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Leo32

Complete set: eigenfunctions of Schrödinger equation

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While trying to teach myself a little quantum mechanics in the evenings, I came across the following sequence of formula's and text. Although the subject is quantum mechanics, the question I have is a mathematical one.

 

[math]\int_{-\infty}^{+\infty}dx\psi_m(x)\psi_n(x)=\delta_{mn}[/math]

 

these functions form a complete set

 

[math]\sum_{n=1}^{\infty}\psi_n(x)-\psi_n(x')=\delta(x-x')[/math]

 

 

The question I have is: are both statements required for the set of [math]\psi_n(x)[/math] to be complete ? Or which of both is sufficient ?

 

Greetz,

 

Leo

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Funnily enough, I seem to be doing a similar sort of thing (in regards to solving the Schrodinger equation using boundary problems, simple PDE stuff) although what I have doesn't immediately look like that. I'll have a look at it later, just browing atm :)

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unfortunately i am going to ditch my module on quantum mechanics for a Stats modules. altough quantum mechanics is interesting, i doubt it will come handy in my workplace.

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unfortunately i am going to ditch my module on quantum mechanics for a Stats modules. altough quantum mechanics is interesting, i doubt it will come handy in my workplace.

 

=> what about scientific curiosity ?

(Isnt "usefullness" a rather subjective notion and moreover how do you determine if something will or will not be usefull in the future ?)

 

Just a general remark, => no offense meant

 

Mandrake

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The first relation merely states they are orthogonal. It is not necessary nor for a complete set to be orthogonal, nor is it necessarily true that an orthogonal set is complete. In particular that relation remains true if I allow only one eigenfunction in the set.

 

I don't understand what the second equation states, really, cos I'm not a quantum physicist.

 

 

Whether the eigenfunctions are complete is a property of the linear operator of which they are eigenfunctions. (the eigenfunctions of any hermitian operator are orthogonal and real, though not necessarily complete)

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=> what about scientific curiosity ?

(Isnt "usefullness" a rather subjective notion and moreover how do you determine if something will or will not be usefull in the future ?)

 

Just a general remark' date=' => no offense meant

 

Mandrake[/quote']

well, i am definitely not gonna go into reserach (me thinks) . Most likely I will join the Financial sector, i.e Actuary, Economist. etc... So Stats would be more helpful than Quantum Mechanics.

 

As for Scientific curiosity, I could always do quantum mechanics in my spare time....

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Back at home, I see that the second formula indicates that the eigenfunctinos form a complete set... Anybody any idea ?

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You could try proving that if S is a complete set of eigenfunctions for some function space that the sum over all elements of S is a delta fucntion. I don't think that that is a necessary condition, but I've not proved it. It also depends on how liberal you are with convergence issues, obviously, but thinking about L^2(T^1) and fourier expansions I don't think that you can meaningfully even do that sum.

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Sweet.

 

Well, this is an interesting question...

The first equation is just what was said in post #5. All it says is that your set of Psi functions are orthoganol. It is called a delta function and what it means is this;

 

when

n=m the answer for the integral is 1

when

n != m the solution to the integral is 0

 

cool

 

the second equation actually says something similare to the first. It is known as the dirac delta function and it is used to say basically the same thing, but for the discrete case (integrals are used in the continous case).

 

It is a bit complicated to understand, but the idea is that if you were to try to represent the funcitons used in the summation by a curve the the area under the curve would be 1 if the functions were identical (x=x'), but 0 if they were not (x != x').

 

This is not entirely correct, but i think the dirac delta function is sort of hard to understand and explain (at least for me!). I am not quite sure what the equation means, physically, since it is hard to know what it is referring to when taken out of context :( Though i wish i could help.

 

perhaps if you let us know what you were reading about in the first place with regards to q.m. then someone would have a better idea?

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The 1st equation simply sais that the states [math] \{ \psi_n \} [/math] form an orthonormal base (ignoring that you forgot the complex cunjugation on the 1st factor). This, however, does not say it´s a base for the whole vector space so it does not show completeness.

 

The 2nd equation is flawed, at least because for x' = x: [math] \sum_n \psi_n(x) - \psi_n(x) = \sum_n 0 = 0 \neq \delta(0) [/math].

I´ll try to figure out the correct expression because I´m also interested in that but it´ll take a minute.

 

----------------------------------------------

 

Hopefully final EDIT:

 

The 2nd equation is (staying within your notation) supposed to be:

[math] \sum_n \psi_n(x) \psi^*_n(x') = \delta(x-x') [/math] (*)

 

The base [math] \{ \psi_n(x) \} [/math] is complete if any state [math] \phi(x) [/math] can be written as: [math] \phi = \sum_n c_n \psi_n(x) [/math]. Or in words: You must be able to write any vector as a linear combination of base-vectors.

 

Proof that a set of states satisfying (*) is a complete base:

[math] \phi(x) = \int dx' \delta(x-x') \phi(x') [/math]

[math] = \int dx' \left(\sum_n \psi_n(x) \psi^*_n(x') \right) \phi(x') [/math]

[math] = \sum_n \psi_n(x) \int dx' \psi_n^*(x') \phi(x') [/math]

[math] = \sum_n <\psi_n | \phi> \psi_n(x) [/math]

[math] = \sum_n c_n \psi_n(x) [/math]

 

Remarks:

a) To sum it up: The first equation sais that the base is orthonormal. The second one sais the the base is complete. The first equation is certainly a hidden condition in above proof but I´m too lazy to double-check it to find out where.

b) Please take more time to check your equations next time. Both had more or less serious flaws in it and it can take others (me in this case) quite some time to find them. Just because you saved two minutes on double-checking your post.

c) One of the more interesting questions, I´ve read here.

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Vending Menace, Atheist,

 

Thank you for the in depth replies. Atheist, reason why I didn't include the complex conjugation was that the text I'm trying to read at this point didn't introduce it yet...

 

I promise not to ask any questions anymore trying to remember the formula's I read before going to sleep the night before LOL. The minus should have been a multiplication...

 

Boy that Dirack Delta function is something strange and new! Just to make sure that I kind of get it, is following expression correct ?

 

[math]\int_{-\infty}^{+\infty}dx\delta(x) = 1[/math]

 

Greetz,

Leo

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Yes. The dirac-delta is also not a function in a mathematical sense. Think of it as a ... aehrm ... method (wanted to write "function" but that would have sounded curious in this context) that gives you the value of a function at a certain variable in the sense of

[math] \int_{-\infty}^{+\infty} dx f(x) \delta(x-x') = f(x') [/math]

 

In your example above that would be f(x)=1 and x'=0.

 

From a technical standpoint it´s more or less the same as the normal delta which is, for example, used to extract certain terms from sums. Just that the sums are integrals here (which are also quite similar to sums).

 

The missing complex conjugation was the minor problem as it´s obvious. I don´t know if it´s really worth remembering these conditions as for most problems you usually simply assume you have a complete base. At least I didn´t know the exact conditions (and the way they´re usually written down) out of my head - would have had less trouble with your question if I had, though :rolleyes:

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