A particle of mass m is confined to a box of length`. Its initial wave function is identical to that of the displacement of the string in th

Question

A particle of mass m is confined to a box of length`. Its initial wave function is identical to that of the displacement of the string in the problem above, Boas Ch. 13, Sec. 4, #4.Find the solution of the Schrodinger equation

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King 5 months 2021-09-04T17:28:07+00:00 1 Answers 6 views 0

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  1. 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

                   e^{ikx} = 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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