Time Evolution Operators In Physics

Classical

In classical mechanics we will use Hamiltonian dynamics relating the time evolution of real vectors of position and momentum $\vec{q}, \vec{p}$ .

\frac{d q_{i}}{d t} = \frac{\partial H (\vec{q}, \vec{p}, t)}{\partial p_{i}}

\frac{d p_{i}}{d t} = - \frac{\partial H (\vec{q}, \vec{p}, t)}{\partial p_{i}}

Quantum

In quantum mechanics we use Schrodinger's equation that relates the time evolution of a vector in a Hilbert space for a Hamiltonian operator.

\frac{\partial}{\partial t} ψ (\vec{x}, t) = - \frac{i}{ℏ} \hat{H} (t) ψ (\vec{x}, t)

Time evolution of state

Classical

Define an operator for the derivative of the state vector of positions and mometum.

k (t) (⟨ \vec{q} (t), \vec{p} (t) ⟩) = \frac{d ⟨ \vec{q} (t), \vec{p} (t) ⟩}{d t} = ⟨ \frac{\partial H (t)}{\partial p_{i}}, \dots, \frac{\partial H (t)}{\partial q_{i}} ⟩

An evolution for a small change in $Δ t$ of the state vector can be approximated as

⟨ \vec{q} (t + Δ t), \vec{p} (t + Δ t) ⟩ = ⟨ \vec{q} (t), \vec{p} (t) ⟩ + (\int_{t}^{t + Δ t} k (t^{'}) d t^{'}) (⟨ \vec{q} (t), \vec{p} (t) ⟩)

= (1 + \int_{t}^{t + Δ t} k (t^{'}) d t^{'}) (⟨ \vec{q} (t), \vec{p} (t) ⟩)

so in the limit of many evolutions it should become exact.

⟨ \vec{q} (t + Δ t), \vec{p} (t + Δ t) ⟩ = \lim_{n \to \infty} (1 + \frac{1}{n} \int_{t}^{t + Δ t} k (t^{'}) d t^{'})^{n} (⟨ \vec{q} (t), \vec{p} (t) ⟩)

Call this operator exponentiation.

⟨ \vec{q} (t + Δ t), \vec{p} (t + Δ t) ⟩ = e^{\int_{t}^{t + Δ t} k (t^{'}) d t^{'}} (⟨ \vec{q} (t), \vec{p} (t) ⟩)

Define this as the "time evolution operator"

ℳ (t_{a}, t_{b}) = e^{\int_{t_{a}}^{t_{b}} k (t^{'}) d t^{'}}

so that it evolves a state from time $t_{a}$ to $t_{b}$ .

⟨ \vec{q} (t_{b}), \vec{p} (t_{b}) ⟩ = ℳ (t_{a}, t_{b}) ⟨ \vec{q} (t_{a}), \vec{p} (t_{a}) ⟩

Quantum

An evolution for a small change in time $Δ t$ can be approximated as

ψ (t + Δ t) \approx ψ (t) + (\int_{t}^{t + Δ t} - \frac{i}{ℏ} \hat{H} (t^{'}) d t^{'}) ψ (t)

= (1 + \int_{t}^{t + Δ t} - \frac{i}{ℏ} \hat{H} (t^{'}) d t^{'}) ψ (t)

so in the limit of many evolutions it should become exact.

ψ (t + Δ t) = \lim_{n \to \infty} (1 + \frac{1}{n} \int_{t}^{t + Δ t} - \frac{i}{ℏ} \hat{H} (t^{'}) d t^{'})^{n} ψ (t)

Call this operator exponentiation.

ψ (t + Δ t) = e^{- \frac{i}{ℏ} \int_{t}^{t + Δ t} \hat{H} (t^{'}) d t^{'}} ψ (t)

Define this as the "time evolution operator"

ℳ (t_{a}, t_{b}) = e^{- \frac{i}{ℏ} \int_{t_{a}}^{t_{b}} \hat{H} (t^{'}) d t^{'})}

so that it evolves a state from time $t_{a}$ to $t_{b}$ .

ψ (t_{b}) = ℳ (t_{a}, t_{b}) ψ (t_{a})

Operator exponentiation

Operator exponentiation can be also defined like below.

e^{𝒪} = \lim_{n \to \infty} \sum_{k = 0}^{n} \frac{𝒪^{k}}{k!}

Observables

Classical

Observables functions of the position and momentum.

\hat{A} (\vec{q}, \vec{p}) \in ℝ^{n} \times ℝ^{n} \to ℝ

They are measured by applying giving the function it's arguments

A (t) = \hat{A} (\vec{q} (t), \vec{p} (t))

Quantum

Observables are self-adjoint operators on the Hilbert space.

\hat{A} \in ℋ \to ℋ

\hat{A} = {\hat{A}}^{†}

They are measured by applying the state vector to both sides

A (t) = ⟨ ψ (t) | \hat{A} | ψ (t) ⟩

Heisenberg Picture

Instead of measuring observables by applying a modified state, instead change how the function measures it with time, while the state stays constant. We define ${\hat{A}}_{H}$ to the below.

Classical

A (t) = {\hat{A}}_{H} (t) (\vec{q} (t_{0}), \vec{p} (t_{0}))

{\hat{A}}_{H} (t) = \hat{A} ℳ (t_{0}, t)

Quantum

A (t) = ⟨ ψ (t_{0}) | {\hat{A}}_{H} (t) | ψ (t_{0}) ⟩

{\hat{A}}_{H} (t) = ℳ^{†} (t_{0}, t) \hat{A} ℳ (t_{0}, t)

Time evolution of observables

Classical

Using the chain rule we get

\frac{d A}{d t} (t) = (\sum_{i} \frac{\partial \hat{A}}{\partial q_{i}} \frac{d q_{i}}{d t} + \frac{\partial \hat{A}}{\partial p_{i}} \frac{d p_{i}}{d t}) (\vec{q} (t), \vec{p} (t))

then we can substitute in for the left side the Heisenberg picture and the right side the equations of motion for the time derivatives.

(\frac{d {\hat{A}}_{H}}{d t} (t)) (\vec{q} (t_{0}), \vec{p} (t_{0})) = (\sum_{i} \frac{\partial \hat{A}}{\partial q_{i}} \frac{\partial H}{\partial p_{i}} - \frac{\partial \hat{A}}{\partial p_{i}} \frac{\partial H}{\partial q_{i}}) (\vec{q} (t), \vec{p} (t))

The time evolution of the Heisenberg observable is the Heisenberg version of the right side's operator.

\frac{d {\hat{A}}_{H}}{d t} (t) = (\sum_{i} \frac{\partial \hat{A}}{\partial q_{i}} \frac{\partial H}{\partial p_{i}} - \frac{\partial \hat{A}}{\partial p_{i}} \frac{\partial H}{\partial q_{i}})_{H} (t)

Then we expand the definition of the Heisenberg observable.

\frac{d {\hat{A}}_{H}}{d t} (t) = (\sum_{i} \frac{\partial \hat{A}}{\partial q_{i}} \frac{\partial H}{\partial p_{i}} - \frac{\partial \hat{A}}{\partial p_{i}} \frac{\partial H}{\partial q_{i}}) ℳ (t_{0}, t)

\frac{d {\hat{A}}_{H}}{d t} (t) = \sum_{i} \frac{\partial {\hat{A}}_{H}}{\partial q_{i}} \frac{\partial H_{H}}{\partial p_{i}} - \frac{\partial {\hat{A}}_{H}}{\partial p_{i}} \frac{\partial H_{H}}{\partial q_{i}}

Quantum

Using the chain rule we get

\frac{d A (t)}{d t} = ⟨ \frac{d ψ}{d t} (t) | \hat{A} | ψ (t) ⟩ + ⟨ ψ (t) | \hat{A} | \frac{d ψ}{d t} (t) ⟩

then we can substitute in for the right side the equations of motion for the time derivatives.

⟨ ψ (t) | (\frac{i}{ℏ} {\hat{H}}^{†}) \hat{A} | ψ (t) ⟩ + ⟨ ψ (t) | \hat{A} (- \frac{i}{ℏ} \hat{H}) | ψ (t) ⟩

Remember that $\hat{H}$ is self adjoint by definition of an observable.

= ⟨ ψ (t) | (\frac{i}{ℏ} \hat{H}) \hat{A} | ψ (t) ⟩ + ⟨ ψ (t) | \hat{A} (- \frac{i}{ℏ} \hat{H}) | ψ (t) ⟩

= - \frac{i}{ℏ} ⟨ ψ (t) | \hat{A} \hat{H} - \hat{H} \hat{A} | ψ (t) ⟩

The measured time evolution of the Heisenberg picture is equal to this equation.

⟨ ψ (t_{0}) | \frac{d {\hat{A}}_{H}}{d t} (t) | ψ (t_{0}) ⟩ = - \frac{i}{ℏ} ⟨ ψ (t) | \hat{A} \hat{H} - \hat{H} \hat{A} | ψ (t) ⟩

The time evolution of the Heisenberg observable is the Heisenberg version of the right side's operator.

\frac{d {\hat{A}}_{H}}{d t} (t) = - \frac{i}{h} (\hat{A} \hat{H} - \hat{H} \hat{A})_{H} (t)

Then we can expand the definition of the Heisenberg operator.

\frac{d {\hat{A}}_{H}}{d t} (t) = - \frac{i}{h} ℳ^{†} (t_{0}, t) (\hat{A} \hat{H} - \hat{H} \hat{A}) ℳ (t_{0}, t)

\frac{d {\hat{A}}_{H}}{d t} (t) = - \frac{i}{h} (ℳ^{†} (t_{0}, t) \hat{A} \hat{H} ℳ (t_{0}, t) - ℳ^{†} (t_{0}, t) \hat{H} \hat{A} ℳ (t_{0}, t))

And since the time evolution is just an exponentiation of an observable it is also an observable, and thus self adjoint, so $ℳ^{†} ℳ = 1$ , and we can insert them as an identity.

\frac{d {\hat{A}}_{H}}{d t} (t) = - \frac{i}{h} (ℳ^{†} (t_{0}, t) \hat{A} ℳ^{†} (t_{0}, t) ℳ (t_{0}, t) \hat{H} ℳ (t_{0}, t) - ℳ^{†} (t_{0}, t) \hat{H} ℳ^{†} (t_{0}, t) ℳ (t_{0}, t) \hat{A} ℳ (t_{0}, t))

\frac{d {\hat{A}}_{H}}{d t} (t) = - \frac{i}{h} ({\hat{A}}_{H} {\hat{H}}_{H} - {\hat{H}}_{H} {\hat{A}}_{H})

Possion Bracket

If we define a Possion bracket for two observables

{\hat{A}, \hat{B}} = \sum_{i} \frac{\partial \hat{A}}{\partial q_{i}} \frac{\partial \hat{B}}{\partial p_{i}} - \frac{\partial \hat{A}}{\partial p_{i}} \frac{\partial \hat{B}}{\partial q_{i}}

then we can see

\frac{d {\hat{A}}_{H}}{d t} (t) = {{\hat{A}}_{H} (t), H_{H} (t)}

Commutator Bracket

If we define a Commutator bracket for two observables

[\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A}

then we can see

\frac{d {\hat{A}}_{H}}{d t} = - \frac{i}{ℏ} [{\hat{A}}_{H} (t), H_{H} (t)]

Liouville operator

Classical

Define an operator on observables.

ℒ (t) = {\cdot, H_{H} (t)}_{H}

When the operator is applied to a Heisenberg picture observable it should give the derivative.

\frac{d {\hat{A}}_{H}}{d t} (t) = ℒ (t) {\hat{A}}_{H}

Classical

Define an operator on observables.

ℒ (t) = - \frac{i}{ℏ} [\cdot, H_{H} (t)]

When the operator is applied to a Heisenberg picture observable it should give the derivative.

\frac{d {\hat{A}}_{H}}{d t} = ℒ (t) {\hat{A}}_{H}

Time evolution operator

The time evolution can be approximated as

{\hat{A}}_{H} (t + Δ t) \approx {\hat{A}}_{H} (t) + (\int_{t}^{t + Δ t} ℒ (t^{'}) d t^{'}) {\hat{A}}_{H} (t) = (1 + (\int_{t}^{t + Δ t} ℒ (t^{'}) d t^{'})) {\hat{A}}_{H} (t)

making it be exact in the limit.

{\hat{A}}_{H} (t + Δ t) = \lim_{n \to \infty} (1 + (\int_{t}^{t + Δ t} ℒ (t^{'}) d t^{'}) \frac{1}{n})^{n} {\hat{A}}_{h} (t)

The limit becomes exact.

{\hat{A}}_{H} (t + Δ t) = e^{\int_{t}^{t + Δ t} ℒ (t^{'}) d t^{'}} {\hat{A}}_{H} (t)

Use this to define a time evolution operator on observables.

𝒰 (t_{a}, t_{b}) = e^{\int_{t_{a}}^{t_{b}} ℒ (t^{'}) d t^{'}}

Such that when applied will transform an observable from $t_{a}$ to $t_{b}$ .

{\hat{A}}_{H} (t_{b}) = 𝒰 (t_{a}, t_{b}) {\hat{A}}_{H} (t_{a})