Minimum energy and five lowest energies for particle in 2D box

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Homework Statement


It can be shown that the allowed energies of a particle of mass m in a two dimensional square box of side L are

E =h2(n2+m2)/8mL2

The energy depends on two quantum numbers, n and m, both of which must have an integer value 1, 2, 3...

What is the minumum energy for a particle in a two dimensional quare box of side L?

What are the five lowest allowed energies (give as multiples of Eminimum)?

Homework Equations



given above

The Attempt at a Solution



no idea, substitute 1 for n and m? that's my only thought
 
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hy23 said:
What is the minumum energy for a particle in a two dimensional quare box of side L?

What are the five lowest allowed energies (give as multiples of Eminimum)?


The Attempt at a Solution



no idea, substitute 1 for n and m? that's my only thought

Yes, for the lowest energy.

ehild
 
what about for the other 5 lowest energies? there's two quantum numbers n and m, which one do I change? and what does n and m represent anyways?
 
See: http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/pbox.html

The quantum numbers are related to the number of nodes and antinodes of the wave function inside the box. n=1 and m=1 means one antinode in the middle of the box. As they increase, the more the number of the nodes.

Find the energies when (n,m) is equal to (1,1), (1, 2), (2,1), (2,2), .. . and so on.

ehild
 
I still don't get it
are u saying n and m represent nodes and antinodes respectively? then it can't be equal to (1,2) or (1, 1) because a particle in a box must have two nodes?
 
hy23 said:
I still don't get it
are u saying n and m represent nodes and antinodes respectively? then it can't be equal to (1,2) or (1, 1) because a particle in a box must have two nodes?
No. There can be infinite number of nodes. n represents the antinodes in the x direction and m is the number of antinodes in the y direction. There are nodes at the walls and the nodes are separated by the antinodes.
The attachment shows the wavefunction in a 2D-box for n=4 and m=4
http://en.wikipedia.org/wiki/Particle_in_a_box.

ehild
 
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oh ok thanks a lot, the part that I didn't realize until now was that each quantum number stood for a direction
 
wait a minute!

if n,m are quantum numbers, then if n=1
so l = (n-1) =0

m= -l to l = 0 again!

how is the (n,m) pair (1,1)?

_____________________________________________

instead use,

n=3 so that m = -1,0,1

use m= -1

you get negative energy ...

but there is a problem, you can have n=infinity!
and then m can be also more negative... we need to do a little math here!
 
cupid.callin: I don't understand what you speak about. Read the first post. ehild
 
Wooooooops!

Thats why people say that read complete post before you start writing your answer!

But still if n=1, m cannot be 1 !

is the m in denominator 'mass' and in numerator is quantum number 'm' ?

For lowest, m=0 which is not allowed ... therefore lowest n=2 and m=1 ...

Right?
 
@cupid.callin: You seem to think that the n and m here are the quantum numbers describing for example the electrons around the atoms (with quantum numbers n for shell, l for angular momentum, m as the magnetic quantum number and s as spin). This is not correct. Here, the n and m:s are just integers describing the number of nodes in the x- and y-directions respecitvely.

So there is NO restriction on m depending on what value n has. They can both be any number.
 
ok I've read that wikipedia page

it seems that in a square box, there are some wavefunctions that can have the same energy level, for example (2,1) and (1,2) works out to have the same energy, so should this energy level only be written once, that is write out the energy for (2,1) and neglect (1,2)?

and the other thing I don't get is why is it that n and m can be different integers? To me it seems intuitive that the wavelengths in both dimensions should be symmetrical and therefore the number of antinodes in both directions in a square should be the same. Comments?
 
As the problem asked the five lowest energy values, you need to include duplicate values only once, I think.

The wavelengths need not be symmetrical in the x and y directions, why should they? The wavefunction is of the form

F(x.y)= A sin(ax) sin(by)

a, b are arbitrary for the time being, but the function must fulfil the condition that F(x,y)=0 at the sides of the rectangle, that is, at x=0 or y=0 or x=L or y=L. From this condition it follows that
a*L=n*pi and b*L= m*pi, that is

F(x,y) = A sin(n*pi/L) sin(m*pi/L).

If you plot such function you find integer number of half-waves in both the x and y direction. Attached is the wafunction for (1.2)


ehild
 
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