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QM2 Concept Test 10.1 Consider the Hamiltonian

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1 QM2 Concept Test 10.1 Consider the Hamiltonian 𝐻 0 + 𝐻 β€²= 𝑉 0 1βˆ’πœ€ πœ€ 0 πœ€ 1 πœ€ 0 πœ€ 2 , where πœ€β‰ͺ1. The basis vectors for the matrix π‘Ž , 𝑏 , and 𝑐 are the energy eigenstates of the unperturbed Hamiltonian 𝐻 0 (πœ€=0). Choose all of the following statements that are correct. π‘Ž is a β€œgood” state for the perturbation 𝐻 β€². 𝑐 is a β€œgood” state for the perturbation 𝐻 β€². The perturbation matrix in the degenerate subspace of 𝐻 0 is 𝑉 0 βˆ’πœ€ πœ€ πœ€ 0 . A. 1 only B. 2 only C. 1 and 3 only D. 2 and 3 only E. All of the above

2 QM2 Concept Test 10.2 Consider the Hamiltonian 𝐻 0 + 𝐻 β€²= 𝑉 0 1βˆ’πœ€ πœ€ 0 πœ€ 1 πœ€ 0 πœ€ 2 , where πœ€β‰ͺ1. The basis vectors for the matrix π‘Ž , 𝑏 , and 𝑐 are the energy eigenstates of the unperturbed Hamiltonian 𝐻 0 (πœ€=0). Choose all of the following statements that are correct. The first order correction to the energy for the state π‘Ž is βˆ’πœ€ 𝑉 0 . The first order correction to the energy for the state 𝑏 is zero. The first order correction to the energy for the state 𝑐 is zero. A. 1 only B. 3 only C. 1 and 2 only D. 1 and 3 only E. All of the above.

3 QM2 Concept Test 10.3 A perturbation 𝐻 β€² acts on a hydrogen atom with the unperturbed Hamiltonian 𝐻 0 =βˆ’ ℏ 2 2π‘š 𝛻 2 βˆ’ 𝑒 2 4πœ‹ πœ€ π‘Ÿ . To calculate the perturbative corrections, we use 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 (the eigenstates of ( 𝐻 0 , 𝐿 2 , and 𝐿 𝑧 ) as the basis vectors. Choose all of the following statements that are correct. If 𝐻 β€² =𝛼 𝐿 𝑧 , where 𝛼 is a suitable constant, we can calculate the first order corrections as 𝐸 1 = 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 𝐻 β€² 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 . If 𝐻 β€² =𝛼𝛿(π‘Ÿ), the first order correction to energy is 𝐸 1 = 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 𝐻 β€² 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 . If 𝐻 β€² =𝛼 𝐽 𝑧 (𝑧 component of 𝐽 = 𝐿 + 𝑆 ) we can calculate the first order correction as 𝐸 1 = 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 𝐻 β€² 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 . A. 1 only B. 2 only C. 1 and 2 only D. 1 and 3 only E. All of the above

4 QM2 Concept Test 10.4 A perturbation 𝐻 β€² acts on a hydrogen atom with the unperturbed Hamiltonian 𝐻 0 =βˆ’ ℏ 2 2π‘š 𝛻 2 βˆ’ 𝑒 2 4πœ‹ πœ€ π‘Ÿ . To calculate the perturbative corrections, we use the coupled representation 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 as the basis vectors. Choose all of the following statements that are correct. If 𝐻 β€² =𝛼 𝐿 𝑧 , where 𝛼 is a suitable constant, we can calculate the first order corrections as 𝐸 1 = 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 𝐻 β€² 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 . If 𝐻 β€² =𝛼𝛿(π‘Ÿ), the first order correction to energy is 𝐸 1 = 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 𝐻 β€² 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 . If 𝐻 β€² =𝛼 𝐽 𝑧 (𝑧 component of 𝐽 = 𝐿 + 𝑆 ) we can calculate the first order correction as 𝐸 1 = 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 𝐻 β€² 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 . A. 1 only B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only E. All of the above

5 QM2 Concept Test 10.5 A perturbation 𝐻 β€² acts on a hydrogen atom with the unperturbed Hamiltonian 𝐻 0 =βˆ’ ℏ 2 2π‘š 𝛻 2 βˆ’ 𝑒 2 4πœ‹ πœ€ π‘Ÿ . Choose all of the following statements that are correct. If 𝐻 β€² =𝛼( 𝐿 𝑧 + 𝑆 𝑧 ) we can calculate the first order correction as 𝐸 1 = 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 𝐻 β€² 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 . If 𝐻 β€² =𝛼( 𝐿 𝑧 + 𝑆 𝑧 2 ) we can calculate the first order correction as 𝐸 1 = 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 𝐻 β€² 𝑛,𝑙,𝑠,𝑗, π‘š 𝑗 . If 𝐻 β€² =𝛼( 𝐿 𝑧 + 𝑆 𝑧 2 ) we can calculate the first order correction as 𝐸 1 = 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 𝐻 β€² 𝑛,𝑙, π‘š 𝑙 , 𝑠, π‘š 𝑠 . A. 1 only B. 1 and 2 only C. 1 and 3 only D. 2 and 3 only E. All of the above

6 QM2 Concept Test 10.6 Suppose the eigenstates π‘Ž , 𝑏 , and 𝑐 of 𝐻 0 are 3-fold degenerate with energy 𝐸 0 . A perturbation 𝐻 ’ acts on this system and the perturbation matrix is 𝐻 β€² =𝑉 if π‘Ž , 𝑏 , and 𝑐 are used as basis vectors. Note: The matrix has eigenvectors and βˆ’1 with eigenvalues 2 and 0, respectively. Choose all of the following statements that are correct. The perturbation 𝐻 β€² splits the original energy level 𝐸 0 into three distinct energy levels because the eigenvalues of the matrix are 0, 1, and 2. π‘Ž is a β€œgood” state for this system. 𝑏 + 𝑐 and 𝑏 βˆ’ 𝑐 are β€œgood” 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.

7 QM2 Concept Test 10.7 Suppose the unperturbed Hamiltonian 𝐻 0 is two-fold degenerate, i.e., 𝐻 0 πœ“ π‘Ž 0 = 𝐸 πœ“ π‘Ž 0 , 𝐻 0 πœ“ 𝑏 0 = 𝐸 πœ“ 𝑏 0 , πœ“ π‘Ž πœ“ 𝑏 0 =0. A perturbation 𝐻 β€² acts on this system and a Hermitian operator 𝐴 commutes with both 𝐻 0 and 𝐻 β€² . Choose all of the following statements that are correct. 𝐻 0 and 𝐻 β€² must commute with each other. If πœ“ π‘Ž 0 and πœ“ 𝑏 0 are degenerate eigenstates of 𝐴 , they must be β€œgood” states for finding perturbative corrections to the energy and wavefunction due to 𝐻 β€² . If πœ“ π‘Ž 0 and πœ“ 𝑏 0 are non-degenerate eigenstates of 𝐴 , they must be β€œgood” states. A. 1 only B. 2 only C. 3 only D. 1 and 2 only E. 1 and 3 only


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