QUANTUM OPERATORS
hence formally an eigenstate ofL^2 with= 0, all three components of angular
momentum could be (and are) known to be zero.
Working in one dimension, show that theminimum value of the product∆px×∆xfor
aparticleis^12 . Find the form of the wavefunction that attains this minimum value for a
particle whose expectation values for position and momentum are ̄xand ̄p, respectively.We have already seen, in (19.23) that the commutator ofpxandxis−i,aconstant.
Therefore, irrespective of the actual form of|ψ〉, the RHS of (19.38) is^14 ^2 (see observation
(ii) above). Thus, since all quantities are positive, taking the square roots of both sides of
the equation shows directly that
∆px×∆x≥^12 .Returning to the derivation of the Uncertainty Principle, we see that the inequality becomes
an equality only when
(〈u|v〉+〈v|u〉)^2 =4〈u|u〉〈v|v〉.The RHS of this equality has the value 4||u||^2 ||v||^2 and so, by virtue of Schwarz’s inequality,
we have
4 ‖u‖^2 ‖v‖^2 =(〈u|v〉+〈v|u〉)^2
≤(|〈u|v〉|+|〈v|u〉|)^2
≤(‖u‖‖v||+‖v‖‖u‖)^2
=4‖u‖^2 ‖v‖^2.Since the LHS is less than or equal to something that has the same value as itself, all of
the inequalities are, in fact, equalities. Thus〈u|v〉=‖u‖‖v‖, showing that|u〉and|v〉are
parallel vectors, i.e.|u〉=μ|v〉for some scalarμ.
We now transform this condition into a constraint that the wavefunctionψ=ψ(x)
must satisfy. Recalling the definitions (19.37) of|u〉and|v〉in terms of|ψ〉, we have
(
−id
dx− ̄p)
ψ=μi(x− ̄x)ψ,dψ
dx+
1
[μ(x− ̄x)−i ̄p]ψ=0.The IF for this equation is exp
[
μ(x− ̄x)^2
2 −
i ̄px
]
, givingd
dx{
ψexp[
μ(x− ̄x)^2
2 −
i ̄px
]}
=0,
which, in turn, leads to
ψ(x)=Aexp[
−
μ(x− ̄x)^2
2 ]
exp(
i ̄px
)
.
From this it is apparent that the minimum uncertainty product ∆px×∆xis obtained when
the probability density|ψ(x)|^2 has the form of a Gaussian distribution centred onx ̄.The
value ofμis not fixed by this consideration and it could be anything (positive); a large
value forμwould yield a small value for ∆xbut a correspondingly large one for ∆px.