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Power series method help

I would like to get only the first order terms in h-bar, but I am not sure how to go about this...
I used j=0 for the first term, but since j can only be an integer, (based on the physical situation) I'm not sure what to do about the second term. Help!
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Original post by IDontKnowReally
I would like to get only the first order terms in h-bar, but I am not sure how to go about this...
I used j=0 for the first term, but since j can only be an integer, (based on the physical situation) I'm not sure what to do about the second term. Help!

When you say 'first order terms', which variable's order are you referring to? Is it h?
Yes, h sorry, not h-bar!
Original post by IDontKnowReally
Yes, h sorry, not h-bar!

Ok so at leading order your equation reduces to

−iℏ2mS0′′(x)+12m[S0′(x)]2+V(x)+E=0-\dfrac{i \hslash}{2m} S_0''(x) + \dfrac{1}{2m} [S_0'(x)]^2 + V(x) + E = 0

At order hh we have;

−iℏ2mS1′′(x)+1m[S0′(x)S1′(x)]=0-\dfrac{i \hslash}{2m} S_1''(x) + \dfrac{1}{m} [S_0'(x)S_1'(x)] = 0
Original post by RDKGames
Ok so at leading order your equation reduces to

−iℏ2mS0′′(x)+12m[S0′(x)]2+V(x)+E=0-\dfrac{i \hslash}{2m} S_0''(x) + \dfrac{1}{2m} [S_0'(x)]^2 + V(x) + E = 0

At order hh we have;

−iℏ2mS1′′(x)+1m[S0′(x)S1′(x)]=0-\dfrac{i \hslash}{2m} S_1''(x) + \dfrac{1}{m} [S_0'(x)S_1'(x)] = 0

Yes, I got the same for the zero-th order, but for the first order, how do you get this term:
1m[S0′(x)S1′(x)]=0 \dfrac{1}{m} [S_0'(x)S_1'(x)] = 0
Original post by IDontKnowReally
Yes, I got the same for the zero-th order, but for the first order, how do you get this term:
1m[S0′(x)S1′(x)]=0 \dfrac{1}{m} [S_0'(x)S_1'(x)] = 0

It's because of the squaring.

You are multiplying two brackets;

(S0′+hS1′+h2S2′+…)(S0′+hS1′+h2S2′+…)(S_0' + h S_1' + h^2 S_2' + \ldots)(S_0' + h S_1' + h^2 S_2' + \ldots)

The overall coefficient of h term will come in when we either:

-- Take hS1′hS_1' from first bracket and multiply it by S0′S_0' in second bracket.
-- Take S0S_0 from first bracket and multiply it by hS1′hS_1' in second bracket.

Thus, this squared sum at order h is simply hS1′⋅S0′+S0′⋅hS1′hS_1' \cdot S_0' + S_0' \cdot hS_1' i.e. 2hS0′S1′2hS_0'S_1'
Original post by RDKGames
It's because of the squaring.

You are multiplying two brackets;

(S0′+hS1′+h2S2′+…)(S0′+hS1′+h2S2′+…)(S_0' + h S_1' + h^2 S_2' + \ldots)(S_0' + h S_1' + h^2 S_2' + \ldots)

The overall coefficient of h term will come in when we either:

-- Take hS1′hS_1' from first bracket and multiply it by S0′S_0' in second bracket.
-- Take S0S_0 from first bracket and multiply it by hS1′hS_1' in second bracket.

Thus, this squared sum at order h is simply hS1′⋅S0′+S0′⋅hS1′hS_1' \cdot S_0' + S_0' \cdot hS_1' i.e. 2hS0′S1′2hS_0'S_1'


Ohhh, that makes so much more sense thanks! I kept thinking of the second sum as being h2jh^{2j} which confused me.
Original post by IDontKnowReally
Ohhh, that makes so much more sense thanks! I kept thinking of the second sum as being h2jh^{2j} which confused me.

Oh no, rookie error!

(∑jaj)2≠(∑jaj2)\displaystyle \bigg( \sum_{j} a_j \bigg)^2 \neq \bigg( \sum_{j} a_j ^2\bigg)

otherwise we would say (x+y)2=x2+y2(x+y)^2 = x^2 + y^2 !

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