phy-4600/solutions/exam1/prob4

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

 ≐                   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⎠.

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 ⎞ ⎛ α₁ ⎞   ⎛    b α₁ ⎞       ⎛ α₁ ⎞
    ⎜ 0   0 -ιb ⎟ ⎜ β₁ ⎟ = ⎜ -ι b γ₁ ⎟ =  b  ⎜ β₁ ⎟
    ⎝ 0  ιb   0 ⎠ ⎝ γ₁ ⎠   ⎝  ι b β₁ ⎠       ⎝ γ₁ ⎠

reveals -ι γ₁ = β₁, which, combined with the normalization condition, will allow the determination of two eigenstates of the B̂.