Recursion Formula

Viona

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Nov 22, 2018
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Hello,

I was reading the solution of Schrödinger equation for one dimensional harmonic oscillator. Please see the picture below. Every thing was going fine till equation (0.24) I could not understand how equation (0.23) can be written as equation (0.24). please help
Thanks

recursion formula.jpg
 
Hello,

I was reading the solution of Schrödinger equation for one dimensional harmonic oscillator. Please see the picture below. Every thing was going fine till equation (0.24) I could not understand how equation (0.23) can be written as equation (0.24). please help
Thanks

I think it's just a very rough approximation. If you suppose that in fact aj+2 = aj/(j/2) for all even j, and if a2 = C, then a4 = a2/1, a6 = a4/2 = a2/(1*2) = C/(1*2) = C/2, and so on. Can you see that this would be C/(j/2)! ? Well, not quite; this would be aj+2, not aj. But I think that's all they're doing.
 
I think it's just a very rough approximation. If you suppose that in fact aj+2 = aj/(j/2) for all even j, and if a2 = C, then a4 = a2/1, a6 = a4/2 = a2/(1*2) = C/(1*2) = C/2, and so on. Can you see that this would be C/(j/2)! ? Well, not quite; this would be aj+2, not aj. But I think that's all they're doing.

Thanks for you reply. But since aj+2 = aj/(j/2) this means that a0+2 = a2 =a0/(0/2) = infinity ? or Iam doing thin wrong?
 
I think it's just a very rough approximation. If you suppose that in fact aj+2 = aj/(j/2) for all even j, and if a2 = C, then a4 = a2/1, a6 = a4/2 = a2/(1*2) = C/(1*2) = C/2, and so on. Can you see that this would be C/(j/2)! ? Well, not quite; this would be aj+2, not aj. But I think that's all they're doing.


Thank you. But since aj+2 = aj/(j/2) this means that a0+2 = a2 =a0/(0/2) = infinity? or I am doing something wrong?
 
Thank you. But since aj+2 = aj/(j/2) this means that a0+2 = a2 =a0/(0/2) = infinity? or I am doing something wrong?
 
Thank you. But since aj+2 = aj/(j/2) this means that a0+2 = a2 =a0/(0/2) = infinity? or I am doing something wrong?

That's why I started with a2 rather than a0! I made no comment about it because that's not what you asked about ...

Since at this point they are talking about a limit (large j), the recursion doesn't have to work for small values like j=0. In fact, since they are only claiming it is approximately true for large j, it would have been better if I had not even started at 2, but at, say, 2 million. You could still come to the same conclusion.

All I can say is that they are not trying to be particularly careful. I'm guessing that these are physicists, not mathematicians, who would not be so sloppy. Or at least, this part doesn't read like a proof, just a hand-waving argument.
 
That's why I started with a2 rather than a0! I made no comment about it because that's not what you asked about ...

Since at this point they are talking about a limit (large j), the recursion doesn't have to work for small values like j=0. In fact, since they are only claiming it is approximately true for large j, it would have been better if I had not even started at 2, but at, say, 2 million. You could still come to the same conclusion.

All I can say is that they are not trying to be particularly careful. I'm guessing that these are physicists, not mathematicians, who would not be so sloppy. Or at least, this part doesn't read like a proof, just a hand-waving argument.

Now I understood. Thank you very much.
 
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