Thanks for this. My formula is tailored to work for the first 70 quantum numbers and just for one value of mu (but I have rederived
it for other values of mu, hbar and m and the exponent does not depend on mu, hbar or m as long as you stick to the first 70 qn's).
This is food for thought, perhaps for both of us (definitely for me), thanks again. This is quite a surprise.
Rich
----- Original Message -----
From: "Barton Willis" <willisb at unk.edu>
To: "Richard Hennessy" <richhen2008 at gmail.com>
Cc: <Maxima at math.utexas.edu>
Sent: Monday, August 03, 2009 9:56 AM
Subject: Re: [Maxima] Quantum number and energy of electron
Here is a Maxima calculation of the WKB energy of the mu * x^4 oscillator.
One hitch: Maxima handles integrate(sqrt(e - ,mu * x^4),x,0, (e/mu)^(1/4))
OK, but not integrate(sqrt(e - mu * x^4),x,-(e/mu)^(1/4), (e/mu)^(1/4)).
Similarly, Maxima can compute the next two terms in the WKB approximation.
(%i1) assume(e > 0, mu > 0)$
(%i2) 2* integrate(sqrt(e - mu * x^4),x,0, (e/mu)^(1/4)) = %pi * hbar * (n
+ 1/2)/sqrt(2*m);
(%o2) (beta(1/4,3/2)*e^(3/4))/(2*mu^(1/4))=(%pi*hbar*(n+1/2))/(sqrt(2)*sqrt
(m))
(%i3) solve(%,e);
Is hbar*m*(2*n+1) positive, negative, or zero?pos;
(%o3) [e=((sqrt(3)*%pi^(4/3)*%i-%pi^(4/3))*hbar^(4/3)*mu^(1/3)*(2*n+1)^
(4/3))/(2*2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3)),e=-((sqrt(3)*%pi^(4/3)*%i
+%pi^(4/3))*hbar^(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2*2^(2/3)*beta(1/4,3/2)^
(4/3)*m^(2/3)),e=(%pi^(4/3)*hbar^(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2^
(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3))]
Barton
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