$ is defined by $$\begin{align}\left\langle \Phi | L^\dagger \Psi \right\rangle \equiv \left\langle L\Phi | \Psi \right\rangle\end{align}$$ The adjoint of antilinear operator $A$ is defined by $$\begin{align}\left\langle \Phi | A^\dagger \Psi \right\rangle \equiv \left\langle A\Phi | \Psi \right\rangle^* = \left\langle \Psi | A \Phi \right\rangle \end{align}$$ With this definition, the conditions for unitarity or antiunitarity both take the form $$\begin{align} U^\dagger = U^{-1}\end{align}$$

There is always a tricial symmetry transformation $\mathcal{R}\rightarrow\mathcal{R}$, represented by the identity operator $U=1$. Most physical operators can be changed continuously to became identity operators. They are Unitary. The antiunitary operator appears related to time symmetry.

Unitarity and Observable

A symmetry transformation that is infinitesimally close to being tricial can be represented by a linear unitary operator that is infinitesimally close to the identity: $$\begin{align}U=1+i\epsilon t\end{align}$$ where $\epsilon$ a real infinitesimal. For this to be Hermitian and linear, so it is a candidate for an observarble.

The set of symmetry transformations has certain property that define it as a group. If $T_1$ is a transformation that takes rays $\mathcal{R}_n$ into $\mathcal{R}'_n$, and $T_2$ is another transformation that takes $\mathcal{R}'_n$ into $\mathcal{R}''_n$, then the result of performing both transformations is another symmetry transformation, which we write $T_2T_1$, that takes $\mathcal{R}_n$ into $\mathcal{R}''_n$. Also, a symmetry transformation $T$ which takes rays $\mathcal{R}_n$ into $\mathcal{R}'_n$ has an inverse, written $T^{-1}$, which takes $\mathcal{R}'_n$ into $\mathcal{R}_n$, and there is an identity transformation, $T=1$, which leaves rays unchanged.

Acting operator on Hilbert Space

The unitary or antiunitary operators $U(T)$ corresponding to these symmetry transformations have properties that mirror this group structure, but with a complication due to the fact that, unlike the symmetry transformations themselves, the operators $U(T)$ act on vectors in the Hilbert space, rather than on rays.

If $T_1$ takes $\mathcal{R}_n$ into $\mathcal{R}'_n$, then acting on a vector $\Psi_n$ in the ray $\mathcal{R}_n$, $U(T_1)$ must yield a vector $U(T_1)\Psi$ in the ray $\mathcal{R}'_n$, and if $T_2$ takes this ray into $\mathcal{R}''_n$, then acting on $U(T_1)\Psi_n$ in the ray $\mathcal{R}'_n$, and if $T_2$ takes this ray into $\mathcal{R}''_n$, then acting on $U(T_1)\Psi_n$ it must yield a vector $U(T_1)U(T_1)\Psi_n$ in the ray $\mathcal{R}''_n$. But $U(T_2T_1)\Psi_n$ is also in this ray, so these vectors can differ only by a phase $\phi_n(T_2,T_1)$ $$\begin{align} U(T_2)U(T_1) \Phi_n = e^{i\phi_n (T_2,T_1)}U(T_2T_1)\Psi_n\end{align}$$

Phases are independent of tha state

Furthermore, with one significant exception, the linearity (or antilinearity) of $U(T)$ tells us that these phases are independent of the state $\Psi_n$. Here is the proof. Consider any two different vectors $\Psi_A$,$\Psi_B$, which are not proportional to each other. Then, applying Eq. (10) to the state $\Psi_{AB}\equiv \Psi_A+\Psi_B$, we have $$\begin{align}e^{i\phi_{AB}} U(T_2 T_1) (\Psi_A + \Psi_B) = U(T_2)U(T_1) (\Psi_A + \Psi_B) = U(T_2) U(T_1) \Psi_A + U(T_2) U(T_1) \Psi_B  = e^{\phi_A} U(T_2 T_1) \Psi_A + e^{i\phi_B}U(T_2 T_1)\Psi_B\end{align}$$

Any unitary or antiunitay operator has an inverse (its adjoint) which is also unitary or antiunitary. Multiplying (11) on the left with $U^{-1}(T_2T_1)$, we have then $$\begin{align} e^{\pm i\phi_{AB}}(\Psi_A + \Psi_B)=e^{\pm i\phi_A}\Psi_A + e^{\pm i \phi_B}\Psi_B,\end{align}$$ the upper and lower signs referring to $U(T_2T_1)$ unitary or antiuniary, respectively. Since $\Psi_A$ and $\Psi_B$ are linearly independent, this is only possible if $$\begin{align}e^{i\phi_{AB}}=e^{i\phi_A}=e^{i\phi_B}.\end{align}$$

So as promised, the phase in Eq. (10) is independent of the state-vector $\Psi_n$, and therefore this can be written as an operator relation $$\begin{align}U(T_2)U(T_1)=e^{i\phi(T_2,T_1)}U(T_2T_1).\end{align}$$

Projective Representation

For $\phi=0$, this would say that the $U(T)$ furnish a representation of the group of symmetry transformations. For general phases $\phi(T_2,T_1)$, we have what is called a projective representation, or a representation 'up to a phase'. The structure of the Lie group cannot by itself tell us whether physical state-vectors furnish an ordinary or a projective representation, but as we shall see, it can tell us whether the group has any intrinsically projective representations at all.

Superselection Rule

The exception to the argument that that led to Eq. (14) is that it may not be possible to prepare the system in a state represented by $\Psi_A+\Psi_B$. For instance, it is widely believed to be impossible to prepare a system in a superposition of two states whose total angular momenta are integers and half-integers, respectively. In such cases, we say that there is a 'superselection rule' between different classes of states, and the phases $\phi(T_2,T_1)$ may depend on which of these classes of states the operators $U(T_2)U(T_1)$ and $U(T_2,T_1)$ act upon. We will have more to say about these phases and projective representations in Section 2.7. As we shall see there, any symmetry group with projective representations can always be enlarged (without otherwise changing its physical implications) in such a way that its representations can all be defined as non-projective, with $\phi=0$. Until Section 2.7, we will just assume that this has been done, and take $\phi=0$ in Eq. (14).

Lie algebra of Connected Lie group

There is a kind of group, known as a connected Lie group, of special importance in physics. These are groups of transformations $T(\theta)$ that are described by a finite set of real continuous parameters, say $\theta^a$, with each element of the group connected to the identity by a path within the group. 

The group multiplication law then takes the form $$\begin{align}T(\bar{\theta}) T(\theta) = T\left( f(\bar{\theta}, \theta)\right) \end{align}$$ with $f^a(\bar{\theta}, \theta)$ a function of the $\bar{\theta}$s and $\theta$s. Taking $\theta^a=0$ as the coordinates of the identity, we must have $$\begin{align}f^a(\theta, 0) = f^a (0, \theta) = \theta^a.\end{align}$$

As already mentioned, the transformations of such continuous groups must be represented on the physical Hilbert space by unitary (rather than antiunitary) operators $U(T(\theta))$. For a Lie group, these operators can be represented in at least a finite neighborhood of the ideneity by a power series \begin{align}U\left( T(\theta)\right) = 1 + i\theta^a t_a + \frac{1}{2} \theta^b \theta^c t_{bc} + \cdots,\end{align} where $t_a$, $t_{bc}\ (=t_{cb})$, etc. are Hermitian operators independent of the $\theta$s. Suppose that the $U(T(\theta))$ form an ordinary (i.e., not projective) representation of this group of transformations, i.e., $$\begin{align}U\left( T(\bar{\theta})\right) U\left( T(\theta)\right) = U\left( T(f(\bar{\theta}, \theta))\right) .\end{align}$$

Let us see what this condition looks like when expanded in powers of $\theta^a$ and $\bar{\theta}^a$. According to Eq. (16), the expansion of $f^a(\bar{\theta},\theta)$ to second order must take the form $$\begin{align}f^a(\bar{\theta}, \theta) = \theta^a + \bar{\theta}^a + f^a_{bc} \bar{\theta}^b \theta^c + \cdots\end{align}$$ with real coefficients $f^a_{bc}$. (The presence of any terms of order $\theta^2$ or $\bar{\theta}^2$ would violate Eq. (16)) Then Eq. (18) reads $$\begin{align}\left[ 1 + i \bar{\theta}^a t_a + \frac{1}{2} \bar{\theta}^b \bar{\theta}^c t_{bc} +\cdots  \right] \times \left[ 1 + i \theta^a t_a + \frac{1}{2} \theta^b \theta^c t_{bc} + \cdots\right] = 1 + i\left( \theta^a + \bar{\theta}^a + f^a_{bc} \bar{\theta}^b \theta^c + \cdots\right) t_a + \frac{1}{2} \left( \theta^b + \bar{\theta}^b + \cdots\right) \left( \theta^c + \bar{\theta}^c + \cdots\right) t_{bc} + \cdots\end{align}$$

The terms of order $1,\theta,\bar{\theta},\theta^2$, and $\bar{\theta}^2$ automatically match on both sides of Eq. (20), but from the $\bar{\theta}\theta$ terms we obtain a non-trivial condition $$\begin{align}t_{bc} = -t_b t_c - i f^a_{bc} t_a.\end{align}$$

This shows that if we are given the structure of the group, i.e., the function $f(\theta,\bar{\theta})$, and hence its quadratic coefficient $f^a_{bc}$, we can calculate the second-order terms in $U(T(\theta))$ from the generators $t_a$ appearing in the first-order terms. However, there is a consistency condition: the operator $t_{bc}$ must be symmetric in $b$ and $c$ (because it is the second derivative of $U(T(\theta))$ with respect to $\theta^b$ and $\theta^c$) so Eq. (21) requires that $$\begin{align}[t_b, t_c] = i C^a_{bc} t^a\end{align}$$ where $C^a_{bc}$ are a set of real constants known as structure constants $$\begin{align}C^a_{bc} \equiv -f^a_{bc} + f^a_{cb}.\end{align}$$

Such a set of commutation relations is known as a Lie algebra. In Section 2.7 we will prove in effect that the commutation relation (22) is the single condition needed to ensure that this process can be continued: the complete power series for $U(T(\theta))$ may be calculated from an infinite sequence of relations like Eq. (21), provided we know the first-order terms, the generators $t_a$. This does not necessarily mean that the operaors $U(T(\theta))$ are uniquely determined for all $\theta^a$ if we know the $t_a$, but it does mean that the $U(T(\theta))$ are uniquely determined in at least a finite neighborhood of the coordinates $\theta^a=0$ of the identity, in such a way that Eq. (15) is satisfied if $\theta,\bar{\theta}$, and $f(\theta,\bar{\theta})$ are in this neifhborhood. The extension to all $\theta^a$ is discussed in Section 2.7.

Abelian case

There is a special case of some importance, that we will encounter again and again. Suppose that the function $f(\theta,\bar{\theta})$ (perhaps just for some subset of the coordinates $\theta^a$ is simply additive $$\begin{align}f^a(\theta, \bar{\theta}) = \theta^a + \bar{\theta}^a\end{align}$$

This is the case for instanc for translations in spacetime, or for rotations about any one fixed axis (though not for both together). Then the coefficients $f^a_{bc}$ in Eq. (19) vanish, and so do the structure constants (23). The generators then all commute $$\begin{align}[t_b, t_c] = 0.\end{align}$$

Such a group is called Abelian. In this case, it is easy to calculate $U(T(\theta))$ for all $\theta^a$. From Eqs. (18) and (24), we have for any integer $N$ $$U(T(\theta)) = \left[U\left( T\left( \frac{\theta}{N}\right) \right) \right] ^N.$$

Letting $N\rightarrow\infty$, and keeping only the first-order term in $U(T(\theta/N))$, we have then $$U(T(\theta)) = \lim_{N\rightarrow \infty} \left[ 1+ \frac{i}{N} \theta^a t_a\right] ^N $$ and hence $$\begin{align} U\left( T(\theta)\right) = \exp(it_a \theta^a).\end{align}$$