Consider the observablesA=x2andB=Lz .

(a) Construct the uncertainty principle forσAσB

(b) EvaluateσB in the hydrogen stateψn/m .

(c) What can you conclude about<xy>in this state?

Short Answer

Expert verified

(a) The uncertainty principle for σAσBħ|xy|..

(b) in the hydrogen state ψnimisσB=0

(c) The conclusion about xy in the given state is 0.

Step by step solution

01

Definition of the uncertainty

The uncertainty principle is a set of mathematical inequalities that establish a basic limit on the precision with which the values for certain pairs of physical properties of a particle, such as position and momentum, can be anticipated from initial conditions.

ΔxΔph4π

02

(a) Determination of the uncertainty principle for σAσB

Calculate the commutator of the two operators, A and B in the following way,

A,B=X2,Lz=x2,ypx-xpy=x2,ypx-x2,xpy=x2,ypx+yx2,px-x2,xpy-xx2,py

Evaluate the above expression further.

A,B=0+yxx,px+x,pxx-0-xxx,py+x,pyx=yx-ih+-ihx-x0+0=-2ihyx

It is known that σA2σB212i[A,B].

Apply it in expression.

σA2σB212i(-2)xy2oorσAσBħ|xy|

Thus, the uncertainty principle for σAσBħ|xy|.

03

(b) Determination of σB  the hydrogen state

Now get the expectation values of both Lz and Lz2, in some arbitrary hydrogenic normalized state|ψnlm .

Evaluate the variation in B,σB,

B=Lz=ψLzψ=ψmhψ=mhψψ=mh

Similarly, evaluate the variation in B2,σB

role="math" localid="1658137986506" B2=Lz2=ψLz2ψ=ψm2h2ψ=m2h2ψψ=m2h2

Find the value ofσB.

σB=B2-B2=m2h2-m2h2=0

Thus,σB in the hydrogen state ψnim is σB=0.

04

(c) Conclusion about <xy>  in the given state

Since the variation inB,σB, is zero and the right hand side of the uncertainty principle, |xy, is always positive , then the right hand side must also be zero. Hence,it is concluded that xy is also 0.

Thus, the conclusion about xy in the given state is 0.

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