Prove the nth energy eigenfunction has n-1 zeros

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How do we show that ##\psi_n(x)## has ##(n-1)## zeros for all ##n\in Z^+##?

Assuming ##\psi_k(x)## has ##(k-1)## zeros for some ##k\in Z^+##, by oscillation theorem, we can only get ##\psi_{k+1}(x)## has ##\geq k## zeros.

Also, how do we show that ##\psi_1(x)##, the eigenfunction corresponding to ##E_1##, has no zero? Oscillation theorem may permit ##\psi_1(x)## to have 3 zeros, ##\psi_2(x)## to have 5 zeros, ##\psi_3(x)## to have 20 zeros, etc.
 
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I've never heard of the oscillation theorem. Could you post the text of problem 3.9?

Edit: I just saw your other thread, where the problem is stated.
 
@DrClaude, it's Quantum Mechanics by B H Bransden and C J Joachain, 2nd edition, page 119.
 
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stevendaryl said:
Here is a paper that I found giving an argument for the nth eigenfunction having n-1 nodes:
http://arxiv.org/pdf/quant-ph/0702260.pdf

His argument is a little hard to follow, but it is short.

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Does he mean "As we separate the walls, N number of the wave functions (##\psi_1, \psi_2, ..., \psi_N##) go to zero at ##x=\infty## and at ##x=-\infty##, and the positive energy states (##\psi_{N+1}## onwards) become the continuum spectrum"?
 
Happiness said:
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Does he mean "As we separate the walls, N number of the wave functions (##\psi_1, \psi_2, ..., \psi_N##) go to zero at ##x=\infty## and at ##x=-\infty##, and the positive energy states (##\psi_{N+1}## onwards) become the continuum spectrum"?

Now I guess he mean "As we separate the walls, the wave functions for n>N (##\psi_{N+1}## onwards) become the everywhere-zero function" since all these wave functions have a finite number of zeros but the potential can only support an ##N## number of bound states (which have a finite number of zeros) and the other admissible wave functions are positive-energy states and thus have infinite number of zeros.

So I guess he should have said "all of these wave functions for n>N ..." instead of "some of the wave functions ..." .