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Consider a particle in a quantum mechanical infinite square well. At t = 0, the particle is in the ground state, with a wave function given by Show explicitly that this wave function is an eigenstate of the operator, and find its eigenvalue. Now consider a particle given by the wave function Show explicitly that this wave function is not an eigenstate of the same operator. Consider the statements about this second wave function given below. Student 1: "The wave function is a solution of the Schrodinger Equation. So there must be a definite energy associated with it." student 2: "At times the wave function is like the ground state, and at other times it is like the first excited state. So, sometimes it has energy Ei, and at other times E2.H Student 3: "When written as a sum of eigenstates, the coeff icients are always the same. So the probabilities of measuring Ei and E2 don't change, even though the wave function looks Each student is incorrect. Identify the flaw in each student's reasoning. Explain.

Explanation / Answer

a) so d Psi/dt = psi1(x) (- i E1/hbar) e^(-i E t/hbar)

= Psi*(-i E1/hbar)

so i hbar * dPsi/dt = i hbar * -i E1/hbar dPsi/dt = E1 dPsi/dt

so eigenvalue is E1

b)

now dPsi/dt = 1/sqrt(2) ( psi1 e^(-i E1t/hbar) -i E1/hbar + psi2 e^(-i E2t/hbar) -i E2/hbar)

so i hbar dPsi/dt = 1/sqrt(1)( E1 psi1 e^(-i E1t/hbar) + E2 psi2 e^(-i E2 t/hbar))

no way to get Psi out of that

so does not return Psi so not eigenstate

c)

1) is wrong because it is not an eigenfunction of i hbar dPsi/dt it doesnt have a set energy associated with it

2) No it doesnt oscillate between the two it is a combination of both energies at all times

3) if the coefficient in front of them is zero the prob of being in that state is zero,

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