Find the time-dependent wave function Ψ(x, t).

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SUMMARY

The discussion focuses on finding the time-dependent wave function Ψ(x, t) using the standard method for solving the Schrödinger equation. The initial wave function is given as Ψ(0, x) = F(x)(a + bx), which needs to be expressed in terms of energy eigenstates. Participants emphasize the importance of recognizing the decomposition into energy eigenstates, specifically using the general form Ψ(0, x) = Σ c_n ψ_n(x). The explicit formulation of the first few eigenstates of the harmonic oscillator is crucial for this analysis.

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gabriellelee
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Homework Statement
Find the time-dependent wave function Ψ(x, t).
Relevant Equations
Please see below for the full question.
Screen Shot 2020-01-29 at 10.48.31 PM.png

I thought I could start somewhere along the lines of ##\psi(x,t)= \psi(x,0)e^{-iE_nt/\hbar}##, but I'm not sure what ##E_n## would be.
I also thought about doing the steps listed below in the picture, but I'm not sure how to decompose ##\psi(x,0)## like it says to in the first step.
Any help would be very much appreciated.
Screen Shot 2020-01-29 at 10.57.10 PM.png
 
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This is the standard method for solving the Schroedinger equation. In this case the decomposition into energy eigenstates has been done for you! You need to look more closely at the initial wavefunction:

gabriellelee said:

Do you recognise anything there?
 
Are the right two terms ##\psi(x)## and the first term is ##C_n##?
 
gabriellelee said:
Are the right two terms ##\psi(x)## and the first term is ##C_n##?
No.

Look at the explicit formulation of the first few eigenstates of the harmonic oscillator.
 
... the general form of the initial wavefunction in terms of energy eigenstates is:$$\Psi(0, x) = \sum c_n \psi_n(x) = c_0\psi_0(x) + c_1\psi_1(x) + \dots$$
In this problem you are given the initial wavefunction in the form:$$\Psi(0, x) = F(x)(a+bx) = aF(x) + bxF(x)$$
You have to try to express the functions ##F(x)## and ##xF(x)## in terms energy eigenstates. Hint (again): the problem setter has more or less done this for you!
 

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