logo

Solution of Schrödinger Equation for Finite Square Well Potential 📂Quantum Mechanics

Solution of Schrödinger Equation for Finite Square Well Potential

Overview

5D5500710.jpg

Let’s examine how a particle moves when the potential takes the shape of a finite square well as shown in the figure above. The potential UU is

U(x)={0x<−aU0−a<x<a0a<x U(x) = \begin{cases} 0 & x<-a \\ U_{0} & -a < x <a \\ 0 &a<x \end{cases}

When the potential is U(x)U(x), the time-independent Schrödinger equation is

d2u(x)dx2+2mℏ2[E−U(x)]u(x)=0 \dfrac{d^2 u(x)}{dx^2}+\frac{2m}{\hbar ^2} \Big[ E-U(x) \Big]u(x)=0

Solution

E<−U0E<-U_{0}

If the energy is less than the potential, then a solution does not exist and therefore does not need to be considered.

■

−U0<E<0-U_{0} < E < 0

  • Part 2-1. x<−ax<-a

    In this region, the time-independent Schrödinger equation is d2udx2+2mℏ2Eu=0 \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0 Since 2mℏ2E\frac{2m}{\hbar^2}E is negative, if we substitute with −κ2-\kappa ^2, d2udx2−κ2u=0 \dfrac{d^2 u}{dx^2}-\kappa^2u=0 It is a very simple second-order ordinary differential equation. Solving the differential equation, the solution is u1(x)=A+eκx+A−e−κx u_{1}(x)=A_{+}e^{\kappa x} + A_{-}e^{-\kappa x} Since the wave function must be square-integrable, it is A−=0A_{-}=0.

  • Part 2-2. −a<x<a-a<x<a

    In this region, the time-independent Schrödinger equation is d2udx2+2mℏ2(E+U0)u=0 \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}(E+U_{0})u=0 Since E+U0>0E+U_{0}>0, substituting with 2mℏ2(E+U0)=k2\frac{2m}{\hbar^2}(E+U_{0})=k ^2, d2udx2+k2u=0 \dfrac{d^2 u}{dx^2}+k^2 u=0 Solving the equation, u2(x)=B+eikx+B−e−ikx u_{2}(x) = B_{+}e^{i k x}+B_{-}e^{-ik x}

  • Part 2-3. a<xa<x

    In this region, the time-independent Schrödinger equation is d2udx2+2mℏ2Eu=0 \dfrac{d^2 u}{dx^2}+\frac{2m}{\hbar^2}Eu=0 The solution has the same form as in Part 2-1. u3(x)=C+eκx+C−e−κx u_{3}(x)=C_{+}e^{\kappa x} + C_{-}e^{-\kappa x} Since the wave function must be square-integrable, it is C+=0C_{+}=0.

    Part 2-4. Boundary Conditions

    Assuming the wave function is smooth, it is continuous at x=−ax=-a and x=ax=a, and the derivative (slope) of the wave function is also continuous at x=−ax=-a and x=ax=a. Therefore, {u1(−a)=u2(−a)u2(a)=u3(a)  ⟹  {A+e−κa+A−eκa=B+e−ika+B−eika⋯(1)B+eika+B−e−ika=C+eκa+C−e−κa ⋯(2) \begin{cases}u_{1}(-a)=u_{2}(-a) \\ u_{2}(a)=u_{3}(a) \end{cases} \quad \implies \begin{cases} A_{+}e^{-\kappa a}+A_{-}e^{\kappa a} = B_{+}e^{-ik a}+B_{-}e^{ ik a} \quad \cdots (1) \\ B_{+}e^{ik a}+B_{-}e^{-ik a} = C_{+}e^{\kappa a}+C_{-} e^{-\kappa a} \ \quad \cdots (2) \end{cases}

    {u1′(−a)=u2′(−a)u2′(a)=u3′(a)  ⟹  {κA+e−κa−κA−eκa=ikB+e−ika−ikB−eika⋯(3)ikB+eika−ikB−e−ika=κC+eκa−κC−eκa⋯(4) \begin{cases}u_{1}^{\prime}(-a)=u_{2}^{\prime}(-a) \\ u_{2}^{\prime}(a)=u_{3}^{\prime}(a) \end{cases} \quad \implies \begin{cases} \kappa A_{+}e^{-\kappa a}-\kappa A_{-}e^{\kappa a} = ik B_{+}e^{-ik a}-ik B_{-}e^{ik a} \quad \cdots (3) \\ ik B_{+}e^{ik a}-ik B_{-}e^{-ik a} = \kappa C_{+}e^{\kappa a}-\kappa C_{-} e^{\kappa a} \quad \cdots (4) \end{cases}

    Expressing (1)(1) and (3)(3) in matrix form,

    (e−κaeikaike−ika−ikeika)(A+A−)=(e−κaeκaκe−κa−κeκa)(B+B−)⋯(5) \begin{pmatrix} e^{-\kappa a} & e^{ika} \\ ike^{-ika} & -ike^{ika} \end{pmatrix} \begin{pmatrix} A_{+} \\ A_{-} \end{pmatrix} = \begin{pmatrix} e^{-\kappa a} & e^{\kappa a} \\ \kappa e^{-\kappa a} & -\kappa e^{\kappa a} \end{pmatrix} \begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix}\quad \cdots (5)

    Expressing (2)(2) and (4)(4) in matrix form,

    (eκae−κaκeκa−κe−κa)(B+B−)=(eikae−ikaikeika−ike−ika)(C+0)⋯(6) \begin{pmatrix} e^{\kappa a} & e^{-\kappa a} \\ \kappa e^{\kappa a} & -\kappa e^{-\kappa a} \end{pmatrix} \begin{pmatrix} B_{+} \\ B_{-} \end{pmatrix} = \begin{pmatrix} e^{ik a} & e^{-ik a} \\ ik e^{ik a} & -ik e^{-ika } \end{pmatrix} \begin{pmatrix} C_{+} \\ 0 \end{pmatrix}\quad \cdots (6)