
#1
Dec1112, 08:23 AM

P: 991

In my book I need to prove the general theorem that I can always pick the eigenstates in the time independent Schrödinger equation to be either even or odd ψ(x). This provides that the potential is an even function and is not hard to show if you use that if ψ(x) is a solution then too is ψ(x) and you can write any solutions a superposition of an even and odd solutions. But my problem is that ψ are eigenstates. So if a general ψ which is neither even nor odd is found to be an eigenstate by solving the equation. How can you then say that the even and odd solution forming ψ are also eigenstates? For me that just doesn't make sense in terms of the discrete picture. Imagine you have an eigenvector a. You can scale it and pick a scaling of a as an eigenvector yes. But how would you possibly be able to write that eigenstate up as a linear combination of two other eigenstates?  the eigenstates are linearly independent for a hermitian operator!!!
Thus I think I am misunderstanding something somewhere. 



#2
Dec1112, 09:23 AM

P: 123

You claim that if ψ(x) is a solution then ψ(x) is also a solution but your problem is how even and odd solutions appear as solutions. Well, after the above claim it is pretty easy to see that. First consider that, since SE is linear, linear combinations of solutions are also a solution. Then try to figure out what kind of linear combinations of ψ(x) and ψ(x) you can make, in order to construct an even or an odd solution. For example, the function f(x) = ψ(x) + ψ(x) is an even function, whatever ψ(x) is, and since it is a linear combination of solutions, then it will be also a solution. With the same argument, try to find what kind of combination would make an odd solution.




#3
Dec1112, 11:12 AM

P: 991

hmm I may be wrong but I think we are talking about two different things. I say suppose f1 and f2 energy eigenstates so they solve the SE then so too does f3 = f1 + f2 but is it guaranteed to be an eigenstate?
Now suppose f3 is also a solution to the SE but is neither even nor odd. Since it solves it, it must be an eigenstate. But we also know in the discrete case that there are exactly enough eigenvectors of a hermitian operator (like the hamiltonian) to span the space. So shouldn't this nonodd/even solution be linear independent of the odd and even solutions..? I guess you can say my trouble comes with the words "that we can pick the eigenstates to he odd or even". For me that is nonsense because if you look at a hermitian matrix it has n eigenvectors and it is not like we can somehow be smart and choose linear combinations of these that are still eigenstates.. Like if A is a matrix with eigenvectors a and b then A(a+b) = λ1a + λ2b ≠ λ3 (a+b) 



#4
Dec1112, 11:20 AM

P: 123

Odd and even solutions to the SE
@aaaa202
"Since it solves it, it must be an eigenstate" Nope. Consider the ground and first excited state of the infinite square well. Their sum is not an eigenstate (to what energy should it belong?). The sum of two eigenfunctions to different eigenstates is not an eigenstate itself. You are probably making the mistake of looking only at teh timeimdependent SGL, but that's not o.k., if you add two different eigenvectors with different time dependence. Not every solution of the timedependent SGL is an energy eigenstate. You can find some animations of sums of eigenfunctions on my website (the site is in German, but you'll understand the pictures): http://scienceblogs.de/hierwohnend...lesimkasten/ 



#5
Dec1112, 11:24 AM

P: 991

ahh you are right. But still this doesn't ruin my point. Suppose you find an eigenfunction to the hamiltonian which is neither and even nor odd function. Then by the virtue of the mentioned theorem I should be able to write this eigenstate as a linear combination of odd and even eigenstates. But that can't be right since this nonodd/even eigenfunction is linearly independent of any other eigenstates. So how is it that you can always choose the eigenfunction to be either even or odd? Please show me how if you found a function which is neither of the two to be an eigenfunction to the SE :(




#6
Dec1112, 12:08 PM

P: 123

But can you find a function that is neither even nor odd?
And even if you can, doesn't this imply that you have a degenerate eigenvalue and can thus write it as a linear combination? Perhaps it would be helpful if you copied here exactly what is in your book. 



#7
Dec1112, 01:13 PM

Sci Advisor
P: 1,185

The point is that all eigenfunctions are either even or odd. If this were not true, the arguments already given show that the eigenvalue would have to be degenerate. But eigenvalues in one dimension are not degenerate. (This is not hard to prove.)




#8
Dec1112, 01:13 PM

P: 991

what does degenerate mean? sorry I'm not so strong in english terms :(




#9
Dec1112, 02:04 PM

P: 123

Degenerate means that you have two (or more) eigenfunctions with the same eigenvalue. Then every linear combination is also an eigenfunction.




#10
Dec1112, 02:37 PM

P: 991

The problem statement is: Show that ψ(x) can always be taken to be either even or odd. Maybe I am confused by the language: Does the term "can be taken to be" that it must be either even or odd? Because that is what you guys seem to say. Either way I attached the answer from my solutions to it. I don't understand how it shows that the eigenfunctions MUST be either even or odd  how was that again?




#11
Dec1112, 04:34 PM

Sci Advisor
P: 1,185

You have shown that, if the eigenvalue is nondegenerate (that is, there is only one eigenfunction for that eigenvalue), then the eigenfunction must be even or odd.
To show that it must be even or odd, you need to show that the eigenfunction is unique. For a potential that rises to infinity as x goes to plus or minus infinity, this follows from general properties of the differential equation. 



#12
Dec1112, 04:36 PM

P: 991

uhh but thats what I am not understanding. Where have I shown that it must be even or odd for the eigenvalue to be nondegenerate?




#13
Dec1112, 04:59 PM

Mentor
P: 10,799

If ψ(x) is an eigenfunction and neither even nor odd, then ψ(x) is a different eigenfunction with the same eigenvalue, and you have degenerate eigenvalues.
In general: If ψ(x) is an eigenfunction and neither even nor odd, ##\phi_+ = \frac{1}{\sqrt{2}}\left(\psi(x) + \psi(x)\right)## and ##\phi_ = \frac{1}{\sqrt{2}}\left(\psi(x)  \psi(x)\right)## are eigenfunctions with the same eigenvalue and even/odd, respectively. Your state ψ(x) can be written as ##\psi(x) = \frac{1}{\sqrt{2}}\left(\phi_+(x) + \phi_(x)\right)## 



#14
Dec1112, 05:14 PM

Sci Advisor
P: 1,185

And if it's nondegenerate, then either ##\phi_+(x)## or ##\phi_(x)## must vanish. (Because if one of them didn't vanish, then there would be two eigenfunctions for the same eigenvalue, and the eigenvalue would be degenerate.)




#15
Dec1112, 05:16 PM

P: 991

thanks. That clarified things alot. I suppose that the general statement you mention doesn't apply to the SE since a hermitian operator does not have degenerate eigenvalues.
But since your first line basically explains why the eigenstates are even and odd, why is it you have to go through the proof of how any solution can be written as a superposition of odd and even eigenstates? What does that bit show me? 


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