# [Maxima] QM

Richard Hennessy rvh2007 at comcast.net
Sat Jun 14 12:35:48 CDT 2008

```Thanks for that.  So if the complex one is square integrable then so is the real one and if not then not. I was getting confused over that point.

Thanks,

Rich

------------Original Message------------
From: Leo Butler <l.butler at ed.ac.uk>
To: "Richard Hennessy" <rvh2007 at comcast.net>
Date: Sat, Jun-14-2008 11:24 AM
Subject: Re: [Maxima] QM

If a,b are square integrable, then a+b are also. This is a simple application of
the triangle inequality:

|a+b| <= |a|+|b|.

Leo

On Sat, 14 Jun 2008, Richard Hennessy wrote:

< I have been using Griffith's "Introduction to Quantum Mechanics", 2nd edition.  In the book there is a proof that the time independent Shrodinger equation can be real or complex and that it is always possible to find a real valued solution to this equation, so the complex solutions are not really necessary.  The idea is that if there are two functions that satisfy Shrodinger's equation then even if they are complex you can always make them real by taking a linear combination of the two independent solutions.  That can always be done because one linear combination is
<
< f(z)+conjugate(f(z))= f'(x) which is real
<
< and the other is
<
< %i*(g(z)-conjugate(g(z)))= g'(x)
<
< which is always real.  So you don't need the complex values ones.
<
< I noticed that in Maxima the contrib_ode package and the odelin function return for the x^4 potential always a complex valued combination of 2 functions.  I have been trying to find the corresponding real valued versions.  This is a problem.  The solutions have to be square integrable and it looks to me like f(z) and g(z) are but not so for f'(x) and g'(x) so how would you do this change of form?  Is this a flaw in the proof?
<
< f(z)+conjugate(f(z)) may not be square integrable.
<
< If someone knows I might be able to get some sleep!
<
< Thanks,
<
< Rich
<
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