1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two dimensional Square well and parity

  1. Jan 10, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle is placed in the potential (a 2 dimensional square well)

    V(x) = (0 for -a/2 <= x =< a/2 and -a/2 <= y =<a/2, infinity for x>a/2, x<-a/2 and y>a/2, y<-a/2)

    The hamiltonian commutes with the parity operator P, Pψ(x,y) = ψ(-x,-y) = λψ(x,y), where the eigenvalue λ can take two possible values +(-)1 Write down the eigenstates corresponding to the four lowest energies in such a way that they are also eigenfunctions of the parity operator P. What is the parity of these states?

    2. Relevant equations

    I calculated the eigenfunctions and i got:

    ψ(x) = √(2/a)*sin(nπx/a), n=2,4,6,.. (odd)

    ψ(x) = √(2/a)*cos(nπx/a), n=1,3,5 (even)



    3. The attempt at a solution

    E(n,m) = E(n) + E(m)

    E(n,m) = (π^2*(h-bar)^2)/(2*M*a^2)*(n^2+m^2)

    In the solution manual it says (n^2+m^2) = 2,5,8,10

    How did they come up with those numbers ?
    the odd is 2,4,6 and the even is 1,3,5, so how can (n^2+m^2) = 2,5,8,10 ?

    and this also in the solution manual:

    E(1,1) = one state (n^2+m^2) = 2, odd*odd = even

    E(1,2)=E(2,1) = two states (n^2+m^2) = 5, odd*even = odd

    E(2,2) = one state (n^2+m^2) = 8, even*even = even

    E(1,3) = E(3,1) = two states (n^2+m^2)=10, odd*odd = even

    What do they mean with E(1,1), E(1,2) etc ?. How can E(1,1) be one state and E(2,2) be two states ? Where did they get odd*even from ?
     
  2. jcsd
  3. Jan 10, 2014 #2
    The eigenfunctions you have written down in your relevant equations are those in 1D but we have a 2D system, so an eigenfunction of the full problem has the form
    [itex]\psi_{mn}(x,y)=\psi_{m}(x)\psi_{n}(y)[/itex]
    where m and n label the solutions you have given.
    Now, the energy of such a solution is just the sum of the energies in 1D and they are labeled by the two integers m and n. The energies are quadratic in n and m, so the full energy depends on the sum of the squares [itex]n^2+m^2[/itex]. The smallest sums you can get this way are 1+1=2, 1+4=5, 4+4=8 and 1+9=10.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted