Question

If you were to use a trial function of the form φ(x):: (1 + cax 2)r"r/2, where α (kμ/h2)1/2 and c is a variational parameter, to calculate the ground-state energy of a harmonic oscillator, what do you think the value ofc will come out to be? Why? 8-6.

Answer 1

Wave function for the harmonic oscillator in nth state is, y (x)= H, (ax) 2nt 11a Here, a = 1 and H, (ax) is the Hermite polynomial. For ground state wave function substitute n=0 in v,(x) and and for ground state the value for Hermite polynomial H, (ax) is H, (ax) and is equal to 1. The given trial wave function is $(x) =(1-cax?)eż**, if we substitute c = 0 in the trial function and do the normalization we get the required ground state wave function. (6(x) 6(x)) = 1 $(x)= Ae Jax Lep je se že dr = 1 Lap Ježe dr=1 je**d=(n) then the resultant wave Now substitute the above result in equation (x)= Ae 2 function is a ground state function. \.(x)=(" So, the given trial wave function is $(x)= (1-cax?)ezar can be written as $(x)=-(cax? – 1)e žu If we substitute c = 2 in the above equation $(x)=-(cax? - 1)e 2** then $(x) will be, $(x)=-(2ax-1)e 32 Which will be equal to w. (x) in the given wave function that is in the 2nd excited state wave function.