Linear Combination of 2 particular solutions of the TDSE

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Homework Help Overview

The discussion revolves around the time-dependent Schrödinger equation (TDSE) and its solutions, particularly focusing on linear combinations of specific solutions derived from the time-independent Schrödinger equation (TISE). Participants are exploring the implications of these combinations and their properties in quantum mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the validity of a linear combination of two solutions of the TDSE and its relationship to the TISE. They question how to demonstrate that the linear combination satisfies the TDSE but not the TISE. There are inquiries about the implications of the Hamiltonian's eigenvectors and energy eigenvalues on the state of the system over time.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting substitutions into the relevant equations. Others are exploring the relationship between the solutions of the TDSE and TISE, and there is ongoing clarification regarding the steps needed to demonstrate the properties of the solutions.

Contextual Notes

Participants express uncertainty about specific steps in the problem-solving process, particularly in relation to the equations involved and the conditions for recurrence in the system's state. There is also mention of the need to clarify assumptions regarding the time evolution of the quantum states.

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Homework Statement



1. A particular solution \Psin(x,t) of the TDSE, -iℏ∂\Psi/∂t=\hat{H}\Psi, can be constructed by taking \Psin(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\Psi1(x,t) + c2\Psi2(x,t) is also a solution of the TDSE but not a solution of the TISE.

b) Consider a Hamiltonian \hat{H} with eigenvectors \phi1 and \phi-1 and corresponding energy eigenvalues E1=ℏω and E-1=-ℏω. If at time t=0, the state of the system is \Psi(t=0)=c1\phi1(t=0)+c-1\phi-1(t=0), give the state of the system at an arbitrary later time t. This system demonstrates recurrences, where \Psi(t)=\Psi(0). At what time will this occur?

c) What's the expectation value of the energy \langleE\rangle at an arbitrary later time t for the case c1=c-1.

d) Consider now the 2 (also orthonormal) quantum states: v±(t)=(\phi1(t)±\phi-1(t))/√2. Find \phi1(t) and \phi-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

\Psi(x,t)=c1ψ1(x)exp(-iE1t/ℏ)+c2ψ2(x)exp(-iE2t/ℏ)

|\Psi(x,t)|2= |c1|21|2+|c2|22|2+c1*c2ψ12exp(-i(E2-E1)t/ℏ)+c1c21ψ2*exp(-i(E1-E2)t/ℏ)

Not a stationary state since |\Psi(x,t)|2≠ |\Psi(x,t=0)|2 ∴ solution of TDSE but not of TISE

b) Not sure how to finish the question

\Psi(x,t)=c1\phi1(x)exp(-iE1t/ℏ)+c-1\phi-1(x)exp(-iE-1t/ℏ)

=c1\phi1(x)exp(-iωt)+c-1\phi-1(x)exp(iωt)
 
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For a), you merely need to substitute the linear combination into the equations, and show that one is satisfied, and the other is not.

For b), what is the relation between the solutions of the time-dependent and time-independent equations? Can you use the result of a)?
 
voko said:
For a), you merely need to substitute the linear combination into the equations, and show that one is satisfied, and the other is not.

For b), what is the relation between the solutions of the time-dependent and time-independent equations? Can you use the result of a)?

For a) what equations do I substitute the linear combination into? I tried to sub it into the TDSE but it didn't seem to work :s
 
For a) you have to sub the combo into both equations, and TDSE must be satisfied identically, while TISE must not. Unless you show exactly what you do, it is hard to say what goes wrong.
 
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For (b), you just need to find when the complex exponentials have argument equal to a multiple of 2*pi, and then they will equal 1. It's very simple from where you got.
 
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