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QM2 Concept Test 10.1 Consider the Hamiltonian π» 0 + π» β²= π 0 1βπ π 0 π 1 π 0 π 2 , where πβͺ1. The basis vectors for the matrix π , π , and π are the energy eigenstates of the unperturbed Hamiltonian π» 0 (π=0). Choose all of the following statements that are correct. π is a βgoodβ state for the perturbation π» β². π is a βgoodβ state for the perturbation π» β². The perturbation matrix in the degenerate subspace of π» 0 is π 0 βπ π π 0 . A. 1 only B. 2 only C. 1 and 3 only D. 2 and 3 only E. All of the above
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QM2 Concept Test 10.2 Consider the Hamiltonian π» 0 + π» β²= π 0 1βπ π 0 π 1 π 0 π 2 , where πβͺ1. The basis vectors for the matrix π , π , and π are the energy eigenstates of the unperturbed Hamiltonian π» 0 (π=0). Choose all of the following statements that are correct. The first order correction to the energy for the state π is βπ π 0 . The first order correction to the energy for the state π is zero. The first order correction to the energy for the state π is zero. 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 10.3 A perturbation π» β² acts on a hydrogen atom with the unperturbed Hamiltonian π» 0 =β β 2 2π π» 2 β π 2 4π π π . To calculate the perturbative corrections, we use π,π, π π , π , π π (the eigenstates of ( π» 0 , πΏ 2 , and πΏ π§ ) as the basis vectors. Choose all of the following statements that are correct. If π» β² =πΌ πΏ π§ , where πΌ is a suitable constant, we can calculate the first order corrections as πΈ 1 = π,π, π π , π , π π π» β² π,π, π π , π , π π . If π» β² =πΌπΏ(π), the first order correction to energy is πΈ 1 = π,π, π π , π , π π π» β² π,π, π π , π , π π . If π» β² =πΌ π½ π§ (π§ component of π½ = πΏ + π ) we can calculate the first order correction as πΈ 1 = π,π, π π , π , π π π» β² π,π, π π , π , π π . A. 1 only B. 2 only C. 1 and 2 only D. 1 and 3 only E. All of the above
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QM2 Concept Test 10.4 A perturbation π» β² acts on a hydrogen atom with the unperturbed Hamiltonian π» 0 =β β 2 2π π» 2 β π 2 4π π π . To calculate the perturbative corrections, we use the coupled representation π,π,π ,π, π π as the basis vectors. Choose all of the following statements that are correct. If π» β² =πΌ πΏ π§ , where πΌ is a suitable constant, we can calculate the first order corrections as πΈ 1 = π,π,π ,π, π π π» β² π,π,π ,π, π π . If π» β² =πΌπΏ(π), the first order correction to energy is πΈ 1 = π,π,π ,π, π π π» β² π,π,π ,π, π π . If π» β² =πΌ π½ π§ (π§ component of π½ = πΏ + π ) we can calculate the first order correction as πΈ 1 = π,π,π ,π, π π π» β² π,π,π ,π, π π . 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 10.5 A perturbation π» β² acts on a hydrogen atom with the unperturbed Hamiltonian π» 0 =β β 2 2π π» 2 β π 2 4π π π . Choose all of the following statements that are correct. If π» β² =πΌ( πΏ π§ + π π§ ) we can calculate the first order correction as πΈ 1 = π,π,π ,π, π π π» β² π,π,π ,π, π π . If π» β² =πΌ( πΏ π§ + π π§ 2 ) we can calculate the first order correction as πΈ 1 = π,π,π ,π, π π π» β² π,π,π ,π, π π . If π» β² =πΌ( πΏ π§ + π π§ 2 ) we can calculate the first order correction as πΈ 1 = π,π, π π , π , π π π» β² π,π, π π , π , π π . 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 10.6 Suppose the eigenstates π , π , and π of π» 0 are 3-fold degenerate with energy πΈ 0 . A perturbation π» β acts on this system and the perturbation matrix is π» β² =π if π , π , and π are used as basis vectors. Note: The matrix has eigenvectors and β1 with eigenvalues 2 and 0, respectively. Choose all of the following statements that are correct. The perturbation π» β² splits the original energy level πΈ 0 into three distinct energy levels because the eigenvalues of the matrix are 0, 1, and 2. π is a βgoodβ state for this system. π + π and π β π are βgoodβ states for this system. 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 10.7 Suppose the unperturbed Hamiltonian π» 0 is two-fold degenerate, i.e., π» 0 π π 0 = πΈ π π 0 , π» 0 π π 0 = πΈ π π 0 , π π π π 0 =0. A perturbation π» β² acts on this system and a Hermitian operator π΄ commutes with both π» 0 and π» β² . Choose all of the following statements that are correct. π» 0 and π» β² must commute with each other. If π π 0 and π π 0 are degenerate eigenstates of π΄ , they must be βgoodβ states for finding perturbative corrections to the energy and wavefunction due to π» β² . If π π 0 and π π 0 are non-degenerate eigenstates of π΄ , they must be βgoodβ states. A. 1 only B. 2 only C. 3 only D. 1 and 2 only E. 1 and 3 only
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