130_notes.dvi

14.5.2 The Expectation Value ofLx

What is the expectation value ofLxin the stateψ(~r) =R(r)(

√

2

3 Y^11 (θ,φ)−

√

1

3 Y^10 (θ,φ))?

We needLx= (L++L−)/2.

〈ψ|Lx|ψ〉 =

1

2

〈√

2

3

Y 11 −

√

1

3

Y 10 |L++L−|

√

2

3

Y 11 −

√

1

3

Y 10

〉

=

1

2

〈√

2

3

Y 11 −

√

1

3

Y 10 |

√

2

3

L−Y 11 −

√

1

3

L+Y 10

〉

=

̄h

2

〈√

2

3

Y 11 −

√

1

3

Y 10 |

√

2

3

√

2 Y 10 −

√

1

3

√

2 Y 11

〉

=

̄h

2

2

3

(− 1 −1) =−

2

3

̄h

14.6 Sample Test Problems

1. Derive the commutators [L^2 ,L+] and [Lz,L+]. Now show thatL+Yℓm=CYℓ(m+1), that is,
   L+raises theLzeigenvalue but does not change theL^2 eigenvalue.
   Answer

L+=Lx+iLy

SinceL^2 commutes with bothLxandLy,

[L^2 ,L+] = 0.

[Lz,L+] = [Lz,Lx+iLy] = [Lz,Lx] +i[Lz,Ly] =i ̄h(Ly−iLx) = ̄h(Lx+iLy) = ̄hL+

We have the commutators. Now we apply them to aYℓm.

[L^2 ,L+]Yℓm=L^2 L+Yℓm−L+L^2 Yℓm= 0

L^2 (L+Yℓm) =ℓ(ℓ+ 1) ̄h^2 (L+Yℓm)

So,L+Yℓmis also an eigenfunction ofL^2 with the same eigenvalue.L+does not changeℓ.

[Lz,L+]Yℓm=LzL+Yℓm−L+LzYℓm= ̄hL+Yℓm

Lz(L+Yℓm)−m ̄h(L+Yℓm) = ̄h(L+Yℓm)

Lz(L+Yℓm) = (m+ 1) ̄h(L+Yℓm)

So,L+raises the eigenvalue ofLz.

2. Write the (normalized) state which is an eigenstate ofL^2 with eigenvalueℓ(ℓ+1) ̄h^2 = 2 ̄h^2 and
   also aneigenstate ofLxwith eigenvalue 0 ̄hin terms of the usualYℓm.
   Answer