Mean Value and RMS Deviation

The expression for the mean value of an observable A in the normalized state |y> is <A>=<y|A|y>.  If |y> is not normalized then .

.

.

The root mean square deviation DA characterizes the dispersion of the measurement around <A>.

DA 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   
Here a and p0 are real constants and N is a normalization constant.
(a) Find N so that y(x) is normalized.
(b) If the position of the particle is measured, what is the probability of finding the particle between and ?
(c) Calculate the mean value of the momentum of the particle.
Solution:

(a)
.

(b)

(c)



                                      0, odd function

The generalized uncertainty relation

Let A and B be two observables (Hermitian operators). In any state of the system

In particular, for any two Hermitian operators Q and P for which [Q,P]=,

Proof:

Let |y> be any state vector and let A1=A - <A>I and B1=B - <B>I.  Let |f>=A1|y> + ixB1|y> with x an arbitrary real number.  Then <f|=<y|A1 - ix<y|B1.

<f|f>=<y|A12|y>+x2<y|B12|y>-ix<y|B1A1|y>+ix<y|A1B1|y>

=<A12>+x2<B12>+x<y|i[A1,B1]y>=(DA)2+x2(DB)2+x<i[A,B]>

for all x.  But <f|f> ³ 0.  Therefore =(DA)2+x2(DB)2+x<i[A,B]> ³ 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 .  If the equation has no roots.  If b2=4ac or x=-b/2a  the equation is zero.  We therefore need . Therefore