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Published byΞΞ΅ΟΞ½Ξ―Ξ΄Ξ±Ο ΞΡοδΟΟΞ―Ξ΄Ξ·Ο Modified over 5 years ago
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QM2 Concept Test 11.1 In a 3D Hilbert space, π» 0 is the unperturbed Hamiltonian. π and π are the 2-fold degenerate energy eigenstates with energy πΈ 1 and π is the energy eigenstate with energy πΈ 2 . If π and π are not βgoodβ states for the perturbation π» β², choose all of the following statements that must be correct. π»β² ππ = π»β² ππ β 0, where π π» β π = π»β² ππ . π»β² ππ = π» β² ππ β β 0 If π» β² does not commute with π» 0 , we can never find a proper set of coefficients πΌ 1 , π½ 1 , πΌ 2 , π½ 2 to diagonalize both π» 0 and π» β² completely using the basis vectors πΌ 1 π + π½ 1 π , πΌ 2 π + π½ 2 π , and π . A. 2 only B. 3 only C. 1 and 2 only D. 2 and 3 only E. all of the above
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QM2 Concept Test 11.2 Choose all of the following statements that are correct. If π» 0 and π» β commute with each other , we can always find a basis to diagonalize the matrices for both π» 0 and π» β simultaneously. If π» 0 and π»β do not commute with each other , we cannot find a basis to diagonalize the matrices for both π» 0 and π» βsimultaneously. In perturbation theory, π» 0 is chosen to be a diagonal matrix and the basis vectors are chosen as the orthonormal eigenstates of π» 0 . A. 1 only B. 3 only C. 1 and 2 only D. 1 and 3 only E. All of the above.
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QM2 Concept Test 11.3 Suppose π» 0 and π» β² commute with each other. Choose all of the following statements that are correct. If π» 0 is diagonal in a given basis and there is no degeneracy in the eigenvalue spectrum of π» 0 and π» β², then π» β²must be diagonal in that basis. If π» 0 is diagonal in a given basis and there is a degeneracy in the eigenvalue spectrum of π» 0 , then π» β²must be diagonal in that basis. We can always find a special basis in which both π» 0 and π» β² are diagonal simultaneously. A. 1 only B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only E. All of the above
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QM2 Concept Test 11.4 Choose all of the following statements that are correct about the spin-orbit coupling term π» β² ππ = π 2 8π π π 2 π 2 π 3 π β πΏ in the Hamiltonian of the hydrogen atom (including the fine structure correction). π» β² ππ commutes with πΏ π§ . π» β² ππ commutes with π½ π§ = πΏ π§ + π π§ π» β² ππ commutes with πΏ 2 . A. 1 only B. 2 only C. 1 and 2 only D. 2 and 3 only E. All of the above
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QM2 Concept Test 11.5 The fine structure correction for the hydrogen atom is πΈ ππ = πΈ ππ + πΈ π = πΈ π π π 2 3β 4π π , where π=π+π , π+π β1, β¦ πβπ is the quantum number corresponding to the total angular momentum and π=1,2,3β¦. Choose all of the following statements that are correct including fine structure. πΈ ππ is always negative for any possible value of π and π. When π=2, there are 2 distinct values of π. There is no degeneracy left for the energy level with π=3, π=3/2 after we account for fine structure correction. A. 1 only B. 2 only C. 1 and 2 only D. 2 and 3 only E. All of the above.
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QM2 Concept Test 11.6 Without considering the fine structure, the energy for a hydrogen atom is πΈ π = β13.6 ππ π Choose all of the following statements that are correct. Ignoring spin, the energy level n=2 is four-fold degenerate corresponding to (π=1, π π =1), (π=1, π π =0), (π=1, π π =β1), and (π=0, π π =0). Including the fine structure, when the electron is in the state (π=2, π=1/2), we will definitely obtain zero if we measure the square of the magnitude of the angular momentum πΏ 2 . Including the fine structure correction, when the electron is in the state (π=2, π=3/2), we will definitely obtain 2 β 2 if we measure the square of the magnitude of the angular momentum πΏ 2 . A. 1 only B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only E. All of the above
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QM2 Concept Test 11.7 For hydrogen atom, the Zeeman term in the perturbation is given by π»β² π = π 2π πΏ +2 π β π΅ ππ₯π‘ . Choose all of the following statements that are true about the intermediate field Zeeman effect, where neither the Zeeman term π»β² π nor the fine structure term π»β² ππ dominates. The βgoodβ basis states for the perturbation are the coupled states π, π, π , π, π π . The βgoodβ basis states for the perturbation are the uncoupled states π, π, π π , π , π π . Both the coupled and uncoupled states are equally βgoodβ states for the perturbation. 1 only B. 2 only C. 3 only D. Not enough information E. None of the above
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