Proving Angular Momentum Magnitudes: Binomial Expansion Method

[perplexabot]

Homework Statement

The allowed magnitudes of angular momentum are L = √(l(l+1)) * hbar. Use the binomial expansion to prove that when l is large, L ≈ (l + .5) * hbar.

Homework Equations

Binomial expansion formula: (1 + z)^n = 1 + nz + [n*(n-1)*z^2] / 2 + ...

The Attempt at a Solution

Ok. First I did:
L = √(l(l+1)) * hbar = [hbar * √l] * [1 + l]^.5
Then doing Bin. expan. apprx. for [1 + l]^.5,
I get a final answe, L ≈ hbar * √l * (1 + l/2)

Which is wrong. When compared to L ≈ (l + .5) * hbar, I can tell that I have an extra factor of √l, I think it has something to do with l being very large. Can someone please help me out? Thanks

 

[Dick]

Factor the l out of (l(l+1))^(1/2)=(l^2+l)^(1/2)=(l^2*(1+1/l))^(1/2).

 

[perplexabot]

Thank you so much. I understand it. So:

[l * (l + 1)] ^ (1/2) = (l + l^2) ^ (1/2) = [l^2 * (l + 1/l)] ^ (1/2) = l * (1+ 1/l)^(1/2) ≈ l * (1 + 1/2l).

Great, that achieves the correct answer. Thanks again

 

[Dick]

Thank you so much. I understand it. So:

[l * (l + 1)] ^ (1/2) = (l + l^2) ^ (1/2) = [l^2 * (l - 1/l)] ^ (1/2) = l * (1- 1/l)^(1/2) = l * (1 + 1/2l).

Great, that achieves the correct answer. Thanks again

In the last step it isn't equal, right? It's just an approximation. I'm also not sure where the minus signs came from and then disappeared to. Just a typo, right?

 

[perplexabot]

In the last step it isn't equal, right? It's just an approximation. I'm also not sure where the minus signs came from and then disappeared to. Just a typo, right?

Yes, you are right, this is an approximation and that minus sign is a typo (I will edit my post). Thanks again.
One last optional question though, would you by any chance know why my first trial was wrong (my first post)?

 

[Dick]

Yes, you are right, this is an approximation and that minus sign is a typo (I will edit my post). Thanks again.
One last optional question though, would you by any chance know why my first trial was wrong (my first post)?

Sure. (1+x)^(1/2)~(1+x/2) is only a good approximation if x is small. So (1+1/l)^(1/2)~(1+1/(2l)) is a good approximation because 1/l is small. (1+l)^(1/2)~1+l/2 is a bad approximation because l is large.

 

[perplexabot]

Sure. (1+x)^(1/2)~(1+x/2) is only a good approximation if x is small. So (1+1/l)^(1/2)~(1+1/(2l)) is a good approximation because 1/l is small. (1+l)^(1/2)~1+l/2 is a bad approximation because l is large.

Thank you for your time.