mirror of
https://asciireactor.com/otho/phy-4600.git
synced 2024-12-12 20:55:07 +00:00
48 lines
2.1 KiB
Plaintext
48 lines
2.1 KiB
Plaintext
|
Two operators' matrix representations are known in the |1〉, |2〉, |3〉 basis, where and b are real numbers:
|
|||
|
|
|||
|
A ≐ B ≐
|
|||
|
⎛ 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 common eigenstates. Therefore, if the eigenstates for one operator can be determined, they are determined for both operators. Since the problem states that a new set of orthonormal kets need be determined, and the given set are the eigenstates related to the operator A, then the eigenstates of the operator B should be determined.
|
|||
|
|
|||
|
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, and
|
|||
|
ι b β₁ = γ₁
|
|||
|
|
|||
|
|