Answer to Question #93970 in Quantum Mechanics for Bornface

Question #93970
Consider a particle of mass m confined in an infinite 1-D potential well of width a;
V (x)=0 for 0 less or equal to x and x less than or equal to a,
V(x)=infinite otherwise
Find the eigenstates of the Hamiltonian and the corresponding eigen energies Using three possibilities considering the energy E,
E>0
E<0
E=0
1
Expert's answer
2019-09-10T13:25:43-0400

The eigenvalues of the Hamiltonian are:


Hψ=EψH|\psi\rangle=E|\psi\rangle

for every eigenvector.

Write the Hamiltonian:


H=22md2dx2,H=-\frac{\hbar^2}{2m}\frac{\text{d}^2}{\text{d}x^2},

substitute this to the first equation:


22md2dx2ψ=Eψ,-\frac{\hbar^2}{2m}\frac{\text{d}^2}{\text{d}x^2}|\psi\rangle=E|\psi\rangle,

the solution of this equation gives the set of eigenstates:


ψ=C sin(πnxa).|\psi\rangle=C\space\text{sin}\Big(\frac{\pi n x}{a}\Big).

The corresponding energies (eigenvalues):


E=n2π222ma.E=\frac{n^2\pi^2\hbar^2}{2ma}.



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