Superposition of the solutions of the Shrodinger equation

[smstry]

We all know that any superposition of two solutions of the shrodinger equation is still a solution .
but what about the case when Psi1 is a solution related to a potential V1 and Psi2 is a solution related to a potential V2 .?
how these solutions are related to the solution of the equation with V1+V2 ?

 

[Demystifier]

They are not related.

 

[SpectraCat]

We all know that any superposition of two solutions of the shrodinger equation is still a solution .
but what about the case when Psi1 is a solution related to a potential V1 and Psi2 is a solution related to a potential V2 .?
how these solutions are related to the solution of the equation with V1+V2 ?

Hmmm ... that first sentence looks wrong, or at least imprecise ... a linear combination of degenerate eigenstates is also an eigenstate of the Hamiltonian. For non degenerate states, $$\left|\psi_{1}\right\rangle$$ and $$\left|\psi_{2}\right\rangle$$, with energy eigenvalues $$(\epsilon_{1})$$ and $$(\epsilon_{2})$$, we have:

$$\hat{H}\left(\left|\psi_{1}\right\rangle + \left|\psi_{2}\right\rangle\right) = \epsilon_{1}\left|\psi_{1}\right\rangle + \epsilon_{2}\left|\psi_{2}\right\rangle$$

but that is not an eigenfunction-eigenvalue relation unless $$(\epsilon_{1}=\epsilon_{2})$$, i.e. the degenerate case

With regards to your question, the solutions will be "related" if and only if the potentials are separated in space such that they do not interact. Otherwise, you need to solve the more complicated problem for the particle in the full V1 + V2 potential.

 

[conway]

Something doesn't have to be an eigenfunction to be a solution.

 

[SpectraCat]

Something doesn't have to be an eigenfunction to be a solution.

How does that square with the following statement?

"Solving the Schrodinger equation is equivalent to diagonalizing the Hamiltonian matrix."

Seems like that says a solution to the Schrodinger equation has to be an eigenfunction of the Hamiltonian. Is there some semantic point I am missing here about what "solution" means?

 

[conway]

Whenever you "solve" a differential equation, the "solutions" are a set of eigenfunctions. You are always allowed to take linear combinations of these eigenfunctions to create new functions which are also solutions of the original differential equation.

That's why you can take the sinusoidal eigenfunctions of a guitar string and combine them to show what a plucked string with a triangular profile does.

 

[SpectraCat]

Whenever you "solve" a differential equation, the "solutions" are a set of eigenfunctions. You are always allowed to take linear combinations of these eigenfunctions to create new functions which are also solutions of the original differential equation.

That's why you can take the sinusoidal eigenfunctions of a guitar string and combine them to show what a plucked string with a triangular profile does.

But then the OP's first statement is really trivial, because any function can be written as a linear combination in a complete basis.

I also wonder if you are correct about the linear combinations being solutions of the original differential equation, at least in the general case ...

Take the following example:

$$\frac{d^{2}f}{dx^{2}} + cf(x) = 0$$

two solutions of this are:

$$f_{1}=cos(ax)$$, which solves it for $$c=a^{2}$$, and

$$f_{2}=sin(bx)$$, which solves it for $$c=b^{2}$$

However, if we construct the arbitrary linear combination $$g=\alpha f_{1} + \beta f_{2}$$, as far as I can tell, there are no choices of non-zero values for both $$\alpha$$, $$\beta$$ and c for which g is a solution of the original diff. eq., except for in the degenerate case where a=b, in which case any values of $$\alpha$$ and $$\beta$$ will work.

More generally, the whole point of finding the eigenfunctions is that they are a complete set of linearly independent functions. That statement by itself precludes the possibility of finding other solutions that can be written as linear combinations of the eigenfunctions, since then it would not be a complete set, and any such solution would be linearly dependent.

 

[conway]

I'm pretty sure what I've said is correct. If you want I can try to explain it better. It's true as you said that any function can be expressed as a linear combination of the eigenfunctions. It's the time evolution that makes them different and unique.

 

[SpectraCat]

Ok, I guess I see where my confusion arose, since it is clear that you are considering the time-dependent case. I was talking about the stationary states only (as I guess you know).

So, I realize that (of course) the OP's original statement that "any linear combination of two solutions to the Schrodinger equation is also a solution" is correct for the time-dependent case. I am so used to working with the TISE that that is what I naturally think of when I see "Schrodinger equation".

EDIT: I was composed this before I saw your reply ... I think we are in accord now. Thanks for being patient with me

EDIT2: I fixed my erroneous statements from last night (it was late) ...

 

[sweet springs]

Hi, smstry. I am afraid I cannot give you anything new, but..

We all know that any superposition of two solutions of the shrodinger equation is still a solution .
but what about the case when Psi1 is a solution related to a potential V1 and Psi2 is a solution related to a potential V2 .?
how these solutions are related to the solution of the equation with V1+V2 ?

Let F ψ=0 be a homogeneous linear differential equation.
F ψ1=0 and Fψ2=0 then F (aψ1+bψ2)=0. Superposition of solutions is also a solution.

F=H-E gives time-independent shrodinger equation. Eigenvalue E is constant here. According to different values of E, F1=H-E1 and F2=H-E2 are two different operators. F1 ψ1=0 and F2 ψ2=0 but F 1(aψ1+bψ2) or F 2(aψ1+bψ2) does not have to be zero.
Superposition of the solutions should not be a solution anymore.

Moreover in your case F1=H1-E1 and F2=H2-E2 where not only eigenvalue but Hamiltonian are different.

Regards.

PS
Thanks for the dialiogue of conway and Spectracat, F=H-i hbar d/dt gives time dependent Shrodinger equation.
F ψ1=0 and Fψ2=0 then F (aψ1+bψ2)=0. Superposition of solutions is also a solution.
For different hamiltonian F1=H1-i hbar d/dt and F2=H2-i hbar d/dt. F1 ψ1=0 and F2 ψ2=0 but F 1(aψ1+bψ2) or F 2(aψ1+bψ2) does not have to be zero. Superposition of the solutions should not be a solution anymore.

 

[SpectraCat]

Hi, smstry. I am afraid I cannot give you anything new, but..
Let F ψ=0 be a homogeneous linear differential equation.
F ψ1=0 and Fψ2=0 then F (aψ1+bψ2)=0. Superposition of solutions is also a solution.

Right, this corresponds to the case of degenerate states (same eigenvalue).

F=H-E gives time-independent shrodinger equation. Eigenvalue E is constant here. According to different values of E, F1=H-E1 and F2=H-E2 are two different operators. F1 ψ1=0 and F2 ψ2=0 but F 1(aψ1+bψ2) or F 2(aψ1+bψ2) does not have to be zero.
Superposition of the solutions should not be a solution anymore.

That has been my point all along

Moreover in your case F1=H1-E1 and F2=H2-E2 where not only eigenvalue but Hamiltonian are different.

That was what deMystifier was getting at .. the solutions are not related in any general way

Thanks for the dialiogue of conway and Spectracat, F=H-i hbar d/dt gives time dependent Shrodinger equation.
F ψ1=0 and Fψ2=0 then F (aψ1+bψ2)=0. Superposition of solutions is also a solution.
For different hamiltonian F1=H1-i hbar d/dt and F2=H2-i hbar d/dt. F1 ψ1=0 and F2 ψ2=0 but F 1(aψ1+bψ2) or F 2(aψ1+bψ2) does not have to be zero. Superposition of the solutions should not be a solution anymore.

Right ... all of my comments were only referring to the case where the two solutions are generated with the same Hamiltonian.

 

[peteratcam]

This thread is a lesson in why the words "Schrodinger equation" should always be prefixed by "time dependent" or "time independent".

 

[SpectraCat]

This thread is a lesson in why the words "Schrodinger equation" should always be prefixed by "time dependent" or "time independent".

Heh ... lesson learned ... next time I'll ask for clarification before posting a detailed response