Let |Ψ_{S}(t_{0})> 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,t_{0})|Ψ_{S}(t_{0})>
= U^{†}(t_{0},t)Ψ_{S}(t_{0})> .

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}(t_{0})> = U(t_{0},t)|Ψ_{S}(t)>,
|Ψ_{H}> = U^{†}(t,t_{0})|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(t_{0},t)A_{s}U^{†}(t_{0},t) =
U^{†}(t,t_{0})A_{s}U(t,t_{0}).

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(t_{0},t)|Ψ_{S}(t)>, |Ψ_{H}> =
U^{†}(t,t_{0})|Ψ_{S}(t)> = |Ψ_{S}(t_{0})>.

<Ψ_{S}(t)|A_{S}|Φ_{S}(t)> = <Ψ_{S}(t)|U(t,t_{0})U^{†}(t,t_{0})A_{S}U(t,t_{0})U^{†}(t,t_{0})|Φ_{S}(t)>

= <Ψ_{H}(t)|U^{†}(t,t_{0})A_{S}U(t,t_{0})|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) = p^{2}/(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,t_{0})|Ψ_{S}(t)> =
|Ψ_{S}(t_{0})>.

Q_{H }= U(t_{0},t)Q_{s}U^{†}(t_{0},t) =
U^{†}(t,t_{0})Q_{s}U(t,t_{0}).

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,t_{0}) = exp(-iH(t - t_{0})/ħ), U(t_{0},t) = exp(-iH(t_{0} - t)/ħ)

If H = H^{†} then U^{†}(t,t_{0}) = exp(+iH(t - t_{0})/ħ)
= U(t_{0},t),

and U(t,t_{0})U^{†}(t,t_{0}) = U(t_{0},t)U^{†}(t_{0},t)
= I. Therefore U
is unitary.

<Ψ_{S}(t)|Q_{S}|Φ_{S}(t)> = <Ψ_{S}(t)|U(t,t_{0})U^{†}(t,t_{0})Q_{S}U(t,t_{0})U^{†}(t,t_{0})|Φ_{S}(t)>

= <Ψ_{H}(t)|U^{†}(t,t_{0})Q_{S}U(t,t_{0})|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_{S}H_{s}
- H_{s}A_{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,t_{0})(∂A_{S}/∂t)U(t,t_{0})|Φ_{H}>
+ (iħ)^{-1}<Ψ_{S}(t)|A_{S}H_{s} - H_{s}A_{s}|Φ_{S}(t)>

= <Ψ_{H}|(∂A_{S}/∂t)_{H}|Φ_{H}> + (iħ)^{-1}<Ψ_{H}(t)|U^{†}A_{S}UU^{†}H_{s}U
- U^{†}H_{s}U^{†}UA_{s}U|Φ_{H}(t)>

= <Ψ_{H}|(∂A_{S}/∂t)_{H}|Φ_{H}> + (iħ)^{-1}<Ψ_{H}|A_{H}H_{H}
- H_{H}A_{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) = P^{2}/(2m) + U(Q).

In the Heisenberg picture we write H_{H}(P_{H},Q_{H})
= P_{H}^{2}/(2m) + U(Q_{H}).
We have

dQ_{H}/dt = (iħ)^{-1}[Q_{H},H_{H}] =
(iħ)^{-1}[Q_{H},P_{H}^{2}/(2m)] = (iħ2m)^{-1}[Q_{H},P_{H}^{2}]
= 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,∑_{n}f_{n}Q^{n}] = ∑_{n}f_{n}[P,Q^{n}]
= -∑_{n}f_{n}niħQ^{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_{H}^{2}/(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_{H}^{2}/(2m)]
= (iħ2m)^{-1}[Q_{H},P_{H}^{2}] = P_{H}/m.