Quantum Mechanics for Mathematicians

one has the equality of distributions

qδ(q−q′) =q′δ(q−q′)

soδ(q−q′) is an eigenfunction ofQwith eigenvalueq′.
The operatorsQandPdo not commute, since

[Q,P]f=−iq

d

dq

f+i

d

dq

(qf) =if

and we get (reintroducing~for a moment) the fundamental operator commu-
tation relation
[Q,P] =i~ 1

the Heisenberg commutation relation. This implies thatQand the free parti-
cle HamiltonianH= 21 mP^2 also do not commute, so the position, unlike the
momentum, is not a conserved quantity.
For a finite dimensional state space, recall that the spectral theorem (4.1)
for a self-adjoint operator implied that any state could be written as a linear
combination of eigenvectors of the operator. In this infinite dimensional case,
the formula

ψ(q) =

∫+∞

−∞

δ(q−q′)ψ(q′)dq′ (12.1)

can be interpreted as an expansion of an arbitrary state in terms of a continuous
linear combination of eigenvectors ofQwith eigenvalueq′, theδ-functionsδ(q−
q′). The Fourier inversion formula (11.4)

ψ(q) =

1

√

2 π

∫+∞

−∞

eikqψ ̃(k)dk

similarly gives an expansion in terms of eigenvectors√^12 πeikqofP, with eigen-

valuek.

12.2 Momentum space representation

We began our discussion of the state spaceHof a free particle by taking states
to be wavefunctionsψ(q) defined on position space, thought of variously as
being inS(R),L^2 (R) orS′(R). Using the Fourier transform, which takes such
functions to their Fourier transforms

ψ ̃(k) =Fψ=√^1

2 π

∫+∞

−∞

e−ikqψ(q)dq

in the same sort of function space, we saw in section 11.5 that the state space
Hcan instead be taken to be a space of functionsψ ̃(k) on momentum space.
We will call such a choice ofH, with the operatorPnow acting as

Pψ ̃(k) =kψ ̃(k)