Continuous symmetries and generators · Quantum Mechanics

Intuition

Some symmetries come in families: translations by every distance, rotations by every angle, evolutions over every time. Such a family is built from infinitely many tiny steps, and a tiny step is the identity plus something small — an operator times the size of the step. That operator is the generator of the family, it is always Hermitian, and so it is an observable. This is why momentum, angular momentum and energy are the quantities they are: they generate translations, rotations and the passage of time.

A walk of a kilometre is a thousand steps of a metre, and knowing the direction of one step fixes the whole straight walk. The generator is the direction of the step; the family of transformations is the walk.

One-parameter families

A family of unitary operators with $\hat{U}(0)=1$ and $\hat{U}(s)\hat{U}(t)=\hat{U}(s+t)$ that depends smoothly on $s$ is the exponential of one Hermitian operator, its generator.

\begin{aligned}\hat{U}(s)&=e^{-is\hat{G}/\hbar},\qquad \hat{G}^{\dagger}=\hat{G}\\ \hat{U}(\varepsilon)&=1-\frac{i\varepsilon}{\hbar}\hat{G}+O(\varepsilon^{2})\end{aligned}

Properties

Every such family has a generator — Stone’s theorem, used here without proof. The factor $\hbar$ makes the generator carry the units of a physical quantity.
Unitarity of every $\hat{U}(s)$ forces $\hat{G}$ to be Hermitian, as the theorem shows: the generator is an observable.
A finite transformation is many small ones: .

The generator of a unitary family is Hermitian

\hat{U}(\varepsilon)=1-\frac{i\varepsilon}{\hbar}\hat{G}+O(\varepsilon^{2})\ \text{unitary}\ \Longrightarrow\ \hat{G}^{\dagger}=\hat{G}

Expand $\hat{U}^{\dagger}\hat{U}$ to first order in the small parameter. It must equal the identity for every small $\varepsilon$ , so the first-order term vanishes, and that term is proportional to ${\overset{}{G}}^{}$ .

Proof steps

\hat{U}(\varepsilon)=1-\frac{i\varepsilon}{\hbar}\hat{G}+O(\varepsilon^{2})

A small step of the family.

\hat{U}(\varepsilon)^{\dagger}=1+\frac{i\varepsilon}{\hbar}\hat{G}^{\dagger}+O(\varepsilon^{2})

The adjoint conjugates $-i$ into $+i$ .

\hat{U}^{\dagger}\hat{U}=1+\frac{i\varepsilon}{\hbar}\bigl(\hat{G}^{\dagger}-\hat{G}\bigr)+O(\varepsilon^{2})

Multiply and keep the first order.

\hat{U}^{\dagger}\hat{U}=1\ \text{for every small}\ \varepsilon\ \Rightarrow\ \hat{G}^{\dagger}-\hat{G}=0

A first-order term that vanishes for every $\varepsilon$ is zero.

Applications

Practice

Built From Small Steps

A continuous family of unitary transformations is generated by one Hermitian operator $\hat{G}$ . A small step is $1-i\varepsilon\hat{G}/\hbar$ ; a finite one is the exponential.

\hat{U}(s)=e^{-is\hat{G}/\hbar}

Try it

To first order in a small parameter $\varepsilon$ , what is $\hat{U}(\varepsilon)$ ?