Eigenfunction energy levels in a harmonic well

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Lazy Rat
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Homework Statement


If the first two energy eigenfunctions are
## \psi _0(x) = (\frac {1}{\sqrt \pi a})^ \frac{1}{2} e^\frac{-x^2}{2a^2} ##,
## \psi _1(x) = (\frac {1}{2\sqrt \pi a})^ \frac{1}{2}\frac{2x}{a} e^\frac{-x^2}{2a^2} ##

Homework Equations

The Attempt at a Solution


Would it then be correct to presume
## \psi _3(x) = (\frac {1}{4\sqrt \pi a})^ \frac{1}{2}\frac{4x}{a} e^\frac{-x^2}{2a^2} ##

Thank you for your time in considering this.
 
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No. In terms of ladder operators, the nth eigenfunction is given by

[tex] <br /> |n \rangle \equiv \psi_{n}(x) = \frac{(a^\dagger)^n}{\sqrt{n!}} |0 \rangle<br /> [/tex]
 
the specific question goes as so

For this equation

## \Psi (x,0) = \frac {1}{\sqrt{2}}(\psi_1 (x)-\psi_3 (x)) ##

The system is undisturbed, obtain an expression for ##\psi (x,t)## that is valid for all t ≥ 0. Express in terms of the functions ##\psi_1 (x)##, ##\psi_3 (x)## and ##ω_0##, the classical angular frequency of the oscillator.

I am trying to approach this by simply inputting the eigenfunctions for

##\psi _1(x) = (\frac {1}{2\sqrt \pi a})^ \frac{1}{2}\frac{2x}{a} e^\frac{-x^2}{2a^2}##

And then for

##\psi _3(x)## (which as yet I haven't understood)

And

##a = \sqrt{\frac {\hbar}{ω_0}}##

Would this be the correct approach to express in the terms as stated?

Thank you for assisting me with my problem.
 
So would i use the fact that ## E_1 = \frac {3}{2} \hbar ω_0 ## which would give ## e^ \frac {- 3iω_0t}{2} ##
And ## E_3 = \frac {7}{2} \hbar ω_0 ## which would give ## e^ \frac {- 7iω_0t}{2} ##

Am I on the right track?
 
Lazy Rat said:
So would i use the fact that ## E_1 = \frac {3}{2} \hbar ω_0 ## which would give ## e^ \frac {- 3iω_0t}{2} ##
And ## E_3 = \frac {7}{2} \hbar ω_0 ## which would give ## e^ \frac {- 7iω_0t}{2} ##

Am I on the right track?

Yes.
 
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