Particle in a 2D Box: Min Energy & Lowest Energies

  • Thread starter Thread starter hy23
  • Start date Start date
  • Tags Tags
    2d Box Particle
Click For Summary

Homework Help Overview

The discussion revolves around the energy levels of a particle confined in a two-dimensional square box, specifically focusing on the minimum energy and the lowest allowed energy states based on quantum numbers. The energy expression involves two quantum numbers, n and m, which are both integers representing different modes of the wave function in the box.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of minimum energy by substituting values for n and m, questioning how these quantum numbers relate to the energy levels. There is discussion about the implications of changing n and m, and whether they represent nodes and antinodes in the wave function.

Discussion Status

Participants are actively engaging with the problem, raising questions about the interpretation of quantum numbers and their implications for energy levels. Some guidance has been provided regarding the relationship between quantum numbers and the wave function, though confusion remains about the conditions for n and m.

Contextual Notes

There is uncertainty regarding the definitions of quantum numbers in this context, and participants are questioning the assumptions about symmetry in the wave functions. The discussion includes references to external resources for clarification.

hy23
Messages
60
Reaction score
0

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
 
Last edited:
Physics news on Phys.org
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
 

Attachments

  • 2dbox.JPG
    2dbox.JPG
    11.9 KB · Views: 524
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
 
  • #10
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?
 
  • #11
@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.
 
  • #12
OH!

Then i am wrong!
SORRY PEOPLE! MY BAD!
 
  • #13
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?
 
  • #14
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
    box12.JPG
    13.8 KB · Views: 548
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
9K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
1K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K