Answer:

 φ = √2/L sin (kx),   E = (h² / 8 mL²) n²  

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}

             

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

             

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

             

We change the variable

       sin² kx = ½ (1 – cos 2kx)

       A =√ 2 / L

The definitive function is

          φ = √2/L sin (kx)

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