Quantum Mechanics

From Mind-Brain.org

Quantum mechanics is a physical science dealing with the behaviour of matter and energy on the scale of atoms and subatomic particles. In quantum mechanics, "particles" exhibit both wavelike and particle-like properties, depending on how they are observed. This concept is called wave-particle duality. Quantum also forms the basis for the contemporary understanding of how very large objects such as stars and galaxies, and cosmological events such as the Big Bang.

In quantum mechanics, a state function is a linear combination (a superposition) of eigenstates. In the Schrödinger interpretation, the state of a system evolves with time, where the evolution for a closed quantum system is brought about by an unitary operator called the time-evolution operator. This differs from the Heisenberg interpretation where the states are constant while the observables evolve in time.

1 The Time Evolution Operator
2 Further reading

[edit] The Time Evolution Operator

[edit] Definition

The time evolution operator $U (t, t 0)$ is defined as:

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

That is, this operator when acting on the state ket (using Bra-ket notation) at $t 0$ gives the state ket at a later time $t$ . For bras (using Bra-ket notation), we have:

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

[edit] Properties

[edit] 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,

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

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

[edit] Property 2

Clearly $U (t 0, t 0)$ = I, the Identity operator. As:

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

[edit] Property 3

Also time evolution from $t 0$ to $t$ may be viewed as time evolution from $t 0$ to an intermediate time $t 1$ and from $t 1$ to the final time $t$ . therefore:

U (t, t 0) = U (t, t 1) U (t 1, t 0)

[edit] Differential Equation for Time Evolution Operator

We drop the $t 0$ index in the time evolution operator with the convention that $t 0 = 0$ and write it as $U (t)$ . The Schrodinger 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( it is the state ket at $t = 0$ ), we see that the time evolution operator obeys the Schrodinger 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 a , we get:

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

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 time commute then, the time evolution operator can be written as:

$U(t) = e^{-i/\hbar \int\limits _0^t H(t^')\, dt^'}$

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 interpretation.