The expression for the mean value of an observable A in the normalized state |Ψ> is <A> = <Ψ|A|Ψ>.
If |Ψ> is not normalized then < A> = <Ψ|A|Ψ>/<Ψ|Ψ>.
<x> = <Ψ|X|Ψ> = ∫ d3r <Ψ|r><r|X|Ψ>  = ∫ d3r Ψ*(r) x Ψ(r).
<px> = <Ψ|Px|Ψ> = ∫ d3r <Ψ|r><r|Px|Ψ>  = ∫ d3r Ψ*(r) (ħ/i)∂Ψ(r)/∂x.

The root mean square deviation ΔA characterizes the dispersion of the measurement around <A>.
ΔA = (<(A - <A>)2>)1/2.

ΔA = (<(A - <A>)2>)1/2 = (<Ψ|(A - <A>)2|Ψ>)1/2
=  (<Ψ|A2|Ψ> - 2<A><Ψ|A|Ψ> + <A>2<Ψ|Ψ>)1/2
=  (<A2> - 2<A><A> + <A>2)1/2
= (<A2> - <A>2)1/2.
ΔA is a measure of the spread that one should expect in the result of a measurement of the observable A.

Problem:

Assume the wave function of a particle is Ψ(x) = N exp(ip0x/ħ)/(x2 + a2)1/2.
Here a and p0 are real constants and N is a normalization constant.
(a)  Find N so that ψ(x) is normalized.
(b)  If the position of the particle is measured, what is the probability of finding the particle between  -a/√3  and  +a/√3?
(c)  Calculate the mean value of the momentum of the particle.

Solution:
• Concepts:
The postulates of Quantum Mechanics, the mean value
• Reasoning:
When a physical quantity described by the operator A is measured on a system in a normalized state |ψ>, the probability of measuring the eigenvalue aα is given by dP(aα) = |<uα|ψ>|2dα, where |uα> is the eigenvector corresponding to the eigenvalue aα; we assume aα is a non-degenerate continuous eigenvalue of A.
The expression for the mean value of an observable A in the normalized state |ψ> is <A> = <ψ|A|ψ>.
• Details of the calculation:
(a) <Ψ|Ψ> = 1.  ∫-∞+∞Ψ*(x)Ψ(x) dx = 1.
N2-∞+∞exp(-ip0x/ħ)/(x2 + a2)1/2 exp(ip0x/ħ)/(x2 + a2)1/2
= N2-∞+∞dx/(x2 + a2)  = 1.
-∞+∞dx/(x2 + a2)  = (1/a)tan-1(x/a)|-∞+∞  = (1/a)[π/2 - (-π/2)] = π/a.
N2 = a/π,  N = (a/π)1/2.

(b)  Let P denote the probability of finding the particle between -a/√3 and +a/√3.
Let y = a/√3.
P = ∫-y+y |Ψ(x)|2 dx = (a/π)∫-y+ydx/(x2 + a2)  = (1/π)tan-1(x/a)|-y+y

= (1/π)[tan-1(1/√3) - tan-1(-1/√3)]
= (1/π)[π/6 + π/6] = 1/3.

(c)  Let <P> denote the mean value of the momentum of the particle.
<P> = ∫-∞+∞Ψ*(x) ((ħ/i)∂/∂x) Ψ(x) dx
= (a/π)∫-∞+∞exp(-ip0x/ħ)/(x2 + a2)1/2 (ħ/i)[(ip0/ħ)exp(ip0x/ħ)/(x2 + a2)1/2 - x exp(-ip0x/ħ)/(x2 + a2)3/2]
= (a/π)p0-∞+∞dx/(x2 + a2)  - (a/π)(ħ/i)∫-∞+∞ xdx/(x2 + a2)2 = p0.
(π/a)                                 0, odd function

#### The generalized uncertainty relation Let A and B be two observables (Hermitian operators).  In any state of the system ΔA ΔB ≥ ½|<i[A,B]>|. In particular, for any two Hermitian operators Q and P for which [Q,P] = iħ, ΔQ ΔP ≥ ½|<iiħ>| = ħ/2.

Proof:
Let |Ψ> be any state vector and let A1 = A - <A>I and B1 = B - <B>I.
Let |Φ> = A1|Ψ> + ixB1|Ψ> with x an arbitrary real number.
Then <Φ| = <Ψ|A1 - ix<Ψ|B1.
<Φ|Φ> = <Ψ|A12|Ψ> + x2<Ψ|B12|Ψ> - ix<Ψ|B1A1|Ψ> + ix<Ψ|A1B1|Ψ>
= <A12> + x2<B12> + x<Ψ|i[A1,B1]Ψ> = (ΔA)2 + x2(ΔB)2 + x<i[A,B]>
for all x.  But <Φ|Φ> ≥ 0.  Therefore  = (ΔA)2 + x2(ΔB)2 + x<i[A,B]> ≥ 0 for all x.
For this to be true we either need that the quadratic equation y = ax2 + bx + c either has no zeros, or is equal to zero for only one value of x.  (The equation describes a parabola.  The vertex of the parabola can either touch the x-axis or must lie above the x-axis.)  To find a root we set x = (b ± (b2 - 4ac)1/2)/(2a).  If b2 < 4ac the equation has no roots.  If b2 = 4ac or x = -b/2a  the equation is zero.  We therefore need b2 < 4ac .
Therefore |<i[A,B]>|2 ≤ 4(ΔA)2(ΔB)2.