phy-4600/solutions/exam1/prob4

Two operators' matrix representations are known in the |1〉, |2〉, |3〉 basis, where a and b are real numbers, and ι is the imaginary unit:

 ≐                   B̂ ≐
    ⎛ a  0  0 ⎞           ⎛ b   0   0 ⎞
    ⎜ 0 -a  0 ⎟           ⎜ 0   0 -ιb ⎟
    ⎝ 0  0 -a ⎠  and      ⎝ 0  ιb   0 ⎠.

B exhibits a degenerate spectrum when it has repeated eigenvalues. The eigenvalues of B are obtained from its characteristic equation.
 
|⎛ b-λ  0    0 ⎞|
|⎜ 0   -λ  -ιb ⎟| = 0, i.e.
|⎝ 0   ιb   -λ ⎠|

(b - λ)(λ² + ι²b²) = (b - λ)(λ² - b²) = (b - λ)(b - λ)(b + λ) = 0.

(a) The eigenvalues for this operator are therefore λ = b,b,-b. Since b appears twice, the operator exhibits a degenerate spectrum.

To find if A and B commute, their commutator need be evaluated. They commute if the value is 0. The commutator of two operators is defined as

    [Â,B̂] = Â B̂ - B̂ Â.

For the given operators, then, the commutator is

    ⎛ a  0  0 ⎞ ⎛ b   0   0 ⎞   ⎛ b   0   0 ⎞ ⎛ a  0  0 ⎞
    ⎜ 0 -a  0 ⎟ ⎜ 0   0 -ιb ⎟ - ⎜ 0   0 -ιb ⎟ ⎜ 0 -a  0 ⎟
    ⎝ 0  0 -a ⎠ ⎝ 0  ιb   0 ⎠   ⎝ 0  ιb   0 ⎠ ⎝ 0  0 -a ⎠,

which reduces to

    ⎛ ab    0    0 ⎞   ⎛ ab    0    0 ⎞    
    ⎜ 0     0  ιab ⎟ - ⎜ 0     0  ιab ⎟ = 0.
    ⎝ 0  -ιab    0 ⎠   ⎝ 0  -ιab    0 ⎠

(b) Therefore, these operators commute.


Since the operators commute, they share a set of common eigenstates. The eigenstates of  are apparent from inspection:

|a₁〉 ≐ ⎛1⎞  |a₂〉 ≐ ⎛0⎞  and |a₃〉 ≐ ⎛0⎞     
       ⎜0⎟         ⎜1⎟             ⎜0⎟
       ⎝0⎠;        ⎝0⎠;            ⎝1⎠.

Eigenstates of B̂ must be obtained, and if they are a linear combination of the eigenstates of Â, they will serve as the complete set of eigenstates shared by  and B̂. If not, some subset of those will be the shared basis, and this will need to be determined.

For B̂, the eigenvalues are already known (λ = b,b,-b.), and using the eigenvalue equations, the eigenstates can be determined. The eigenvalue equation

    ⎛ b   0   0 ⎞ ⎛ α ⎞   ⎛    b α ⎞       ⎛ α ⎞
    ⎜ 0   0 -ιb ⎟ ⎜ β ⎟ = ⎜ -ι b γ ⎟ =  b  ⎜ β ⎟ 
    ⎝ 0  ιb   0 ⎠ ⎝ γ ⎠   ⎝  ι b β ⎠       ⎝ γ ⎠

reveals -ι γ = β, which, combined with the normalization condition, will allow the determination of two eigenstates of the B̂. One eigenstate is obvious from inspection,

	|b₁〉 ≐ ⎛1⎞
	       ⎜0⎟
	       ⎝0⎠.

If β = 1 is chosen, then γ = -ι, and if β = ι, then γ = -1. The two additional eigenstates of B̂ are therefore, after normalizing,

	|b₂〉 ≐ 1  ⎛ 0 ⎞     |b₃〉 ≐ 1  ⎛ 0  ⎞
           √2 ⎜ 1 ⎟            √2 ⎜ ι  ⎟
              ⎝-ι ⎠ and           ⎝ -1 ⎠.