Hamiltonian of the Quantum Harmonic Oscillator-Eigenfunction & Eigenvalue

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Calcifur
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Homework Statement


Show that the equation below is an eigenfunction for the Quantum Harmonic Oscillator Hamiltonian and find its corresponding eigenvalue.


Homework Equations


u1(q)=A*q*exp((-q[itex]^{2}[/itex])/2)


The Attempt at a Solution


Ok, so I know that the Quantum Harmonic Oscillator Hamiltonian (H[itex]_{QHO}[/itex]) is :
(H[itex]_{QHO}[/itex])=[itex]\frac{1}{2}[/itex][itex]\hbar[/itex]ω(((-d^2)/(dq^2))+q^2) and I know that:
(H[itex]_{QHO}[/itex])u1(q)=Eu1(q)

but how do I show that it's an eigenfunction? Simply subbing it into the eqn doesn't appear to help.

Many thanks in advance.
 
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Calcifur said:
Simply subbing it into the eqn doesn't appear to help.
It should. Either that or you need to solve the differential equation, which is a much harder task.

Show us what you got when you plugged u1 into the equation.
 
vela said:
Show us what you got when you plugged u1 into the equation.

Ok, so here goes:

HQHO*U1(q)=E*U1(q)

[itex]\frac{1}{2}[/itex][itex]\hbar[/itex]ω([itex]\frac{-d^{2}}{dq^{2}}[/itex]+q[itex]^{2}[/itex]).A.q exp([itex]\frac{-q^{2}}{2}[/itex])=E*U1(q)

[itex]\frac{A}{2}[/itex][itex]\hbar[/itex]ω(-[itex]\frac{d}{dq}[/itex][itex]\frac{d}{dq}[/itex](q.exp([itex]\frac{-q^{2}}{2}[/itex]))+(q[itex]^{3}[/itex])exp([itex]\frac{-q^{2}}{2}[/itex]))=E*U1(q)

which eventually comes to:

(q[itex]^{3}[/itex]+q[itex]^{2}[/itex]-2q-1)exp([itex]\frac{-q^{2}}{2}[/itex])[itex]\frac{A}{2}[/itex][itex]\hbar[/itex]ω=E*U1(q)

So does this mean that : (q[itex]^{3}[/itex])+q[itex]^{2}[/itex]-2q-1) is the corresponding eigenvalue?

Is my method correct?

Many thanks.
 
vela said:
You must have calculated the second derivative incorrectly.

Thanks, my mistake. So I now have:

[itex]\frac{\hbarω}{2}[/itex]A.exp([itex]\frac{-q^{2}}{2}[/itex])(-q[itex]^{3}[/itex]-q[itex]^{2}[/itex]+3q)=E.U[itex]_{1}[/itex](q).

So is -q[itex]^{3}[/itex]-q[itex]^{2}[/itex]+3q the corresponding eigenvalue? Or can I simplify even further?

Many thanks.
 
I think I've figured it out now. Many thanks Vela