Question

I need part c please :)     

2. What makes the operators a and a', defined in problem 1 e, of the last homework, useful, is that they make it easy to manipulate the solutions to the harmonic oscillator. The general behavior of the operators when they operate on harmonic oscillator wavefunctions, yv, is as follows: a' SQRT(v+1) i.c., operating with a' on one of the harmonic oscillator eigenfunctions, ww, converts it to the next highest eigenfunction, ψν+1-Therefore at is called a "raising operator". a ψ.-SQRT() . ψν-1, İ.е., it lowers the quantum number by one, so a is called a "lowering operator". The exception is that if the system is already in the ground state, you can't lower v below zero, therefore: ayo = 0 a. Using the definitions of a and a' given in last week's homework, and the functional form of vo (table 5.3), prove that these relations work for the ground state of the harmonic oscillator. Also, you can rearrange the definitions I gave you for a and a', to write the x and p, operators in terms of a and a' as follows: x sqrt(/(Ho)). 1/sqrt(2) . (a* a) Ps sqrt(uho). i/sqrt(2) . (a' -a) b. These expressions for x and p, turn out to be very useful in working problems with the harmonic oscillator. Use these definitions of x and p, plus the fact that the harmonic oscillator wave functions are orthonormal, to show that xp 0. Why do you think this is reasonable? c. You can also define xand p.2 operators in terms of a and a by just squaring the expressions given in part a. Remember that last week you showed that a and a do not commute. Calculatexandp> for the harmonic oscillator wavefunction y for arbitrary v.

Answer 1