Wave mechanics and the abstract formalism · Quantum Mechanics

Intuition

The course has written one theory in several languages. Schrödinger’s wave mechanics: a wavefunction $\psi(x,t)$ and a differential equation. Heisenberg’s matrix mechanics: fixed states, operators that move, and the commutator $[\hat{x},\hat{p}]=i\hbar$ . Dirac’s notation, in which $\psi(x)=\braket{x|\psi}$ is one set of components of an abstract vector and $\tilde{\psi}(p)=\braket{p|\psi}$ another, joined by a Fourier transform. Feynman’s path integral: the amplitude as a sum over histories. Each makes some questions easy and others hard, and they are one theory because each represents the same objects. The propagator is the position matrix element of the evolution operator; the path integral is a way of computing it; and its Fourier transform in time is the resolvent $\hat{G}(E)=(E-\hat{H}+i\varepsilon)^{-1}$ , whose poles are the energy levels and whose form for a free particle is the outgoing Green function of scattering. The pictures agree on every prediction because predictions are traces and inner products, which do not care which basis is used or whether the state or the operator carries the time.

A city can be given as a street map, an aerial photograph or a list of addresses. Each answers some questions faster than the others, and all describe the same city.

$-\frac{1}{\pi}\operatorname{Im}\operatorname{tr}\hat{G}(E)$ for an oscillator, with across in units of and : a peak of width at each level . As the peaks become the spectrum.

One theory, several representations

The evolution operator in the position representation is the propagator, and its Fourier transform in time, with outgoing waves picked by a small $\varepsilon>0$ , is the resolvent:

\begin{aligned}K(x,x';t)&=\bra{x}e^{-i\hat{H}t/\hbar}\ket{x'}\\ \hat{G}(E)&=\frac{1}{i\hbar}\int_{0}^{\infty}e^{i(E+i\varepsilon)t/\hbar}e^{-i\hat{H}t/\hbar}\,dt\\ &=\frac{1}{E-\hat{H}+i\varepsilon}\end{aligned}

Properties

$\psi(x)=\braket{x|\psi}$ and $\tilde\psi(p)=\braket{p|\psi}$ are the components of one vector in two bases, joined by the Fourier transform.

The resolvent is the Fourier transform of the evolution

\frac{1}{i\hbar}\int_{0}^{\infty}e^{i(E+i\varepsilon)t/\hbar}\,e^{-i\hat{H}t/\hbar}\,dt=\frac{1}{E-\hat{H}+i\varepsilon}

Work in the eigenbasis of $\hat{H}$ , where the evolution operator multiplies each eigenstate by a phase. The time integral of each phase, damped by $\varepsilon$ , is elementary and gives $1/(E-E_{n}+i\varepsilon)$ . Summed over the eigenstates with their projectors, this is the resolvent, and in the position representation the same sum shows its poles at the levels, with the stationary states as residues.

Proof steps

e^{-i\hat{H}t/\hbar}=\sum_{n}e^{-iE_{n}t/\hbar}\ket{n}\bra{n}

The spectral decomposition of the evolution operator.

\int_{0}^{\infty}e^{i(E-E_{n}+i\varepsilon)t/\hbar}\,dt=\frac{i\hbar}{E-E_{n}+i\varepsilon}

The upper limit gives nothing, since $\varepsilon>0$ damps it.

\frac{1}{i\hbar}\sum_{n}\frac{i\hbar\,\ket{n}\bra{n}}{E-E_{n}+i\varepsilon}=\sum_{n}\frac{\ket{n}\bra{n}}{E-E_{n}+i\varepsilon}

Put the terms together.

\sum_{n}\frac{\ket{n}\bra{n}}{E-E_{n}+i\varepsilon}=\frac{1}{E-\hat{H}+i\varepsilon}

The spectral decomposition of the inverse; a continuous spectrum turns the sum into an integral.

\bra{x}\hat{G}(E)\ket{x'}=\sum_{n}\frac{\psi_{n}(x)\,\psi_{n}^{*}(x')}{E-E_{n}+i\varepsilon}

In the position representation: a pole at every level, with the stationary states as residues.

Applications

Practice

Components of One Vector

$\psi(x)=\braket{x|\psi}$ and $\tilde{\psi}(p)=\braket{p|\psi}$ are the components of the same vector in two bases, and the Fourier transform turns one into the other.

Try it

How are the position and momentum wavefunctions of a state related?