Particle in A Well

We will solve the Schroedinger equation for a simple problem which however describes essential ways to solve other more complicated problems.
We consider electron to move from coordinate 0 to a along x-axis absolutely freely, i.e. V(x) = 0 for x < a and x > 0. The potential however is infinite in other two regions; V(x £ 0; x ³ a) ® ¥. Since the potential energy is infinite on the borders a particle can't be delayed on the barrier walls, i.e. we need the following boundary conditions to be fulfilled:
y(0) = 0 and y(a) = 0

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/h²

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.

The lowest energy isn't equal to 0 that isn't impress us due to the uncertainty relation. Since our particle is squeezed into a range 0 < x < a it must have minimal but finite impulse and energy that is greater than 0, correspondingly. The wavefunction of n-fold energy state:
yn(x) = C sin(nπx/a)

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.

Solution for Three Dimensions

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:

Schroedinger equation: −h²/2m(∂²/∂x² + ∂²/∂y² + ∂²/∂z²)y(x,y,z) - Ey(x,y,z) = 0
Separation of variables: y(x,y,z) = yx(x) ·yy(y) ·yz(z) k2 = 2mE/h²
Substitution: yyyz∂²/∂x²yx + yxyz∂²/∂y²yy + yxyy∂²/∂z²yz + k2yx(x)yy(y)yz(z) = 0

Division on (yx·yy·yz) and substitution of k2 = kx2 + ky2 + kz2 :

(1/y x∂²/∂x²yx + kx2) | + | (1/y y∂²/∂y²yy + ky2) | + | (1/y z∂²/∂z²yz + kz2) | = 0
|¾¾¾¾¯¾¾¾¾| |  | |¾¾¾¾¯¾¾¾¾| |  | |¾¾¾¾¯¾¾¾¾| |
= 0 |  | = 0 |  | = 0 |

→ similar to ODE in one-dimensional case since the simple transformation finally gives the following, for instance for x: ∂²/∂x²yx + kx² yx= 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

o∫ao∫bo∫c |y(x,y,z)|2 dxdydz = 1

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.