Expanding Powers of 1/R - Modern Quantum Mechanics by JJ Sakurai

[Master J]

Expansion in Powers...

I wish to expand the following in powers of 1/R :


1/r1 + 1/ |R+r2-r1| - 1/ |R+r2| - 1/ |R-r1|


Each r is a vector with x,y,z components indexed as x1,y1 etc. R lies on the z axis so x and y components.

In Modern Quantum Mechanics by JJ Sakurai, equation 5.3.40 is the required result, yet I cannot obtain it. Is it simply Taylor expansion? Id really appreciate if someone could perhaps go thru this with me, as I encounter expansions like this all the time and never really know where to start.

The result is (1/R^3) . (x1x2 +y1y2 - z1z2)

Cheers

 

[Goongyae]

You want a Taylor series in terms of 1/R. Note that this Taylor series will blow up if R = r1 or R=-r2 or R=r1-r2, which all generate divide by zeroes in the original expression. The series works best for huge |R| - i.e. R is far away from r1, and -r2 - which leads to a quickly converging series.

And I think you meant 1/|R| for the first term in your expression. That way, for huge |R|, the signs will end up cancelling any 1/R's left over in the result. Otherwise there is no possibility of 1/R^3 being the first nonzero Taylor series term!

f(R) = 1/|R| + 1/ |R+r2-r1| - 1/ |R+r2| - 1/ |R-r1|

I'll work on the problem assuming R > 0 since this is difficult to write :(

Each term can be expanded out in terms of R similar to how you can expand (1+x)^p = 1 + px + p(p-1)x^2/2 +...:

1/|R| = 1/R
1/|R+r2-r1| = 1/sqrt(R*R + (r2-r1)*(r2-r1) + 2R*(r2z-r1z))
=1/R * (1+(r2-r1)*(r2-r1)/R^2+2(r2z-r1z)/R)^(-1/2)=1/R - ((r2-r1)*(r2-r1)/R^2+2(r2z-r1z)/R)/(2R) + ...
1/|R+r2| = 1/R - (r2*r2/R^2+2r2z/R)/(2R) + ...
1/ |R-r1| = 1/R - (r1*r1/R^2-2r1z/R)/(2R) + ...

combining, you get
f(R) = 1/R + 1/R - ((r2-r1)*(r2-r1)/R^2+2(r2z-r1z)/R)/(2R) + ...
- 1/R + (r2*r2/R^2+2r2z/R)/(2R) + ...
- 1/R + (r1*r1/R^2-2r1z/R)/(2R) + ...

the 1/R's cancel out to get
f(R) = 1/(2R) [- ((r2-r1)*(r2-r1)/R^2+2(r2z-r1z)/R) + ...
(r2*r2/R^2+2r2z/R) + ...
+ (r1*r1/R^2-2r1z/R) + ...]

The effectively 1/R^2 terms cancel as well leaving 1/R^3 terms:
f(R) = 1/(2R) [- ((r2-r1)*(r2-r1)/R^2) + r2*r2/R^2 + (r1*r1/R^2 + ...]
= 1/(2R^3) [- (r2-r1)*(r2-r1) + r2*r2 + r1*r1+ ...]
= 1/(2R^3) [2r2r1]
=(r2 dot r1)/R^3

Um.. this isn't quite the result you expected but it's very close. I think if you expand the (1+x)^(-1/2) power series above one further term you'll get an additional 1/R^3 contribution and the right result.

Hope this helps.

Edit: BTW the case with R < 0 is easy to handle. If R < 0, then you can perform the transformation R->-R, r1->-r2, r2->-r1 to get to the above problem with R > 0, use the above solution, and then transform back (which leaves the above solution unchanged). Thus you can see the same solution solves the R < 0 case.

 

[Master J]

I cannot thank you enough for expanding that out for me, it was hard work!


Yes, its 1/R, that was a typo. Thanks again, I am extremely greatful!


Now, I understand expansions, but I don't get the "in powers of R" part...what does this entail exactly?

 

[Master J]

I guess I'm stuck at the first part...


Rite, so for example, 1/ | R-r1| is equal to ( R - r1 )^(-1/2), ie. just the length of the vector.

So then I have to expand this out in terms of R right? How do I do this? I can use Taylor series for expanding out differentiable functions around a point, but I just don't see how one applies it here?

 

[Goongyae]

>"in powers of R" part...

I'm not sure what you are asking exactly.

If you use a one-dimensional equivalent to your original problem, it of course becomes much easier:
f(R) = 1/R + 1/ (R+r2-r1) - 1/(R+r2) - 1/(R-r1)

You can use the geometric series to do this expansion. The idea is that (1+x+x^2+x^3+...) * (1-x) = 1. Try mutliplying these yourself and you'll see all the terms drop out except for 1. Thus 1/(1-x) = 1 + x + x^2 + x^3 + ...
And negating x yields 1/(1+x) = 1 - x + x^2 - x^3 + ...

Then expanding each of the four terms, we have
f(R) = 1/R * (1 + 1/ (1+(r2-r1)/R) - 1/(1+r2/R) - 1/(1-r1/R))
= 1/R * [
1 +
1 - (r2-r1)/R + (r2-r1)^2/R^2 - (r2-r1)^3/R^3 + ...
-1 + r2/R + r2^2/R^2 - r2^3/R^3 + ...
-1 - r1/R - r1^2/R^2 - r1^3/R^3 -...
]

and summing all these terms you can see how the expression expands as powers of 1/R: the powers R^-1, R^-2, etc.

Since you are using the vector magnitude (|| symbol), which involves a square root, you can't use the easy technique above based on 1/(1-x). You instead must contend with (1-x)^(-1/2).

Fortunately, there is a Taylor series expansion for (1+x)^p where p is any number. It is 1 + px + p(p-1)x^2/2! + p(p-1)(p-2)x^3/3! + ... That's what I used above, although I only computed it for a couple of terms and not long enough to get the full 1/R^3 result.
You can see that my expansion for (1+x)^p is correct because if you differentiate it, it has the same result as multiplying by p/(1+x).

 

[Goongyae]

>( R - r1 )^(-1/2), ie. just the length of the vector.
>So then I have to expand this out in terms of R right?

You're missing a square there.
It's really ((R - r1)^2) ^ (-1/2) = (RxRx+RyRy+RzRz + r1xr1x + r1yr1y+r1zr1z - 2Rxr1x - 2Ryr1y - 2Rzr1z)^(-1/2)
Since we have Rx=Ry=0 and we're just calling R=Rz, it is
(RR + r1xr1x + r1yr1y-r1zr1z - 2Rzr1z)^(-1/2)
Assume R > 0, then it's
=(1/R) * (1 + r1xr1x/R^2 + r1yr1y-r1zr1z/R^2 - 2zr1z/R)^(-1/2)
We now have something in the form of (1+x)^p = 1 +px +p(p-1)x^2/2 + ...
Unfortunately the x in the case contains powers of 1/R and 1/R^2. Which means computing x^2 and higher powers of x will have a few terms. But it can still be done.