Homework 6 Solutions Temple University PHYSICS 3701

Department of Physics Temple University
Introduction to Quantum Mechanics, Physics 3701 Instructor: Z.-E. Meziani




Solution set for homework # 6
April 16, 2013




Exercise #2, Complement FVI , page 765

Consider an arbitrary physical system whose four-dimensional state space is spanned by a basis of four
eigenvectors |j, mz i common to Jˆ2 and Jz (j = 0 or 1; −j ≤ mz ≤ +j), of eigenvalues j(j + 1)h̄2 and mz h̄,
such that: q
J± |j, mz >= h̄ j(j + 1) − mz (mz ± 1|j, mz ± 1 > (1)
J+ |j, j >= J− |j, −j >= 0 (2)
• a) Express in terms of the kets |j, mz >, the eigenstates common to Jˆ2 and Jˆx to be denoted by
|j, mx >.

We must first form the matrix of the operator Jˆx in the basis {|j, mz >}. If we recall the following
relation
1 ˆ
Jˆx = (J+ + Jˆ− ) (3)
2
then we may use Eqs. ?? and ?? to write the matrix of the Jˆx operator in the given basis. We first
calculate the individual matrix elements. First, the Jˆ+ terms

Jˆ+ |1, 1 > = 0 (4)
√
Jˆ+ |1, 0 > = 2h̄|1, 1 > (5)
√
Jˆ+ |1, −1 > = 2h̄|1, 0 > (6)
Jˆ+ |0, 0 > = 0 (7)

then the J− terms
√
Jˆ− |1, 1 > = 2h̄|1, 0 > (8)
√
ˆ
J− |1, 0 > = 2h̄|1, −1 > (9)
Jˆ− |1, −1 > = 0 (10)
Jˆ− |0, 0 > = 0 (11)

Now we may write the operator Jˆx in matrix form, using Eq. ??. in the basis {|1, 1 >, |1, 0 > and
|1, −1 >, |0, 0 >}
0 1 0 0
 
h̄ 1 0 1 0
Jˆx = √ 

 (12)
2 0 1 0
 0
0 0 0 0



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, To find the eigenvalues of this matrix, it is necessary to diagonalize it.
−λ √h̄ 0 0
2
√h̄ −λ √h̄ 0
2 2 =0 (13)
0 √h̄ −λ 0
2
0 0 0 −λ

The determinant is solved in the usual way, resulting in a characteristic equation given by
−λ √h̄ 0 √h̄ √h̄ 0
2 h̄ 2 2
−λ √h̄ −λ 0 0−√ −λ 0 = 0
2 2
0 0 −λ 0 0 −λ
−λ h̄ √h̄ 0 h̄ h̄ −λ 0
 
0
−λ −λ −√ 2 −√ √ =0
0 −λ 2 0 −λ 2 2 0 −λ
!
2 2 h̄2 h̄2
λ λ − − =0
2 2
 
λ2 λ2 − h̄2 = 0 (14)

This produces four roots, two of which are zero. This λ = 0 eigenvalue is thus two-fold degenerate.
The other two eigenvalues are +h̄ and −h̄. Substitution of these eigenvalues into Eq. ?? allows us to
solve for the eigenvectors.
0 1 0 0 a a
    
h̄ 1 0 1 0b b
Jˆx |φ >= λ|φ >⇒ √ 
   
= λ  (15)
2 0 1 0 0   c   c
0 0 0 0 d d

We are therefore able to write the eigenvectors in the new basis, |j, mx > in terms of the old basis,
|j, mz >. The following are the eigenvalues that correspond to the eigenvectors in both bases, in the
order: |j, mx >, |j, mz >
0 : |0, 0 >x = |0, 0 > (16)
1
0 : |1, 0 >x = √ (|1, −1 > −|1, −1 >) (17)
2
1
+h̄ : |1, 1 >x = √ (|1, 1 > +|1, −1 > +|1, 0 >) (18)
2
1
−h̄ : |1, −1 >x = √ (|1, 1 > +|1, −1 > −|1, 0 >) (19)
2
These are the eigenstates common to the operators Jˆ2 and Jˆx
• b) Consider a system in the normalized state:


|ψ >= α|j = 1, mz = 1 > +β|j = 1, mz = 0 > +γ|j = 1, mz = −1 > +δ|j = 0, mz = 0 > (20)
Note that this state is normalized. Therefore, we must have the following relation between the
coefficients:
|α|2 + |β|2 + |γ|2 + |δ|2 = 1 (21)




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