phy-4600/solutions/exam1/prob4
2016-02-25 00:03:45 -05:00

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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.