Schrödinger picture

The time-evolution operator U(t, t₀) is defined as the operator which acts on the ket at time t₀ to produce the ket at some other time t:

$${\displaystyle |\psi (t)\rangle =U(t,t_{0})|\psi (t_{0})\rangle .}$$

For bras,

$${\displaystyle \langle \psi (t)|=\langle \psi (t_{0})|U^{\dagger }(t,t_{0}).}$$

Unitarity

The time evolution operator must be unitary. For the norm of the state ket must not change with time. That is,

$${\displaystyle \langle \psi (t)|\psi (t)\rangle =\langle \psi (t_{0})|U^{\dagger }(t,t_{0})U(t,t_{0})|\psi (t_{0})\rangle =\langle \psi (t_{0})|\psi (t_{0})\rangle .}$$

Therefore,

$${\displaystyle U^{\dagger }(t,t_{0})U(t,t_{0})=I.}$$

Identity

When t = t₀, U is the identity operator, since

$${\displaystyle |\psi (t_{0})\rangle =U(t_{0},t_{0})|\psi (t_{0})\rangle .}$$

Closure

Time evolution from t₀ to t may be viewed as a two-step time evolution, first from t₀ to an intermediate time t₁, and then from t₁ to the final time t. Therefore,

$${\displaystyle U(t,t_{0})=U(t,t_{1})U(t_{1},t_{0}).}$$

We drop the t₀ index in the time evolution operator with the convention that t₀ = 0 and write it as U(t). The Schrödinger equation is

$${\displaystyle i\hbar {\frac {\partial }{\partial t}}|\psi (t)\rangle =H|\psi (t)\rangle ,}$$

where H is the Hamiltonian. Now using the time-evolution operator U to write ${\displaystyle |\psi (t)\rangle =U(t)|\psi (0)\rangle }$,

$${\displaystyle i\hbar {\partial \over \partial t}U(t)|\psi (0)\rangle =HU(t)|\psi (0)\rangle .}$$

Since ${\displaystyle |\psi (0)\rangle }$ is a constant ket (the state ket at t = 0), and since the above equation is true for any constant ket in the Hilbert space, the time evolution operator must obey the equation

$${\displaystyle i\hbar {\frac {\partial }{\partial t}}U(t)=HU(t).}$$

If the Hamiltonian is independent of time, the solution to the above equation is[note 1]

$${\displaystyle U(t)=e^{-iHt/\hbar }.}$$

Since H is an operator, this exponential expression is to be evaluated via its Taylor series:

$${\displaystyle e^{-iHt/\hbar }=1-{\frac {iHt}{\hbar }}-{\frac {1}{2}}\left({\frac {Ht}{\hbar }}\right)^{2}+\cdots .}$$

Therefore,

$${\displaystyle |\psi (t)\rangle =e^{-iHt/\hbar }|\psi (0)\rangle .}$$

Note that ${\displaystyle |\psi (0)\rangle }$ is an arbitrary ket. However, if the initial ket is an eigenstate of the Hamiltonian, with eigenvalue E:

$${\displaystyle |\psi (t)\rangle =e^{-iEt/\hbar }|\psi (0)\rangle .}$$

The eigenstates of the Hamiltonian are stationary states: they only pick up an overall phase factor as they evolve with time.

If the Hamiltonian is dependent on time, but the Hamiltonians at different times commute, then the time evolution operator can be written as

$${\displaystyle U(t)=\exp \left({-{\frac {i}{\hbar }}\int _{0}^{t}H(t')\,dt'}\right),}$$

If the Hamiltonian is dependent on time, but the Hamiltonians at different times do not commute, then the time evolution operator can be written as

$${\displaystyle U(t)=\mathrm {T} \exp \left({-{\frac {i}{\hbar }}\int _{0}^{t}H(t')\,dt'}\right),}$$

where T is time-ordering operator, which is sometimes known as the Dyson series, after Freeman Dyson.

The alternative to the Schrödinger picture is to switch to a rotating reference frame, which is itself being rotated by the propagator. Since the undulatory rotation is now being assumed by the reference frame itself, an undisturbed state function appears to be truly static. This is the Heisenberg picture.