Schrödinger picture

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In physics, the Schrödinger picture is a formulation of quantum mechanics in which the state vectors evolve in time, but the operators (observables and others) are constant. This differs from the Heisenberg picture which keeps the states constant while the observables evolve in time. The two models are related as active and passive transformations and have the same measurement statistics.

A state function is a linear combination, or a superposition, of eigenstates. In the Schrödinger picture, the state of a system evolves with time. The evolution for a closed quantum system is brought about by a unitary operator called the time evolution operator.

1 The time evolution operator
2 See also
3 Further reading

The time evolution operator

Definition

The time evolution operator U(t,t₀) is defined as:

$| \psi(t) \rangle = U(t,t_0) | \psi(t_0) \rangle$

That is, this operator when acting on the state ket at t₀ gives the state ket at a later time t. For bras, we have:

$\langle \psi(t) | = \langle \psi(t_0) |U^{\dagger}(t,t_0)$

Properties

Property 1

The time evolution operator must be unitary. This is because we demand that the norm of the state ket must not change with time. That is,

$\langle \psi(t)| \psi(t) \rangle = \langle \psi(t_0)|U^{\dagger}(t,t_0)U(t,t_0)| \psi(t_0) \rangle = \langle \psi(t_0) | \psi(t_0) \rangle$

Therefore,

$U^{\dagger}(t,t_0)U(t,t_0)=I.$

Property 2

Clearly U(t₀,t₀) = I, the Identity operator. As:

$| \psi(t_0) \rangle = U(t_0,t_0) | \psi(t_0) \rangle$

Property 3

Also time evolution from t₀ to t may be viewed as time evolution from t₀ to an intermediate time t₁ and from t₁ to the final time t. Therefore:

$U(t,t_0) = U(t,t_1)U(t_1,t_0)\!$

Differential equation for time evolution operator

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 can be written as:

$i \hbar {d \over dt} U(t) | \psi_e (0) \rangle = H U(t)| \psi_e (0)\rangle$

Here H is the Hamiltonian for the system. As $|\psi(0)\rangle$ is a constant ket (the state ket at t = 0), we see that the time evolution operator obeys the Schrödinger equation: i.e.

$i \hbar {d \over dt} U(t) = H U(t)$

If the Hamiltonian is independent of time, the solution to the above equation is:

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

Where we have also used the fact that at t = 0, U(t) must reduce to the identity operator. Therefore we get:

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

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

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

Thus we see that 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:

$U(t) = T\exp\left({-\frac{i}{\hbar} \int_0^t H(t^')\, dt^'}\right) \, .$

where T is time-ordering operator

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.