Math Infinitum

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Intuition

An eigenstate of the Hamiltonian barely evolves: it only turns its phase, at a rate set by its energy. Since a global phase is invisible, nothing measurable about it ever changes — every probability and every average stays exactly what it was. Such states are called stationary, and finding them is most of the work of the rest of the course.

A lighthouse beam sweeps round and round, yet the lighthouse never changes: the turning is all in the phase. A stationary state turns its phase and is otherwise still.

The factor $e^{-iEt/\hbar}$ of a stationary state at three times, as an arrow in the complex plane. It turns clockwise with period $T=2\pi\hbar/E$ and never changes length, so no probability built from the state ever changes.

States of definite energy

If $\hat{H}\ket{E}=E\ket{E}$ , the Schrödinger equation is solved by multiplying $\ket{E}$ by a turning phase. Every expectation value and every probability in such a state is independent of time. The equation for the energy eigenstates themselves is the time-independent Schrödinger equation.

\hat{H}\ket{E}=E\ket{E},\qquad \ket{\psi(t)}=e^{-iEt/\hbar}\ket{E}

What stays still

$\bra{\psi(t)}\hat{A}\ket{\psi(t)}=\bra{E}\hat{A}\ket{E}$ for every observable, because the phase and its conjugate cancel.

Stationary states have constant averages

\ket{\psi(t)}=e^{-iEt/\hbar}\ket{E}\ \Longrightarrow\ \frac{d}{dt}\bra{\psi(t)}\hat{A}\ket{\psi(t)}=0

Check first that the turning phase solves the Schrödinger equation: differentiating brings down $-iE/\hbar$ , which the $i\hbar$ turns into $E$ , exactly what $\hat{H}$ gives. Then in any expectation value the phase from the ket meets its conjugate from the bra and they cancel.

Proof steps

i\hbar\frac{d}{dt}\bigl(e^{-iEt/\hbar}\ket{E}\bigr)=E\,e^{-iEt/\hbar}\ket{E}

Differentiate the phase.

=\hat{H}\bigl(e^{-iEt/\hbar}\ket{E}\bigr)

Because $\hat{H}\ket{E}=E\ket{E}$ : the Schrödinger equation holds.

\bra{\psi(t)}\hat{A}\ket{\psi(t)}=e^{iEt/\hbar}e^{-iEt/\hbar}\bra{E}\hat{A}\ket{E}

The bra carries the conjugate phase.

=\bra{E}\hat{A}\ket{E}

The phases cancel, leaving a number independent of $t$ .

Applications

Practice

Only the Phase Turns

An eigenstate of the Hamiltonian with energy $E$ evolves by being multiplied by a phase that turns at the rate $E/\hbar$ .

\ket{\psi(t)}=e^{-iEt/\hbar}\ket{E}

Try it

A system starts in the energy eigenstate $\ket{E}$ . What is its state at time $t$ ?