Question

Two students have a very pressing homework deadline concerning the application of the variational principle to estimate the ground state energy of the harmonic oscillator. The Hamiltonian operator of such system is î H -12d = 24 d.22 + 2 .2. in which u is the reduced mass of the oscillator and w = (force constant/u)/2 its natural frequency. The correct energies for this system are well known Eo = (v +) , v= 0,1,2, ... As the trial function for the ground state the students are instructed to use a? - r if -a<r <a $(x) = 1 if x>a or I<-a where a is a parameter that needs to be optimized. When calculating the variational integral the students get a different result, and only one of them is correct: a) Student A W(a) = b) Student B 572 W(a) = 411027 HW222 (a) Without calculating W(a) by yourself (i. e., using the results for W(a) reported by the students), determine which one (A or B) is correct. Explain your reasoning, (b) Explain why the proposed trial function $(x) is a sensible choice for the wavefunc- tion of the harmonic oscillator. What is the physical meaning of the variational parameter a?

Answer 1

enessy a) WA (a) and us (a) represent the vcernicational Calenlated by A and B respectively. af Ca) le CA f 2 nofket a, co rew to 7. >> K+ (2494 Seat Recue WA C) = 3h + fear of + Rcco - tw ( feat+243%). = 0.463 tius

Wp (x) = 5 h² + two x² 14 4 6x² A L. - - stone W, 6) er met het en frece ao the greater the 4 zostus we have the actual ground state enersy is a.s tue. Hence, the second choice, what student B reported was Correct WC.). This is because when we use variational principle for computing the eigenvalues, it assumes that the variational enerry must be greater than actual energy. This is the reason for choice made by student is makes sense. (W(x) > hue). Aus: student B's calculation 6) In order to make a care function admissiable, it should well behared at a: Is. Since the care function Vanishes both at He is and its second derivative enists. Due to these two reasons, it is a good trial care function. 4(+4 -0, to
In the case of the variational method, we are not sure about the exact wavefunction, hence we introduce some parameters such that it will account for all the properties of the wavefunction. This is the fundamental assumption in this method. Then after we minimize the variational energy to find the approximate energy. Thus the variational parameter acts as a minimizer to find the closet approximate value.