I have done this calculation too. I had to check my old wxm files to be sure but I did and I found that this is the Born
Sommerfield quantification formula.
I did it for the mu = 1b15 case and got
(%i17) solve([ans], [E[n]])[3];
`rat' replaced -1.864029228766461B-15 by -1/536472274451275 = -1.864029228766462B-15
(%o17) E[n]=(%pi^2*(2*n-1)^(4/3))/(1072944548902550*1072944548902550^(1/3)*gamma(1/4)^(8/3)).
or in the general case
E[n]=(3*3^(1/3)*%pi^2*hbar^(4/3)*mu^(1/3)*(2*n-1)^(4/3))/(2*2^(1/3)*gamma(1/4)^(8/3)*m^(2/3))
This is not the same thing, I have to check my work to see if I made a mistake.
My posted formula works better for small n (< 70) but the constant exponent I cannot derive yet.
Rich
----- Original Message -----
From: "Barton Willis" <willisb at unk.edu>
To: "Richard Hennessy" <rich.hennessy at verizon.net>
Sent: Monday, August 03, 2009 7:49 PM
Subject: Re: [Maxima] Quantum number and energy of electron
A bit nicer:
(%i10) assume(e > 0, mu > 0, %h_bar > 0, n > 0, m > 0)$
(%i11) 2* integrate(sqrt(e - mu * x^4),x,0, (e/mu)^(1/4)) = %pi * %h_bar *
(n+ 1/2)/sqrt(2*m);
(%o11) (beta(1/4,3/2)*e^(3/4))/(2*mu^(1/4))=(%h_bar*%pi*(n+1/2))/(sqrt
(2)*sqrt(m))
(%i12) solve(%,e);
(%o12) [e=((sqrt(3)*%h_bar^(4/3)*%pi^(4/3)*%i-%h_bar^(4/3)*%pi^(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)*%h_bar^(4/3)*%pi^(4/3)*%i+%h_bar^(4/3)*%pi^(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=(%h_bar^(4/3)*%pi^
(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3))]
(%i13) ans : last(%);
(%o13) e=(%h_bar^(4/3)*%pi^(4/3)*mu^(1/3)*(2*n+1)^(4/3))/(2^(2/3)*beta
(1/4,3/2)^(4/3)*m^(2/3))
(%i14) subst([e=e[0], n = 0],ans);
(%o14) e[0]=(%h_bar^(4/3)*%pi^(4/3)*mu^(1/3))/(2^(2/3)*beta(1/4,3/2)^(4/3)*m^(2/3))
(%i15) solve(%,mu);
Is e[0] positive, negative, or zero?pos;
(%o15) [mu=(4*e[0]^3*beta(1/4,3/2)^4*m^2)/(%h_bar^4*%pi^4)]
(%i16) subst(%,ans);
(%o16) e=(4^(1/3)*e[0]*(2*n+1)^(4/3))/2^(2/3)
(%i17) factor(radcan(%));
(%o17) e=e[0]*(2*n+1)^(4/3)
This is a standard QM textbook problem, so I knew the answer before I
did it. Mostly, I did the calculation to see if Maxima could do the
integral.
Barton
-----"Richard Hennessy" <rich.hennessy at verizon.net> wrote: -----
To: "Barton Willis" <willisb at unk.edu>
From: "Richard Hennessy" <rich.hennessy at verizon.net>
Date: 08/03/2009 06:31PM
cc: <Maxima at math.utexas.edu>
Subject: Re: [Maxima] Quantum number and energy of electron
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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