Respuesta :
Answer:
 φ = √2/L sin (kx),  E = (h² / 8 mL²) n² Â
Explanation:
The Schrödinger equation for a particle in a box is, described by a particle within a potential for simplicity with infinite barrier
   V (x) =  ∞       x <0
          0    0 <x <L
          ∞      x> L
This means that we have a box of length L
We write the equation
       (- h’² /2m  d² / dx² + V) φ = E φ
       h’= h / 2π
The region of interest is inside the box, since being the infinite potential there can be no solutions outside the box. The potential is zero
        - h’² /2m d²φ/ dx² = E φ
The solution for this equation is a sine wave,
Because it is easier to work with exponentials, let's use the reaction between the sine function and cook with the exponential
        [tex]e^{ikx}[/tex] = cos kx + i sin kx
Let's make derivatives
       dφ / dx = ika e^{ikx}
       d²φ / dx² = (ik) e^{ikx} = - k² e^{ikx}
      Â
Let's replace
      - h'² / 2m (-k² e^{ikx}) = E e^{ikx}
      E = h'² / 2m   k²
To have a solution this expression
Now let's work on the wave function, as it is a second degree differential bond, two solutions must be taken
       φ = A e^{ikx} + B e^{-ikx}
This is a wave that moves to the right and the other to the left.
Let's impose border conditions
     φ (0) = 0
     φ (L) = 0
For being the infinite potential
With the first border condition
     0 = A + B
     A = -B
They are the second condition
     0 = A e^{ikL}+ B e^{-ikL}
We replace
    0 = A (e^{ikL} - e^{-ikL})
We multiply and divide by 2i, to use the relationship
    sin kx = (e^{ikx} - e^{-ikx}) / i2
    0 = A 2i sin kL
      Â
Therefore kL = nπ
     k = nπ / L
The solution remains
     φ = A sin (kx)
    E = (h² / 8 mL²) n²
To find the constant A we must normalize the wave function
    φ*φ = 1
    A² ∫ sin² kx dx = 1
      Â
We change the variable
    sin² kx = ½ (1 - cos 2kx)
    A =√ 2 / L
The definitive function is
     φ = √2/L sin (kx)
