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Since the potential is zero within 0 < x <
a region and we should equal the ODE −h²/2md²/dx²y
- Ey to 0
d²y/dx²
+ k²y = 0
where k² = 2mE/ |
Having the solution y = Aeikx + Be-ikx (which corresponds to the free particle movement from left to right and vice versa) we fulfill the above-mentioned Schroedinger equation (through the substitution); we must only fulfill the following boundary conditions:
y(x=0) = A + B =
0 → B = −A
y(x) = A(eikx−
e−ikx)
= 2iA sinkx = C sinkx
(C = 2iA is just new constant). From the second boundary condition y(x = a) = C · sinka = 0 follows either sina = 0 or C = 0. C = 0 is absolute nonsense since there is no wavefunction for it. Hence it's immediately clear that sinus should vanish and so k · a = n . π
k = π/a· n n = 1, 2, 3, ....
or after substitution of k one can obtain for energy E:
En = h²/8ma² n² |
For the particle impulse p = (2mE)½ one will have p = h/2a· n
The energy should be quantized since n is integer. The possible energy values are then: E1 = h²/8ma² , E2 = 4E1 , E3 = 9E1 , E4 = 16E1 ,.... This energy quantization happens because of wavefunction is restricted to boundary conditions and potential.
We can obtain C meaning from the normalization condition:
o∫a|yn|2dx = 1 = o∫a C2 sin2(nπx/a)dx = C2 . ½ a = 1
yn(x) = (2/a)½ sin(nπx/a) |
One also can see that the wavefunction holds symmetry in its middle: its value is either the same or it changes its sign upon wavefunction reflection. It's said about positive parity when the wavefunction sign is the same [y(a/2+x) = +y(a/2-x)] upon wavefunction reflection relative to symmetry axis that is x=a/2 and about negative parity whether the wavefunction sign is changed upon reflection [y(a/2+x) = -y(a/2-x)], correspondingly.
Let's solve the problem of a particle in three dimensional potential well having the following size 0<x<a, 0<y<b, 0<z<c:
Division on (yx·yy·yz)
and substitution of k2 = kx2 +
ky2 + kz2 :
(1/yx∂²/∂x²yx + kx2) | + | (1/yy∂²/∂y²yy + ky2) | + | (1/yz∂²/∂z²yz + kz2) | = 0 |
|¾¾¾¾¯¾¾¾¾| | |¾¾¾¾¯¾¾¾¾| | |¾¾¾¾¯¾¾¾¾| | |||
= 0 | = 0 | = 0 |
Therefore we obtain the possible energy states:
E = h²/8m(n1²/a² + n2²/b² + n3²/c²)
where n1, n2 and n3 are quantum numbers for x-,y- and z-directions, correspondingly.
The wavefunction y(x,y,z) = yx(x) . yy(y) ·yz(z) is as follows:
y(x,y,z) = C · sin(n1π x/a)· sin(n2πy/b)· sin(n3πz/c)
where C can be obtained from the normalization condition
The Dirac style for y-function is as follows:
<x,y,z|n1n2n3> = C sin(n1πx/a)·sin(n2πy/b)·sin(n3πz/c)
Sometimes one can write down in a more simple form: |n1n2n3>
= C sin(n1πx/a)
sin(n2πy/b)
sin(n3πz/c).
The normalization condition is as follows in any case:
<n1n2n3|n1n2n3> = 1
And finally we obtain the normalization constant C = (2/a)1/2
(2/b)1/2 (2/c)1/2
= (8/abc)1/2.
If the well measures are the same or a = b = c we have an important case of
different quantum numbers that correspond to one energy level:
E = h²/8ma² (n12 + n22 + n32) = h²/8ma²· K2 = E1 · K2
where E1 is shortened by h²/8ma² in the last formula. It leads us to the Symmetry and Degeneracy of States Description.