The raising and lowering operators change the value of m by one unit:

L±flm=(Alm)flm+1, (4.120).

Where Almare constant. Question: What is Alm, if the Eigen functions are to be normalized? Hint: First show thatL±is the Hermitian conjugate of L±(Since LxandLyare observables, you may assume they are Hermitian…but prove it if you like); then use Equation 4.112.

Short Answer

Expert verified

If the Eigen functions are to be normalized, they are upper and lower signs.

Step by step solution

01

Given.

The raising and lowering operators are:

L+flm=(Alm)flm+1,L-flm=(Blm)flm-1,

02

Eigen functions to be normalized

We start solving the problem by taking the inner product between a hydro genic set acted upon L±and Las following:

Alm=ħl(l+1)-m(m+1)=ħl(l+1)-m(m+1),Blm=ħl(l+1)-m(m+1)=ħl(l+1)-m(m+1),L2=L±L+Lz2±ħLz(4.112).

Note what happens at the top and bottom of the ladder (i.e. when you apply L+L+tofllorL-tofl-lNow,usingEq.4.112,inthefromL±L=L2-Lz2ħLzLL2=L±L+Lz2ħLzL2=L4.112.

flmlLmL±flm=flmlL2-Lz2mhLzflm=flml[h2ll+l-h2m2mh2m]flm=h2ll+1-mm±1flmlflm=h2ll+1-mm±1=L±flmlL±flmUppersigns:ħ2ll+1-mm+1=L+flmlL+flm=Almflm+1lAlmflm+1=Alm2Alm=ħll+1-mm+1Lowersigns:ħ2ll+1-mM-1=L-flmlL-flm=Blmflm-1lBlmflm-1=Blm2

At the top of the ladderm=lwe get All=0wegetAll=0, so there is no higher rung; at the bottom of the ladder m=l we getBl-l=0 , so there is no lower rung.

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Most popular questions from this chapter

(a) Apply S_tolocalid="1656131461017" 10>(Equation4.177), and confirm that you getlocalid="1656131442455" 2h1-1>.

(b) ApplyS+to[00>(Equation4.178), and confirm that you get zero.

(c) Show thatlocalid="1656131424007" 11>andlocalid="1656131406083" 1-1>(Equation4.177) are eigenstates ofS2, with the appropriate eigenvalue

(a) Find〈r〉and〈r²〉for an electron in the ground state of hydrogen. Express your answers in terms of the Bohr radius.

(b) Find〈x〉and (x2)for an electron in the ground state of hydrogen.

Hint: This requires no new integration—note that r2=x2+y2+z2,and exploit the symmetry of the ground state.

(c) Find〈x²〉in the state n=2,l=1,m=1. Hint: this state is not symmetrical in x, y, z. Usex=rsinθcosπx=rsinθcosϕ

(a) Use the recursion formula (Equation 4.76) to confirm that whenI=n-1 the radial wave function takes the form

Rn(n-1)=Nnrn-1e-r/na and determine the normalization constant by direct integration.

(b) Calculate 200a and <r2> for states of the form ψn(n-1)m·

(c) Show that the "uncertainty" in r(δr) is<r>/2n+1for such states. Note that the fractional spread in decreases, with increasing (in this sense the system "begins to look classical," with identifiable circular "orbits," for large ). Sketch the radial wave functions for several values of, to illustrate this point.

Determine the commutator of S2withSZ(1)(whereSS(1)+S(2)) Generalize your result to show that

[S2,S1]=2Ih(S1×S2)

Comment: Because Sz(1)does not commute with S2, we cannot hope to find states that are simultaneous eigenvectors of both. In order to form eigenstates ofS2weneed linear combinations of eigenstates ofSz(1). This is precisely what the Clebsch-Gordan coefficients (in Equation 4.185) do for us, On the other hand, it follows by obvious inference from Equation 4.187that the sumrole="math" localid="1655980965321" S(1)+S(2)does commute withdata-custom-editor="chemistry" S2, which is a special case of something we already knew (see Equation 4.103).

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?

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