1. The problem statement, all variables and given/known data A particle is in a
ID: 1674505 • Letter: 1
Question
1. The problem statement, all variables and given/known dataA particle is in a 1D box of length L on x-axis with0<x<L. What is the shift in the energy of the energy ofeach wavefunction due to the weak spring potential to the firstorder k? The answer involves an integral found on an integraltable.
2. Relevant equations
H=p2/2m
Sin=sqrt(2/L)cos(nx/L)
En=n22hbar2/2mL2
V(x)=kx2/2
3. The attempt at a solution
I'm super confused because the shift in the energy is usuallydetermined by a sum, not an integral. Unless I calculateEn(1) by taking;
En(1)=<n|H|n>
=<n|V|n>
=V * integral(2/Lsin2(nxpi/L)dx) with limits 0 to L
Is this the correct way to do this one? I didn't include kvery well unless I put it in for V at the end...
Explanation / Answer
so, if kL2 has units of energy, then k has units ofenergy over area? - yeah. if you write V=(1/2)kx2 , then k times a length squared hasunits of energy. Also- I wrote my integral with sin even though the wavefunction hadunits of cosine. Is that the correct way to do it? - this depends on how you wrote your wavefunction. The problem saysx is from 0 to L. so you better use sin[nx/L] for yourwavefunction (to satisfy the boundary conditions). Some peoplewrite wavefunctions from -L/2 to L/2 which gives you sines andcosines.
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