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solset1 - PH 216 Problem set 1 Due 4/6 1 In class we...

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PH 216, Problem set 1, Due 4/6 1. In class, we estimated the ground-state energy of the 3D SHO (see Shankar 12.6.42) using a ‘guess’ of ψ 0 ( a ; r ) = R ( a ; r ) Y 0 0 ( θ , φ ) , by varying a and minimizing ψ ( a ) | H | ψ ( a ) . Suppose we try to get the first energy E 1 using the trial function ψ 1 = R ( a ; r ) Y 0 1 , Where the Y enforces orthogonality with our ψ 0 . A trial function with zero nodes that has the right asymptotics ( r +1 as r 0 and 0 as r → ∞ ) is ψ 2 , 1 , 0 = N ( r/a 0 ) exp( r/ 2 a 0 ) cos θ . what then is E 1 , and how does this compare with the first level of the actual 3D SHO? Solution: We have E ( a ) = ψ ( a ) | H | ψ ( a ) = ψ ( a ) | T | ψ ( a ) + ψ ( a ) | V | ψ ( a ) . As in class, we can use the quantum virial theorem, T = 1 2 V T + V = E n = 2 / 2 ma 2 n 2 = T for an energy state of energy E n . So we just have to evaluate V = 1 32 π a 3 0 r 2 dr d ( ω r 2 / 2)( r/a ) 2 exp( r/a ) cos 2 θ , where I have taken the normalization of the wavefunction from Shankar 13.1.27. But since V does nothing to angles, the angular integral just falls out as 4 π , so we get V = m ω 2 48 π a 5 0 dr r 6 exp( r/a ) = m ω 2 48 π a 5 × 720 a 6 = 15 a 2 m ω 2 π . Combining our results leads to E ( a ) written as: E ( a ) = 2 8 ma 2 + 15 a 2 m ω 2 With this in hand, we minimize the energy with respect to a : dE da = 2 4 ma 3 + 30 am ω 2 = 0 , which gives a min = 2 120 m 2 ω 2 1 / 4 . 1
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Thus, we get E 1 = E ( a min ) = 15 2 ω 2 . 74 ω . The exact result for the nth energy level of the 3D SHO is given by E n = ( n + 3 / 2) ω , and with n = 1, E 1 = 2 . 5 ω . This gives E 1 0 . 24 ω which is greater than zero as required by the variational principle. 2. (Shankar 16.2.8) Consider the = 0 radial equation for the three-dimensional Coulomb problem. Since V ( r ) is singular at the turning point r = 0, we cannot use n + 3 4 as in eq. (16.2.42) of Shankar. (a) Will the additive constant be more or less than 3 4 ? (Only heuristic reasoning is required, but you are welcome to derive the result formally if you like.) (b) By analyzing the exact equation near r = 0, it can be shown that the constant equals 1. Using this constant, show that the WKB energy levels agree with the exact results Solution: (a) Heuristically, as discussed in class if the wavefunction must vanish at both ends, the accumulated phase must add up to n half-cycles, so ( n + 1) π . But if the wavefunction need not vanish, then you fit somewhat less, and the WKB formal-
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