From bc9de8cd898252d0ff17ea9c7be64be7dec5b89d Mon Sep 17 00:00:00 2001 From: caes Date: Thu, 25 Feb 2016 00:03:45 -0500 Subject: [PATCH] still exam 1 --- solutions/exam1/prob3 | 22 ++++++++++++++++++---- solutions/exam1/prob4 | 30 +++++++++++++++++------------- 2 files changed, 35 insertions(+), 17 deletions(-) diff --git a/solutions/exam1/prob3 b/solutions/exam1/prob3 index edf5f8d..826e1cf 100644 --- a/solutions/exam1/prob3 +++ b/solutions/exam1/prob3 @@ -24,7 +24,7 @@ and because H is a hermitian operator, 〈a|b〉 = 〈b|a〉, so 〈a|Ĥ|b〉 = δ (〈a|b〉 + 〈a|b〉) = δ 2〈a|b〉; -because of the hermition property, +because of the Hermitian property, 〈b|Ĥ|a〉 = 〈a|Ĥ|b〉 = δ 2〈a|b〉; finally, @@ -60,13 +60,27 @@ This gives the equation α₁ + α₁|〈a|b〉|² + 2β₁〈a|b〉 = E₁α₁ β₁ = ͟α͟₁͟(͟E͟₁͟ ͟-͟ ͟1͟ ͟-͟ ͟|͟〈͟a͟|͟b͟〉͟|͟²͟)͟ 2〈a|b〉 + + + + + + + + + + + + + + Using the normalization condition, the values of each constant can be obtained. Plugging the value for α₁ into the equation reveals a quadratic equation. |α₁|² + |β₁|² = 1, so - |α₁|² + | ͟α͟₁͟(͟E͟₁͟ ͟-͟ ͟1͟ ͟-͟ ͟|͟〈͟a͟|͟b͟〉͟|͟²͟)͟ |² = 1. + |α₁|² + | ͟α͟₁͟(͟E͟₁͟ ͟-͟ ͟1͟ ͟-͟ ͟|͟〈a͟|͟b͟〉͟|͟²͟)͟ |² = 1. | 2〈a|b〉 | @@ -77,7 +91,7 @@ Using the normalization condition, the values of each constant can be obtained. _________ α₁ = ±√1 - |β₁|², so _________ _________ - ±√̅1 - |β₁|² (1 - E₁) + 2β₁〈a|b〉 + ±√1 - |β₁|² |〈a|b〉|² = 0. + ±√1 - |β₁|² (1 - E₁) + 2β₁〈a|b〉 + ±√1 - |β₁|² |〈a|b〉|² = 0. The quadratic formula therefore says that @@ -86,4 +100,4 @@ Using the normalization condition, the values of each constant can be obtained. - \ No newline at end of file + diff --git a/solutions/exam1/prob4 b/solutions/exam1/prob4 index 8088782..d20b746 100644 --- a/solutions/exam1/prob4 +++ b/solutions/exam1/prob4 @@ -1,9 +1,9 @@ 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 ⎠. + ≐ 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. @@ -27,21 +27,25 @@ For the given operators, then, the commutator is which reduces to - ⎛ ab 0 0 ⎞ ⎛ ab 0 0 ⎞ - ⎜ 0 0 ιab ⎟ - ⎜ 0 0 ιab ⎟ = 0. - ⎝ 0 -ιab 0 ⎠ ⎝ 0 -ιab 0 ⎠ + ⎛ 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. + +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 + ⎛ b 0 0 ⎞ ⎛ α₁ ⎞ ⎛ b α₁ ⎞ ⎛ α₁ ⎞ + ⎜ 0 0 -ιb ⎟ ⎜ β₁ ⎟ = ⎜ -ι b γ₁ ⎟ = b ⎜ β₁ ⎟ + ⎝ 0 ιb 0 ⎠ ⎝ γ₁ ⎠ ⎝ ι b β₁ ⎠ ⎝ γ₁ ⎠ - -ι γ₁ = β₁, and - ι b β₁ = γ₁ +reveals -ι γ₁ = β₁, which, combined with the normalization condition, will allow the determination of two eigenstates of the B̂.