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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Most popular questions from this chapter

Use equations 4.27 4.28 and 4.32 to constructy00,y21Check that they are normalized and orthogonal

(a) Starting with the canonical commutation relations for position and momentum (Equation 4.10), work out the following commutators:

[LZ,X]=ihy,[LZ,y]=-ihx,[LZ,Z]=0[LZ,px]=ihpy,[LZ,py]=-ihpx,[LZ,pz]=0

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