Time dependent Schrödinger equation - Separable Solutions

[laser1]

TL;DR

N/A

From Griffiths 3rd edition quantum pg 45

The bottom two lines. Why can the general solution be expressed as that sum of separable solutions? Griffith doesn't seem to explain this step. What am I missing? Thank you.

 

[hutchphd]

Let $L$ be a linear operator and f and g be any solutions to the equation $$Lf=0$$ Then $$Lf+Lg=L(f+g)=0$$ and the Shrodinger equation is of this form. QED

 

[laser1]

Let $L$ be a linear operator and f and g be any solutions to the equation $$Lf=0$$ Then $$Lf+Lg=L(f+g)=0$$ and the Shrodinger equation is of this form. QED

Are you saying that if f is a solution, and g is a solution, then f+g is a solution? I don't have a problem with that. I am just confused why it is the general solution.

 

[Nugatory]

Why can the general solution be expressed as that sum of separable solutions?

Any reasonably well-behaved function can be written as a sum of complex exponentials. Here we’re limiting ourselves to summing discrete exponential terms instead of (as with a Fourier transform) integrating across continuous exponential terms; we handwave our way past that objection by saying that the sum does include all the exponential terms that could be part of any solution to Schrödinger’s equation, excludes only those that cannot be - the $\psi_n$ form a complete basis for the space of solutions to the time-independent Schrödinger equation.

 

[hutchphd]

Suppose h is a general solution. (What does that mean? I don't actually know Lets use the term arbitrary solution) Then, given any other solution f f define g to be h-f. Voila. I guess this assumes there is more than one solution
As @Nugatory alludes, the eigenfunctions of a Sturm Liouville operator can be chosen to be a complete orthoganal set (ICBS if I remember my maths).

 

[Yuras]

I'm not particularly good in mathematical foundations of quantum mechanics, but I think the answer is the spectral theorem. Any solution has spectral decomposition, which is (I think) exactly what Griffith is talking about.

 

[laser1]

Suppose h is a general solution. (What does that mean? I don't actually know Lets use the term arbitrary solution) Then, given any other solution f f define g to be h-f. Voila.

Hmm, I think I follow, but now I am not so sure. If g = h-f, how do we know that f and g are separable solution?

 

[Nugatory]

Hmm, I think I follow, but now I am not so sure. If g = h-f, how do we know that f and g are separable solution?

It’s the other way around. We’ve found that $h$ and $f$ are solutions, therefore by linearity $g=h-f$ will be. This works for any linear combination of the $\psi_n$, and we also know by the argument in post #4, #5, #6 that any general solution can be written as a linear sum of the $\psi_n$.

 

[laser1]

by the argument in post #4, #5, #6 that any general solution can be written as a linear sum of the ψn.

Okay, it seems like there is no solution with my current maths knowledge (basically no linear algebra). I will investigate this!

 

[laser1]

Funnily enough, when studying e-mag just there, I came across this:

which is pretty much my question. Why can we express the general solution as a linear combination of separable solutions?! Time to see this "spectral theorem"... although the wikipedia page looks terrifying!

 

[Nugatory]

Why can we express the general solution as a linear combination of separable solutions?

Have you seen a Fourier transform in classical physics? If you know how that works, the idea that that any solution can be written as a linear combination of the $\psi_n$ will be more intuitive.

Also, although I don't have a copy of Griffith in front of me, I'd expect that somewhere in the next few sections or in the exercises you'll learn how to find the the $c_n$ so will realize the promise of "can be written in this form - it is simply a matter of finding the right constants". Working through that will go a long ways towards convincing you that you really can write any solution as such a sum.

(And if you are not clear on what is meant by "linear combination", learn that. It's not complicated, you've been doing it all all along without noticing, sort of like the guy who had no idea what the word "prose" meant, so was amazed to be told that he had been speaking in prose all his life).

 

[laser1]

Have you seen a Fourier transform in classical physics? If you know how that works, the idea that that any solution can be written as a linear combination of the $\psi_n$ will be more intuitive.

Also, although I don't have a copy of Griffith in front of me, I'd expect that somewhere in the next few sections or in the exercises you'll learn how to find the the $c_n$ so will realize the promise of "can be written in this form - it is simply a matter of finding the right constants". Working through that will go a long ways towards convincing you that you really can write any solution as such a sum.

(And if you are not clear on what is meant by "linear combination", learn that. It's not complicated, you've been doing it all all along without noticing, sort of like the guy who had no idea what the word "prose" meant, so was amazed to be told that he had been speaking in prose all his life).

Fourier series yes, fourier transforms no.

Yes true, I have not found any counterexamples (obviously) but... still.

I am aware of linear combinations. I accept that linear combinations of separable solutions are solutions. But extending this to a general solution just seems wrong.

I had a look at the spectral theorem. I understand what it says now. Basically, all functions can be written as linear combinations of the eigenfunctions (the separable solutions are eigenfunctions because they equal a constant). In the quantum case, the function is the wavefunction, and the eigenfunctions are the separable solutions.

I don't understand the proof (https://math.mit.edu/~dav/spectral.pdf). I think it is a bit beyond me for now. Moving on, for now, I will just accept the spectral theorem as fact, and keep going! Thanks!

 

[Yuras]

Moving on, for now, I will just accept the spectral theorem as fact, and keep going!

Spectral theorem is complicated because it deals with infinite dimensional Hilbert space. If you are looking for intuition (i.e. not particularly interested in mathematical rigor), then I'd suggest to think about the finite dimensional case. If Hilbert space has dimension $N$, then Hamiltonian on the system can be diagonalized, and (in the basis where it's diagonal) vectors of the form $(1,0,0,...)$, $(0,1,0,...)$, $(0,0,1,...)$ etc are eigenvectors of the Hamiltonian, which means they are orthogonal (or can be chosen to be orthogonal). There are exactly $N$ such vectors, so they span the whole Hilbert space, i.e. any state, including any solution of Schrödinger equation, can be represented as their linear combination.
ADDED: I guess the above is worded quite awkwardly. Better check any linear algebra book for why eigenvectors of a Hermitian matrix form a basis.