Let |Ψ_S(t₀)> be a state vector in the Schroedinger picture, i.e. let it evolve in time and let its evolution be described by the Schroedinger equation.  Then
|Ψ_S(t)> = U(t,t₀)|Ψ_S(t₀)> = U^†(t₀,t)Ψ_S(t₀)> .
The Schroedinger picture implies an active unitary transformation.  The state vector is transformed, but all operators are constant in time unless they contain time explicitly.  The basis vectors are not changing.  The operators are defined through their action on the basis vectors.

The Heisenberg picture implies the equivalent passive unitary transformation.  The state vector is constant, 
|Ψ_H> = |Ψ_S(t₀)> = U(t₀,t)|Ψ_S(t)>, |Ψ_H> = U^†(t,t₀)|Ψ_S(t)>.  

However the basis vectors are changing, and therefore the operators are changing.  The operator A_H(t) which does to the new basis vectors what A_S(t) does to the old basis vectors is given by 
A_H = U(t₀,t)A_sU^†(t₀,t) = U^†(t,t₀)A_sU(t,t₀).
The Schroedinger picture and the Heisenberg picture are two different representations.

The Heisenberg picture can be obtained from the Schroedinger picture by a unitary transformation at any time t.
|Ψ_H> = U(t₀,t)|Ψ_S(t)>, |Ψ_H> = U^†(t,t₀)|Ψ_S(t)> = |Ψ_S(t₀)>.
<Ψ_S(t)|A_S|Φ_S(t)> = <Ψ_S(t)|U(t,t₀)U^†(t,t₀)A_SU(t,t₀)U^†(t,t₀)|Φ_S(t)>
= <Ψ_H(t)|U^†(t,t₀)A_SU(t,t₀)|Φ_H(t)> = <Ψ_H(t)|A_H|Φ_H(t)>.
The matrix elements of any operator A are independent of the representation.
The predictions of quantum mechanics are independent of the representation.

Consider the Schroedinger representation, (often called the Schroedinger picture), defined by the Schroedinger equation
iħ∂|Ψ(q,t)>/∂t = H(p,q)|Ψ(q,t)>. 
(a)  Write the transformation equations for the wave function Ψ and arbitrary operators Q to the Heisenberg representation (or picture), in which the operators are time dependent but the wave functions are not.  Show that this transformation is unitary if H is Hermitian and that matrix elements have the same value in the two pictures.
(b)  From the transformation equations in part (a) derive the Heisenberg equation of motion for the operators in the Heisenberg representation.  (Assume the operators in the Schroedinger picture have no explicit time dependence).
(c)  Consider a Hamiltonian operator H(p,q) = p²/(2m) + U(q).   Show that the Heisenberg equation of motion for the operator is  ∂H/∂p = p/m = dq/dt,  which has the same form as the corresponding classical equation of motion.

Solution:
(a) |Ψ_H> = U^†(t,t₀)|Ψ_S(t)> = |Ψ_S(t₀)>.
Q_H = U(t₀,t)Q_sU^†(t₀,t) = U^†(t,t₀)Q_sU(t,t₀).
Note: Q_S is not a function of time unless it contains time explicitly.
If H does not explicitly depend on time then
U(t,t₀) = exp(-iH(t - t₀)/ħ), U(t₀,t) = exp(-iH(t₀ - t)/ħ)
If H  =  H^† then U^†(t,t₀) = exp(+iH(t - t₀)/ħ) = U(t₀,t),
and U(t,t₀)U^†(t,t₀) = U(t₀,t)U^†(t₀,t) = I.  Therefore U is unitary.
<Ψ_S(t)|Q_S|Φ_S(t)> = <Ψ_S(t)|U(t,t₀)U^†(t,t₀)Q_SU(t,t₀)U^†(t,t₀)|Φ_S(t)>
= <Ψ_H(t)|U^†(t,t₀)Q_SU(t,t₀)|Φ_H(t)> = <Ψ_H(t)|Q_H|Φ_H(t)>.

(b) In the Schroedinger picture we write for an arbitrary operator A
(d/dt)<Ψ_S(t)|A_S|Φ_S(t)>
= [(∂/∂t)<Ψ_S(t)|]A_S|Φ_S(t)> + <Ψ_S(t)|∂A_S/∂t|Φ_S(t)> + <Ψ_S(t)|A_S[(∂/∂t)|Φ_S(t)>]
= <Ψ_S(t)|∂A_S/∂t|Φ_S(t)> + (iħ)^-1<Ψ_S(t)|A_SH_s - H_sA_s|Φ_S(t)>
= <Ψ_S(t)|∂A_S/∂t|Φ_S(t)> + (iħ)^-1<Ψ_S(t)|[A_S,H_s]|Φ_S(t)>.
Here we have used the Schroedinger equation,
(iħ∂/∂t)|Ψ_S> = H|Ψ_S>,  (-iħ∂/∂t)<Ψ_S| = H<Ψ_S|.

Matrix elements are independent of the representation.
In the Heisenberg picture this therefore becomes
(d/dt)<Ψ_H(t)|A_H|Φ_H(t)> = <Ψ_H(t)|∂A_H/∂t|Φ_H(t)>
= <Ψ_S(t)|∂A_S/∂t|Φ_S(t)> + (iħ)^-1<Ψ_S(t)|[A_S,H_s]|Φ_S(t)>
= <Ψ_H|U^†(t,t₀)(∂A_S/∂t)U(t,t₀)|Φ_H> + (iħ)^-1<Ψ_S(t)|A_SH_s - H_sA_s|Φ_S(t)>
= <Ψ_H|(∂A_S/∂t)_H|Φ_H> + (iħ)^-1<Ψ_H(t)|U^†A_SUU^†H_sU - U^†H_sU^†UA_sU|Φ_H(t)>
= <Ψ_H|(∂A_S/∂t)_H|Φ_H> + (iħ)^-1<Ψ_H|A_HH_H - H_HA_H|Φ_H>,
for all kets |Ψ_H|> and |Φ_H>.
We therefore have dA_H/dt = (∂A_S/∂t)_H +  (iħ)^-1[A_H,H_H].

If A does not explicitly depend on time then dA_H/dt = (iħ)^-1[A_H,H_H]  describes the evolution of the operators in the Heisenberg picture.

• (i) Unless A_S(t) explicitly depends on time ∂A_S/∂t = 0.
• (ii) If the operator A commutes with the Hamiltonian H and ∂A_S/∂t = 0, then A_H does not depend on time dA_H/dt = 0.  Such an operator is called a constant of motion.  The matrix elements of A, <Φ|A|Ψ>, then do not change with time, i.e. the results of measurements of A do not depend on time.

Consider the Hamiltonian  H(P,Q) = P²/(2m) + U(Q).
In the Heisenberg picture we write  H_H(P_H,Q_H) = P_H²/(2m) + U(Q_H).   We have
dQ_H/dt =  (iħ)^-1[Q_H,H_H] = (iħ)^-1[Q_H,P_H²/(2m)] = (iħ2m)^-1[Q_H,P_H²] = P_H/m.
dP_H/dt =  (iħ)^-1[P_H,H_H] = (iħ)^-1[P_H,U(Q_H)] = -∂U(Q_H)/∂Q_H.
Here we use [P,F(Q)] =  [P,∑_nf_nQⁿ] = ∑_nf_n[P,Qⁿ] = -∑_nf_nniħQ^n-1] = -iħ∂f(Q)/∂Q.
The Heisenberg picture leads to equations similar to the classical equations of motion.  The Heisenberg picture is often used to explore general properties of quantum systems and the formal analogy between classical and quantum theory.
The Schroedinger picture is most often used when doing calculations.  The Schroedinger equation, an equation between vectors, is, in general, easier to solve than the Heisenberg equation, an equation between operators.

(c)  H_H(P_H,Q_H) = P_H²/(2m) + U(Q_H), ∂H/∂P_H = P_H/m = dQ_H/dt,
dQ_H/dt = (iħ)^-1[Q_H,H_H] = (iħ)^-1[Q_H,P_H²/(2m)] = (iħ2m)^-1[Q_H,P_H²] = P_H/m.