Linear Combination of 2 particular solutions of the TDSE

In summary, the conversation discusses the construction of particular solutions to the time-dependent Schrodinger equation using solutions of the time-independent Schrodinger equation. It also examines the properties of linear combinations of these solutions and the recurrence behavior of a quantum system. The expectation value of energy and the probability of finding a particular state are also discussed.
  • #1
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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*c2ψ12exp(-i(E2-E1)t/ℏ)+c1c21ψ2*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)
 
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  • #2
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)?
 
  • #3
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
 
  • #4
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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  • #5
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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1. What is the TDSE?

The TDSE stands for the Time-dependent Schrödinger Equation. It is a fundamental equation in quantum mechanics that describes how a quantum system evolves over time.

2. What is a linear combination of two particular solutions of the TDSE?

A linear combination of two particular solutions of the TDSE is a mathematical operation that combines two solutions of the TDSE in a way that preserves the linearity of the equation. It is often used to find more complex solutions to the TDSE.

3. How is a linear combination of two particular solutions of the TDSE calculated?

To calculate a linear combination of two particular solutions of the TDSE, you simply add the two solutions together, multiplying each by a constant coefficient. This results in a new solution that satisfies the TDSE.

4. Why is a linear combination of two particular solutions of the TDSE useful?

A linear combination of two particular solutions of the TDSE is useful because it allows us to find more complex solutions to the TDSE that cannot be found using single solutions. It also helps us to understand the behavior of quantum systems more comprehensively.

5. Can a linear combination of two particular solutions of the TDSE be used to describe any quantum system?

Yes, a linear combination of two particular solutions of the TDSE can be used to describe any quantum system. This is because the TDSE is a universal equation that accurately describes the evolution of all quantum systems, and the linear combination allows us to find solutions for any given system.

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