Chapter 7. The Eigenvalue Problem

7. The Eigenvalue Problem, December 17, 2009 16
In the Schrödinger representation, the state |n, j, k is (see Metiu, Eq. 8.31)
x, y, z | n, j, k ≡ ψ n,j,k (x, y, z)
2
= sin
L x
nπx
L x
2
L y
sin
jπy 2 kπz
sin
L y L z L z
(78)
For the example of the cube,
x, y, z | 2, 1, 1 =
8 2πx
L sin sin
3 L
πy
sin
L
πz
L
(79)
and
x, y, z | 1, 2, 1 =
8 πx
L sin sin
3 L
2πy
sin
L
πz
L
(80)
A similar equation holds for x, y, z | 1, 1, 2. These three states have the same
energy but they are different states. What is the physical difference between
them
Let us look at the kinetic energy in the x-direction:
ˆK x x, y, z | 2, 1, 1 = − ¯h2
2m ∂x ψ 2,1,1(x, y, z)
2
= ¯h2 2π 2
ψ 2,1,1 (x, y, z) (81)
2m L
I obtained the second line from the first by taking the second derivative of
ψ 2,1,1 (x, y, z) givenbyEq.79. Weseethat|2, 1, 1 is an eigenstate of ˆKx ,
with the eigenvalue ¯h2 2π 2.Inthesameway,youcanshowthat|2,
2m L
1, 1 is
an eigenstate of ˆK y and ˆK
z , with the eigenvalues ¯h2 1π 2.
2m L The total energy
is the sum of those three kinetic energies. In the state |2, 1, 1, the particle
has higher kinetic energy in the x-direction. We can analyze in the same way
the states |1, 2, 1 and |1, 1, 2. Table 1 gives the result. The three degenerate
states have the high (i.e. ¯h2 2π 2)
2m L kinetic energy in different directions.
Can we distinguish these three states by some measurement We can.
The particle in the state |2, 1, 1 will emit a photon in a different direction
than will a particle in |1, 2, 1 or |1, 1, 2.
The degenerate states |2, 1, 1, |1, 2, 1, and|1, 1, 2 are different physical
states, even though they have the same energy.
∂ 2