Combination of wavefunctions in oscillator

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
cfwoods
Messages
2
Reaction score
0

Homework Statement


The equation [itex]\psi(x) = \frac{1}{sqrt(2)}\psi_0 (x) + \frac{i}{sqrt(5)}\psi_1 (x) + \gamma\psi_2 (x)[/itex]

is a combination of the first three eigenfunctions in the 1D harmonic oscillator. So, [itex]\psi_0 = Ae^{-mωx^2 /2\hbar}[/itex] and so on for the first and second excited states. If [itex]\psi_0[/itex], [itex]\psi_1[/itex] and [itex]\psi_2[/itex] are normalised, and [itex]\psi(x)[/itex] is also normalised, determine [itex]|\gamma|[/itex]

The Attempt at a Solution



You can obtain the normalised functions for the ground state and first two excited states from a variety of methods, and you can then expand out [itex]\psi(x)[/itex]. I tried plugging that back into the time dependent Schrödinger equation, but that didn't help (and it also gave a messy derivative) so I am at a loss as to how i can proceed.
 
Physics news on Phys.org
cfwoods said:
You can obtain the normalised functions for the ground state and first two excited states from a variety of methods, and you can then expand out [itex]\psi(x)[/itex].
It is stated that ##\psi_0##, ##\psi_1##, and ##\psi_2## are taken to be normalized.

cfwoods said:
I tried plugging that back into the time dependent Schrödinger equation
How is that realted to normalization?

Let's start from the beginning: what equation does ##\psi(x)## statisfy if it is normalized?
 
Hmm it satisfies [itex]\int^{∞}_{-∞} \psi^{*}\psi dx = 1[/itex] I think
 
cfwoods said:
Hmm it satisfies [itex]\int^{∞}_{-∞} \psi^{*}\psi dx = 1[/itex] I think
Correct. So plug in there the ##\psi(x)## of the problem. Keep the notation ##\psi_0## and so on (i.e., do not write the explicit functions of ##x##) and use the properties of the harmonic oscillator wave functions to simplify the result.