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:

Â ≐                   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 Â 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 Â 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 ⎠

dicates two possible eigenstates (for the eigenvalue b). One eigenstate is obvious from inspection:

    |b₁❭ ≐ ⎛1⎞
           ⎜0⎟
           ⎝0⎠.

The eigenvalue equation also reveals -ι γ = β. If β = 1 is chosen, then γ = -ι, revealing a second eigenstate, after normalizing:

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

Similarly, when the eigenvalue -b is used, the eigenvalue equation reveals ι γ = β. So, if β = ι, γ = 1. The third eigenstate is therefore, after normalizing,

    |b₃❭ ≐ 1  ⎛0⎞
           √2 ⎜ι⎟
              ⎝1⎠

The complete set of eigenstates of the operator B̂ is

    |b₁❭ ≐ ⎛1⎞  |b₂❭ ≐ 1  ⎛ 0 ⎞      |b₃❭ ≐ 1  ⎛0⎞
           ⎜0⎟         √2 ⎜ 1 ⎟             √2 ⎜ι⎟
           ⎝0⎠,           ⎝-ι ⎠, and           ⎝1⎠.

(𝗰) These eigenstates can be expressed as linear combinations of the eigenstates of Â, so they are shared eigenstates between these operators, and so this basis is simultaneously a basis of Â and B̂.