- #1
A9876
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Homework Statement
1. A particular solution [itex]\Psi[/itex]n(x,t) of the TDSE, -iℏ∂[itex]\Psi[/itex]/∂t=[itex]\hat{H}[/itex][itex]\Psi[/itex], can be constructed by taking [itex]\Psi[/itex]n(x,t)=ψn(x)exp(-iEnt/ℏ) where the ψn are solutions of the TISE.
a) Show that a linear combination of 2 such solutions c1[itex]\Psi[/itex]1(x,t) + c2[itex]\Psi[/itex]2(x,t) is also a solution of the TDSE but not a solution of the TISE.
b) Consider a Hamiltonian [itex]\hat{H}[/itex] with eigenvectors [itex]\phi[/itex]1 and [itex]\phi[/itex]-1 and corresponding energy eigenvalues E1=ℏω and E-1=-ℏω. If at time t=0, the state of the system is [itex]\Psi[/itex](t=0)=c1[itex]\phi[/itex]1(t=0)+c-1[itex]\phi[/itex]-1(t=0), give the state of the system at an arbitrary later time t. This system demonstrates recurrences, where [itex]\Psi[/itex](t)=[itex]\Psi[/itex](0). At what time will this occur?
c) What's the expectation value of the energy [itex]\langle[/itex]E[itex]\rangle[/itex] at an arbitrary later time t for the case c1=c-1.
d) Consider now the 2 (also orthonormal) quantum states: v±(t)=([itex]\phi[/itex]1(t)±[itex]\phi[/itex]-1(t))/√2. Find [itex]\phi[/itex]1(t) and [itex]\phi[/itex]-1(t) in terms of the v±(t)
e) If at time t=0, the state of the system is found to be ψ(0)=v-, find the probability as a function of time, that ψ(t) will be found in the state v+
Homework Equations
The Attempt at a Solution
a) Not sure if this is right
[itex]\Psi[/itex](x,t)=c1ψ1(x)exp(-iE1t/ℏ)+c2ψ2(x)exp(-iE2t/ℏ)
[itex]|[/itex][itex]\Psi[/itex](x,t)[itex]|[/itex]2= [itex]|[/itex]c1[itex]|[/itex]2[itex]|[/itex]ψ1[itex]|[/itex]2+[itex]|[/itex]c2[itex]|[/itex]2[itex]|[/itex]ψ2[itex]|[/itex]2+c1*c2ψ1*ψ2exp(-i(E2-E1)t/ℏ)+c1c2*ψ1ψ2*exp(-i(E1-E2)t/ℏ)
Not a stationary state since [itex]|[/itex][itex]\Psi[/itex](x,t)[itex]|[/itex]2≠ [itex]|[/itex][itex]\Psi[/itex](x,t=0)[itex]|[/itex]2 ∴ solution of TDSE but not of TISE
b) Not sure how to finish the question
[itex]\Psi[/itex](x,t)=c1[itex]\phi[/itex]1(x)exp(-iE1t/ℏ)+c-1[itex]\phi[/itex]-1(x)exp(-iE-1t/ℏ)
=c1[itex]\phi[/itex]1(x)exp(-iωt)+c-1[itex]\phi[/itex]-1(x)exp(iωt)