This choice of normalization uniquely determines the completeness relation, and we have^4
I =
∫Rdx|x;X〉〈x;X|
I =
∫Rdk
2 π
|k;P〉〈k;P| (5.59)
Note that the normalization of (5.58) fixes the “norm” of the states, but not their relative
phases, which are left undetermined. The proofs of both completeness relations are similar,
so we shall carry out explicitly only the first one. Denote the result of thex-integral byA,
A=
∫Rdx|x;X〉〈x;X| (5.60)
Now apply the operator to a ket|y;X〉,
A|y;X〉 =
∫Rdx|x;X〉〈x;X|y;X〉
=
∫Rdx|x;X〉δ(x−y) =|y;X〉 (5.61)
In going from the first line to the second, we have used the normalization of the position
eigenstates, defined above. Since this relation holdsfor ally, and|y;X〉spans a basis, it
must be thatA=Iby its very definition.
• The overlaps of the position and momentum eigenstates may be determined by com-
puting the matrix element〈k;P|T(x)|0;X〉in two different ways, namely by lettingT(a) act
either on the left or on the right, and we find,
〈k;P|x;X〉=〈k;P|T(x)|0;X〉=e−ikxφ(k) (5.62)
whereφ(k) =〈k;P|0;X〉. As a result, we have
〈k;P|x;X〉 = φ(k)e−ikx
〈x;X|k;P〉 = φ(k)∗e+ikx (5.63)
The functionφ(k) may be determined, in part, from the completeness relations, as follows,
〈k′;P|k;P〉 =
∫Rdx〈k′;Px;X〉〈x;X|k;P〉
= φ(k′)∗φ(k)
∫Rdxei(k−k
′)x= 2πδ(k′−k)|φ(k)|^2 (5.64)
(^4) The choice of normalization in (5.58) is not unique. One might have multipliedδ(x′−x) by function
f(x), andδ(k′−k) by an independent functionf ̃(k). As a result, the completeness relations of (5.59) will
be replaced bydx→dx/f(x) anddk→dk/f ̃(k). The choice we have adopted in (5.58) is, for almost all
applications, the simplest one, and it is almost universally adopted in the literature. (One exception is when
Rcarries a general Riemannian metric, not just the flat Euclidean metric as we have here.)