Is ψ(x) = a0exp(-βx²) an Eigenfunction of the Hamiltonian?

[phys2]

Homework Statement

A particle moves in a one dimensional potential : V(x) = 1/2(mω^2x

Show that the function ψ(x) = a₀exp(-βx²) is an eigenfunction for the Hamiltonian for a suitable value of β and calculate the value of energy E₁

Homework Equations

The Attempt at a Solution

What I did was take the Hamiltonian, differentiate twice the function ψ(x) and then sub in the twice differentiated function along with the potential into the Hamiltonian. But there I get confused. Apparently taking a peek at the solutions, it argues that you should set the Hamiltonian to zero and then since it is supposed to be a constant, the x^2 terms cancel out and your final answer is β=mω/2hbar. I have no idea what they have done though! Any one care to explain? Thanks!

 

[vela]

Homework Statement

A particle moves in a one dimensional potential : V(x) = 1/2(mω^2x

Show that the function ψ(x) = a₀exp(-βx²) is an eigenfunction for the Hamiltonian for a suitable value of β and calculate the value of energy E₁

Homework Equations

The Attempt at a Solution

What I did was take the Hamiltonian, differentiate twice the function ψ(x) and then sub in the twice differentiated function along with the potential into the Hamiltonian.

Show us what you got when you did that. If $\psi$ is an eigenfunction of $\hat{H}$, what does $\hat{H}\psi$ have to equal?

But there I get confused. Apparently taking a peek at the solutions, it argues that you should set the Hamiltonian to zero and then since it is supposed to be a constant, the x^2 terms cancel out and your final answer is β=mω/2hbar. I have no idea what they have done though! Any one care to explain? Thanks!

 

[phys2]

So I got -h(bar)²/2m (4β²x²⁾ ψ + 1/2mω²x²ψ = Hamiltonian

Hψ=Eψ?

 

[vela]

So I got -h(bar)²/2m (4β²x²⁾ ψ + 1/2mω²x²ψ = Hamiltonian

You're missing a term. You didn't differentiate correctly, perhaps.

Hψ=Eψ?

Right, so after applying the Hamiltonian, you should be able to write the result as a constant times the original wave function. To do that, β will have to take on a specific value.