TDSE, time evolution between states (Check my working please?)

[unscientific]

Homework Statement

Hi guys, I've recently taken up quantum, so it's all very new to me, it would be greatly appreciated if someone could check my working!Let ψ1(x) and ψ2(x) be two orthonormal solutions of the TISE with corresponding
energy eigenvalues E1 and E2. At time t = 0, the particle is prepared in the symmetric
superposition state:

and subsequently allowed to evolve in time. What is the average energy of the system
as a function of time? What is the minimum time ¿ for which the system must evolve
in order to return to its original state (up to an overall phase factor), when it starts in
the state ψ(+) (x)

Determine the probability to ¯nd the system in the antisymmetric superposition
state

as a function of time when it starts in the state ψ(+) (x)

At time t1 the particle is found in the antisymmetric superposition state. What is
the probability to ¯nd the particle in the symmetric superposition state at time t1 + T,
where T is the time found above?

Homework Equations

The Attempt at a Solution

Not sure if the last part is right, as it suggests that the probability of finding the particle in the left half of the box is independent of time! Then again, in an infinite square well the potential doesn't depend on time, so TDSE is reduced to TISE?

 

[mfb]

I think you should express ##\Delta \omega## in terms of the energies.

For the last part, why do you think those mixed terms are zero if you integrate them from 0 to L/2?
The problem statement itself is bad here, I think. The answer depends on the phases used to define the first two states, and those phases are arbitrary.

 

[unscientific]

I think you should express ##\Delta \omega## in terms of the energies.

For the last part, why do you think those mixed terms are zero if you integrate them from 0 to L/2?
The problem statement itself is bad here, I think. The answer depends on the phases used to define the first two states, and those phases are arbitrary.

Because ψ1 and ψ2 are orthogonal states? And to get the answer in the previous parts they were zero...Is the reason why they are non-zero here because the limits of integration are not from -∞ to ∞?

 

[mfb]

Because ψ1 and ψ2 are orthogonal states? And to get the answer in the previous parts they were zero...Is the reason why they are non-zero here because the limits of integration are not from -∞ to ∞?

Right, orthogonality just means that the integral over the whole space is zero.

 

[paranormal]

Hello,

Thank you for sharing your working. From what I can see, your understanding of the concepts is correct. The TDSE (time-dependent Schrodinger equation) is used to describe the time evolution of a quantum system. It is similar to the TISE (time-independent Schrodinger equation), but it takes into account the time dependence of the potential.

In the given problem, the particle is initially in a symmetric superposition state and is allowed to evolve in time. The average energy of the system can be calculated using the time-dependent wave function:

< E > = ∫ ψ*(x) H ψ(x) dx

where H is the Hamiltonian operator. This average energy will change with time as the system evolves.

To determine the minimum time for the system to return to its original state, we need to use the time-dependent wave function and its time evolution operator:

ψ(t) = e^(-iHt/ħ) ψ(0)

where ψ(0) is the initial wave function. We can solve for the time t when ψ(t) = ψ(0) to find the minimum time required for the system to return to its original state.

To determine the probability of finding the system in the antisymmetric superposition state at a given time t, we can use the time-dependent wave function and its time evolution operator to calculate:

P(t) = |< ψ(t) | ψ(-) >|^2

where ψ(-) is the antisymmetric superposition state. This probability will change with time as the system evolves.

Finally, to find the probability of finding the particle in the symmetric superposition state at a later time t1 + T, we can use the time evolution operator to calculate:

P(t1+T) = |< ψ(t1+T) | ψ(+) >|^2

where ψ(+) is the symmetric superposition state. This probability will also change with time as the system evolves.

In conclusion, your understanding and working seem to be correct. Just a small clarification, the probability of finding the particle in the left half of the box will not be independent of time, as it will change with time as the system evolves. It is only independent of time if the potential is time-independent, as you mentioned.

I hope this helps. Keep up the good work in your quantum studies!