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# hw4solutions - HOMEWORK ASSIGNMENT 4 PHYS852 Quantum...

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HOMEWORK ASSIGNMENT 4 PHYS852 Quantum Mechanics I, Spring 2008 Topics covered: Atomic Physics applications: STARK EFFECT, Zeeman effect, spin-orbit coupling 1. [15 pts] Compute the Stark effect to lowest non-vanishing order for the n = 3 level of the hydrogen atoms. Fully evaluate whatever matrix elements, ( nℓm | Z | nℓ m ) , appear. Also remember to identify the ‘good’ states before applying perturbation theory. Include a sketch of the energy levels versus E 0 , with each level labeled by its state(s). Answer : The n = 3 energy level is nine-fold degenerate, containing the states | 300 ) , | 310 ) , | 31 ± 1 ) , | 320 ) , | 32 ± 1 ) , and | 32 ± 2 ) . The selection rules for the Stark effect are = ± 1 and m = m . The states which are mixed are therefore a closed three-level manifold ( m = 0), | 300 ) ↔ | 310 ) ↔ | 320 ) , and two closed two-level manifolds ( m = ± 1), | 31 ± 1 ) ↔ | 32 ± 1 ) . The good eigenstates of the m = 0 manifold are found by diagonalizing the matrix V 0 = 0 z 01 0 z 01 0 z 12 0 z 12 0 with z 01 = eE 0 ( 300 | Z | 310 ) and z 12 = eE 0 ( 310 | Z | 320 ) We can determine z 01 and z 12 by integration z 01 = eE 0 integraldisplay 0 r 2 dr integraldisplay π 0 sin θdθ integraldisplay 2 π 0 ( R 30 ( r ) Y 0 0 ( θ, φ ) ) r cos φR 31 ( r ) y 0 1 ( θ, φ ) = 3 6 ea 0 E 0 z 12 = eE 0 integraldisplay 0 r 2 dr integraldisplay π 0 sin θdθ integraldisplay 2 π 0 ( R 30 ( r ) Y 0 1 ( θ, φ ) ) r cos φR 31 ( r ) y 0 2 ( θ, φ ) = 3 3 ea 0 E 0 This leads to the eigenvalues of V 0 being v 01 = 9 ea 0 E 0 , v 02 = 0 and v 03 = 9 ea 0 E 0 , with correspond- ing eigenvectors | v 01 ) = 1 6 parenleftBig 2 | 300 ) − 3 | 310 ) + | 320 ) parenrightBig | v 02 ) = 1 3 parenleftBig | 300 ) − 2 | 310 ) parenrightBig | v 03 ) = 1 6 parenleftBig 2 | 300 ) + 3 | 310 ) + | 320 ) parenrightBig The good eigenstates of the m = ± 1 manifolds are eigenstates of the matrix V ± = parenleftbigg 0 z 12 z 12 0 parenrightbigg where z 12 = eE 0 ( 31 ± 1 | Z | 32 ± 1 ) = 9 2 ea 0 E 0 1

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Here the eigenvalues are v 11 = 9 2 ea 0 E 0 and v 12 = 9 2 ea 0 E 0 , with corresponding eigenvectors | v 11 ± ) = 1 2 ( | 31 ± 1 ) − | 32 ± 1 ) ) | v 12 ± ) = 1 2 ( | 31 ± 1 ) + | 32 ± 1 ) ) To first-order in perturbation theory we then find that the E 3 level splits into 5 levels, with energy shifts of 9 ea 0 E 0 , (9 / 2) ea 0 E 0 , 0, (9 / 2) ea 0 E 0 , and 9 ea 0 E 0 . The corresponding level manifolds as are | v 01 ) , {| v 11+ ) , | v 11 )} , {| v 02 ) , | 322 ) , | 32 2 )} , {| v 12+ ) , | v 12 )} , and | v 03 ) , respectively. 2. [10 pts] Do the same as the previous problem, but for the Zeeman effect. Answer : This problem is quite simple because the | nℓm m s ) states are already the ‘good’ states for perturba- tion theory. To first-order we find that the level shifts are given by E (1) nℓm m s = μ B B 0 ( m + 2 m s ) The resulting shifts for the n = 3 level are given by: m m s μ B B 0 ( m + 2 m s ) 0 1 / 2 μ B B 0 0,1,2 0 1 / 2 μ B B 0 0,1,2 1 1 / 2 2 μ B B 0 1,2 1 1 / 2 0 1,2 1 1 / 2 0 1,2 1 1 / 2 2 μ B B 0 1,2 2 1 / 2 3 μ B B 0 2 2 1 / 2 μ B B 0 2 2 1 / 2 μ B B 0 2 2 1 / 2 3 μ B B 0 2
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hw4solutions - HOMEWORK ASSIGNMENT 4 PHYS852 Quantum...

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