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Math problem

Blogs > HeaDStrong
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HeaDStrong
Profile Blog Joined January 2009
Scotland785 Posts
Last Edited: 2009-11-26 22:29:37
November 26 2009 22:26 GMT
#1
I had a quite simple homework problem for my quantum mechanics course. I had to solve the infinitely deep square well for two dimensions- easy peasy.

The energies of the system are given by:
E[n,m]=E*(n^2+m^2) where n,m are integers (quantum numbers in x and y)
then for 0 < n, m < 6 we had to find the maximum degeneracy of any state of the system, which means that there are more than one way for the system to arrange to obtain the same energy. In other words in how many different ways you can pick n and m to come up with the same energy.

for the example with limitation on n and m, i could simply draw a table with values of n and m varying from 1 to 5 and calculating n^2+m^2 and then look what is the maximum number of a value (which would correspond to an energy) repeating in the table. I found that it's 2 and it happens when we interchange n and m.

the interesting bit, which was not part of my homework, is whether this is true for all n and m. So one could rephrase the question:
n^2+m^2=l^2+k^2
where none of the n, m, l or k are equal and all positive. are there any integer solutions to this?

i'm not too good with number theory so i dont have any good ideas how to tackle this. the only thing i can think of now is making a computer program which would draw a "bigger" table and count the equal energies, but this doesnt sound too rigorous or mathematically beautiful. (it would be impossible to disprove anything, because it would have to terminate at some size of the table)

any ideas?

ShinyGerbil
Profile Blog Joined June 2008
Canada519 Posts
November 26 2009 22:30 GMT
#2
7^2 + 24^2 = 15^2 + 20^2 = 625 = 25^2
[s]savior[/s] jaedong fighting! // member of LighT eSports
paper
Profile Blog Joined September 2004
13196 Posts
Last Edited: 2009-11-26 22:46:01
November 26 2009 22:44 GMT
#3
pythagorean theorem?

keep c constant, rotate the outer square (and expand/constrict) in the first diagram so that it fits for infinite possibilities

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ShinyGerbil
Profile Blog Joined June 2008
Canada519 Posts
November 26 2009 22:47 GMT
#4
but he was wondering about integer solutions...
[s]savior[/s] jaedong fighting! // member of LighT eSports
HeaDStrong
Profile Blog Joined January 2009
Scotland785 Posts
Last Edited: 2009-11-26 23:00:27
November 26 2009 22:50 GMT
#5
whoa! cool. where did you get this from? did you come up with this yourself or is this like a known problem? (i tried to look somewhere starting from Pythagorean triples and Fermat's Last theorem but couldn't find it.)
can you give me some background on that for my general interest

also if i think about the degeneracies again a viable solution would also be:

n^2+m^2= 2*k^2 (this would mean that for k the x and y quantum numbers are equal) and obviously n!=m (n=/=m)

EDIT: ok it's just Pythagorean triples with equal c's... i'm ashamed i didn't come up with it myself t.t
HeavOnEarth
Profile Blog Joined March 2008
United States7087 Posts
Last Edited: 2009-11-26 23:36:51
November 26 2009 23:17 GMT
#6
lol i tried to guess one
11^2 + 13^2 ~= 8^2 + 15^2 ;o
but meh here's a bunch of them if u want integers
http://www.math.uic.edu/~fields/puzzle/triples.html
everything was said already above actually ;o
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