Unitary Evolution

Postulates of an Isolated System:

  1. U(t2,t1)U(t_2,t_1) should be independent of initial state.
α∣ψ⟩+β∣ϕ⟩→U(t2,t1)(α∣ψ⟩+β∣ϕ⟩)= \alpha\left\lvert \psi \right\rangle+\beta\left\lvert \phi \right\rangle\rightarrow U(t_2,t_1)(\alpha\left\lvert \psi \right\rangle+\beta\left\lvert \phi \right\rangle)=

  1. must preserve probablity (must sum to 1) if isolated which means UU must be unitary if Hilbert space is finite dimensional.

  2. assume UU is unitary in infinite dimensional Hilbert space.

Unpacking 2. we see that

⟨ψ∣ψ⟩=1\left\langle \psi|\psi \right\rangle=1

This is because Born Rule says that, over ∣k⟩\left\lvert k \right\rangle basis states,

∑k∣⟨k∣ψ⟩∣2=1\sum_k |\left\langle k|\psi \right\rangle|^2=1

as probability must sum to 1. Using this lemma, we get

∑k∣⟨k∣ψ⟩∣2=∑k⟨ψ∣k⟩⟨k∣ψ⟩=⟨ψ∣(∑k∣k⟩⟨k∣)∣ψ⟩\sum_k|\left\langle k|\psi \right\rangle|^2=\sum_k\left\langle \psi|k \right\rangle\left\langle k|\psi \right\rangle=\left\langle \psi \right\rvert\left(\sum_k\left\lvert k \right\rangle\left\langle k \right\rvert\right)\left\lvert \psi \right\rangle =⟨ψ∣I∣ψ⟩=⟨ψ∣ψ⟩=\left\langle \psi \right\rvert I\left\lvert \psi \right\rangle=\left\langle \psi|\psi \right\rangle

so if we evolve it over time then

(U∣ψ⟩)†U∣ψ⟩=1∀∣ψ⟩(U\left\lvert \psi \right\rangle)^\dagger U\left\lvert \psi \right\rangle=1\quad\forall\quad\left\lvert \psi \right\rangle ⇒⟨ψ∣U+U∣ψ⟩=1\Rightarrow\quad \left\langle \psi \right\rvert U^+U \left\lvert \psi \right\rangle=1 ⇒U+U=I\Rightarrow\quad U^+U=I

This means that UU is unitary.

Note that

U(t0,t)=U−1(t,t0)=U†(t,t0)U(t_0,t)=U^{-1}(t,t_0)=U^\dagger(t,t_0)