Corresponding Energy Levels for Normalised Eigenfunctions

AI Thread Summary
The discussion revolves around calculating the energy levels for a particle in a 1D infinite square well, using the normalized energy eigenfunction Un(x). The key equation for energy levels is En = (h^2 / 2m) * (n / 2a)^2, where n represents the quantum number. Participants clarify that there is no need to unnormalize the wavefunction, as the problem aims to derive the energy levels directly from the Hamiltonian and the eigenfunction. The correct approach involves applying the Hamiltonian operator to Un(x) to find the corresponding energy levels. This reinforces the application of the time-independent Schrödinger equation in quantum mechanics.
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Homework Statement



Particle of mass m in a 1D infinite square well is confined between 0 ≤ x ≤ a

Given that the normalised energy eigenfunction of the system is:

Un(x) = (\frac{2}{a})\frac{1}{2} sin (\frac{nx\pi}{a})

where n = 1, 2, 3...

what are the corresponding energy levels?

Homework Equations



Normalisation: 1 = \int lψl2 dx (from negative to positive infinity)

Energy Level for infinite well: En = \frac{h^2}{2m} (\frac{n}{2a})2

Where h = reduced Planck constant (sorry couldn't find it)

The Attempt at a Solution



At first I thought, I could just plug in 1, 2, 3 in Un(x)
But then I realized the question said that Un(x) is normalised.
Does this mean I have to reverse normalise Un(x)? If so, is this order right?

Make Un(x) = 1
Differentiate (\frac{2}{a})\frac{1}{2} sin (\frac{nx\pi}{a})
Then square root whatever the result.

Am I going in the right direction or am I missing something out?

Thank you!
 
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You don't have to unnormalize the wavefunction. You already have the answer in the E_n. The intent of the problem is for you to compute E_n from just the wavefunction and the Hamiltonian, instead of just taking the formula from your notes or book.

In other words, remember that
\hat H |U_n \rangle = E_n |U_n \rangle

Because, well, that's the basic content of the time-independent Schrodinger equation.
 
oh... lol. That's simple enough!

So basically apply the hamiltonian to Un(x)
 
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