The Schrodinger and Heisenberg pictures are equivalent

The Schrodinger picture and Heisenberg picture are equivalent, in the sense that we obtain the same results from either. Results are observed through Measurement, so this result means that Measurements are consistent across pictures

Proof

In the Heisenberg picture, we consider the Quantum state $\psi$ fixed and the Quantum observable $A(t)$ evolves in time. In this picture:

• a measurement is given by ${⟨A(t)⟩}_{\psi }:=⟨\psi ,A(t)\psi ⟩$.
• $A(t)$ evolves according to the equation $A(t)=e^{\frac{itH}{\hslash }}A{e}^{-\frac{itH}{\hslash }}$

In the Schrodinger picture, we consider the Quantum observable $A$ to be time independent and the Quantum state $\psi (t)$ to evolve over time. In this picture:

• a measurement is given by ${⟨A⟩}_{\psi (t)}:=⟨\psi (t),A\psi (t)⟩$.
• $\psi (t)$ evolves according to the equation $\psi (t)=e^{-\frac{itH}{\hslash }}\psi (0)$.

As such

$$\begin{array}{rll}{⟨A(t)⟩}_{\psi }& =⟨\psi ,A(t)\psi ⟩& \text{By the Heisenberg picture}\\ & =⟨\psi ,e^{\frac{itH}{\hslash }}A{e}^{-\frac{itH}{\hslash }}\psi ⟩& \text{Moving the operator based on antilinearity}\\ & =⟨-e^{\frac{itH}{\hslash }}\psi ,A{e}^{-\frac{itH}{\hslash }}\psi ⟩& \text{By the Schrodinger picture}\\ & =⟨\psi (t),A\psi (t)⟩& \text{By definition of a measurement}\\ & ={⟨A⟩}_{\psi (t)}& \end{array}$$