Can Different Multiplications of ψ(x) and Φ(t) Result in Solutions to TDSE?

[yosimba2000]

So to get the TISE, we assume a solution Ψ = ψ(x)Φ(t) and plug it into the TDSE.

It will give us two expressions, one of which is the TISE, to find an infinite number of ψ(x) and Φ(t) solutions.

So all solutions to TDSE are ∑₁^∞ ψ(x)_nΦ(t)_n, but couldn't ∑₁^∞ψ_m(x) Σ₁^∞ Φ_n(t) also be a set of solutions to TDSE? Basically I'm using the same ψ(x), namely ψ_m(x), but multiplying it by different Φ_n(t), then moving onto the next ψ_m(x).

 

[Chandra Prayaga]

Easiest way to see the problem in what you are suggesting is to try your second function with just two states. Assuming that the hamiltonian is independent of time, from TISE:
Hψ₁(x) = E₁ψ₁(x)
Hψ₂(x) = E₂ψ₂(x)
The time dependent parts for each of these states should be
Φ₁(t) = e^-iE₁t/ħ
Φ₂(t) = e^-iE₂t/ħ
So the standard solution to TDSE for the superposition state, which is the time evolution of the state (ψ₁(x) + ψ₂(x)) at t = 0, is:
ψ₁(x)Φ₁(t) + ψ₂(x)Φ₂(t), which can be checked by direct substitution into TDSE.
Your suggestion:
(ψ₁(x) + ψ₂(x) )(Φ₁(t) + Φ₂(t))
will not work. It is not a solution of TDSE.

 

[yosimba2000]

Ok. Also, why when writing the most general solution to the TDSE, are the constants Cn the coefficients obtained from the Fourier Transform? Or, why do they have to be the Fourier coefficients? In 2rd order differential equations, we saw that we could multiply the solution X by any constant C1, and the new solution C1X would also be a solution to the differential equation.

 

[Chandra Prayaga]

Ok. Also, why when writing the most general solution to the TDSE, are the constants Cn the coefficients obtained from the Fourier Transform? Or, why do they have to be the Fourier coefficients? In 2rd order differential equations, we saw that we could multiply the solution X by any constant C1, and the new solution C1X would also be a solution to the differential equation.

What is the specific TDSE for which you find the coefficients obtained from the Fourier Transform?

 

[yosimba2000]

What is the specific TDSE for which you find the coefficients obtained from the Fourier Transform?

Uh, the normal TDSE formula? The one that's introduced to an introductory class.

Unless you mean the TDSE solution, then the solution I'm concerned about would be Ψ(x,t) = ψ(x)φ(t).

 

[Leo1233783]

Ok. Also, why when writing the most general solution to the TDSE, are the constants Cn the coefficients obtained from the Fourier Transform? Or, why do they have to be the Fourier coefficients? In 2rd order differential equations, we saw that we could multiply the solution X by any constant C1, and the new solution C1X would also be a solution to the differential equation.

Any linear combination of the eigenstates is also a solution but not all after including the initial conditions and the time dependency. Here, it is a trick to use a Fourier transform to get the final solution to the TDSE.

For the original question, I don't understand why do you think it is the same computation. Could you detail the algorithm you use ?

 

[yosimba2000]

Sorry, I don't know what eigenstates are yet.Basically, we assume that Ψ = ψ(x)Φ(t) is a solution to TDSE, and we find a set of solutions for ψ(x) and Φ(t).

So I would conclude then that I could plug in ψ(x)Φ(t) as my solution to the system. Why do I need to do the Fourier Transform to find some constants if ψ(x)Φ(t) is already my answer?

 

[Leo1233783]

as said above, because it is only a general solution form ; you must now solve the TDSE and consider the initial conditions to get the solution to your peculiar problem.

Do you already know the time-independent Schrodinger Equation ? Il would be difficult ( and a little long ) to expose the generalities and their background. Teacher required for that ... Perhaps is it an exercise centered on something else ? What is the case ?

 

[yosimba2000]

I thought the initial conditions were used to solve for ψ(x) and Φ(t).

The exercise is the solution to an infinite square well.

 

[Chandra Prayaga]

Ah! So it looks like we are getting a little ahead of the game before learning the ropes. As Leo1233783 was pointing out, you need to know some of the basic nomenclature before we get into some of these questions. Is the exercise given to you in the form of a statement of the problem? Is the potential energy given as a function of x, or some independent variable?

 

[yosimba2000]

The potential energy V is infinity when the position X is greater than or equal to A, and when X is less than or equal to 0. The potential energy V(x) when the position is within 0 to a, bounds excluded, is 0.

So the wavefunction ψ(x) exists only when X is within 0 to a, bounds excluded, and 0 outside this range.

I need to find the solution of the wavefunction using the TISE equation to get ψ(x), then stick it with the corresponding Φ(t). So the solution to the TDSE is ψ(x)Φ(t), or it should be.

Then all of a sudden Griffith's says the real solution is actually Cψ(x)Φ(t), where C can be found using "Fourier's Trick". I understand how the trick works, but I'm not getting how or why we introduced C and now calling it part of the solution to the infinite square well.

 

[Leo1233783]

I don't understand how you could understand but perhaps it is another version of "compute and shut up" :)

ok
you have the solution here : http://jila.colorado.edu/chem4541/Time%20Dep%20Sch%20eqn.pdf

Come back to us for any details.

 

[Nugatory]

Then all of a sudden Griffith's says the real solution is actually Cψ(x)Φ(t), where C can be found using "Fourier's Trick". I understand how the trick works, but I'm not getting how or why we introduced C and now calling it part of the solution to the infinite square well.

OK, let's back up for a moment. The problem we're trying to solve is: We have a particle. We know that it's in an infinite square well and we know its initial state at time 0. We want to know how it behaves from there, which means finding its wave function $\Psi(x,t)$.

When the potential is unchanging (as it is in this and many other interesting and important problems) the solutions of the TDSE are all and only those functions that can be written in the form$$\Psi(x,t)=\sum_{n=0}^\infty{c_n}\psi_n(x)\phi_n(t)$$where the $\psi_n$ are the eigenfunctions we found by solving the TISE and $\phi_n(t)=e^{-iE_nt/h}$ where the $E_n$ are the corresponding eigenvalues: $H\psi_n=E_n\psi_n$.

So the general recipe for solving this sort of problem is:
1) Solve the TISE to find the $\psi_n$ and $E_n$; these are the building blocks that go into the summation above. This will require using the boundary conditions on $V$.
2) Find the values of the $c_n$ which will go into the summation above. We know the initial state of the particle; it's $\Psi(x,0)$ so we use the Fourier trick on that function to find them.

And now we have $\Psi(x,t)$, which is what we're looking for: the state of the particle at any time in the future.

 

[Nugatory]

Sorry, I don't know what eigenstates are yet.

In that case you need the equivalent of another semester of differential equations before you can take on Griffiths or any other intro QM text. The overview in my previous post is just an outline of how the various techniques for solving differential equations fit together here; but you have to learn the techniques themselves.