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.

 

[paranormal]

I would like to clarify the steps needed to solve this problem. First, we have to understand the concept of eigenfunctions in Hamiltonian quantum mechanics. Eigenfunctions are special wavefunctions that satisfy the Schrödinger equation and represent the energy states of a system. These eigenfunctions are characterized by their corresponding eigenvalues, which represent the energy of the system. In this problem, we are given a potential function V(x) and we are asked to find the eigenfunction ψ(x) and its corresponding eigenvalue E1.

To solve this problem, we need to use the Schrödinger equation, which is given by:

Hψ(x) = Eψ(x)

Where H is the Hamiltonian operator, ψ(x) is the wavefunction, and E is the energy.

To find the eigenfunction ψ(x), we need to solve the Schrödinger equation for a given potential function V(x). In this case, the potential function is V(x) = 1/2(mω2x). So, we can write the Schrödinger equation as:

-ħ^2/2m(d^2ψ(x)/dx^2) + 1/2(mω2x)ψ(x) = Eψ(x)

Next, we substitute the given wavefunction ψ(x) = a0exp(-βx2) into the Schrödinger equation and solve for the value of β. This step involves differentiating the wavefunction ψ(x) twice and then substituting it into the Schrödinger equation. After simplifying, we get:

β = mω/2ħ

This is the value of β that satisfies the Schrödinger equation for the given potential function V(x).

To calculate the value of the energy E1, we need to substitute the value of β into the Schrödinger equation and solve for E. This will give us the eigenvalue corresponding to the given eigenfunction ψ(x). After solving the equation, we get:

E1 = 1/2ħω

This is the value of the energy corresponding to the eigenfunction ψ(x) = a0exp(-βx2).

In conclusion, Hamiltonian quantum mechanics is a powerful tool that allows us to find the energy states of a system by solving the Schrödinger equation. By using the