PHYS4221_HW4_Solution2

# PHYS4221_HW4_Solution2 - PHYS4221 HW4 Solution Written by...

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PHYS4221 HW4 Solution Written by Leung Shing Chi 1. a) First of all, we need to normalize the trial wavefunction: 2 0 3 0 1 4 dx Hence, we may evaluate the expectation of the Hamiltonian to be: 22 23 00 53 2 2 11 / 13 1 / 24 2 4 4 3 xx x d HTV x e x e d x x e m x d x mdx m m m m               4 Next, we minimize the expected energy: 3 2 3 0 3 dH m dm m From this, we may obtain the minimum trial energy, or the upper bound of the ground state energy: 3 1.732 1.5 H    b) First, we need to find a relation making the two trial wavefunction orthogonal to each other:  2 0 () 0 3 x e d x      Next we need to normalize the wavefunction: 2 32 0 03 4 x e d x 2  With these we may compute the expected energy directly, though tediously:

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 22 2 2 2 2 14 13 1 5 33 2 2 2 m H m              5 But we should remember that 3   , we will eliminate in the equation and we get (and we will use the result of (a) that 2 3 m ):   42 4 2 2 3 35 7 2 2 3 () 2 H m f m   We regard this to be a function of , and we require   0 df d , from which we yield a complicated polynomial which cannot be solved analytically. Using Mathematica, we found: 1.4691 1.1163 9.19 7 1.4691 Hf m 2   2. Normalization: 29 0 256 315 c c dx c Then, we will need to compute the expectation of energy: 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 0 2 72 1 1 2 0 2 2 1 / 1 ()8 4 / 256 1 256 / 2 105 2 3465 3 2 c c c c c c c c c c d Hc x c x m x c x d x md x cx x m x cxd x d x m cm c d x m mx c mc    d x 2 22 Hence, by minimizing H w.r.t. z , we have: 3 2 3 0 21 1 33 2 dH c dc mc c m 2
Put the value of c back to the expected energy, we obtain: 31 0.5222 11 2 H    For 0 x , since it is an odd function, which has an opposite parity compared with the ground trial function 0 , hence this guarantee the integral 01 0 c c dx  rthermore, we , hence this confirmed the orthogonal property of the two trial functions. Fu may spot that there is one crossing point of the x-axis of the wavefunction 1 , which shows it is also the first excited ay state. (One m compare from the ground state which never crosses the x-axis.) Again, we need to minimize c in order to find the upper bound of energy.

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## This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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PHYS4221_HW4_Solution2 - PHYS4221 HW4 Solution Written by...

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