# Particle in a 2D Box

hy23

## 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)?

given above

## The Attempt at a Solution

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

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Homework Helper
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

hy23
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?

Homework Helper
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

hy23
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?

Homework Helper
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

#### Attachments

• 2dbox.JPG
12.1 KB · Views: 411
hy23
oh ok thanks a lot, the part that I didn't realize until now was that each quantum number stood for a direction

cupid.callin
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)?????????

_____________________________________________

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!!!!

Homework Helper
cupid.callin: I don't understand what you speak about. Read the first post.

ehild

cupid.callin
Wooooooops!!!

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???

kloptok
@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.

cupid.callin
OH!!!!!

Then i am wrong!!!

hy23

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?

Homework Helper
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

#### Attachments

• box12.JPG
14 KB · Views: 431
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