International Journal of Innovative Technology and Exploring Engineering (IJITEE) ISSN: 2278-3075, Volume-8 Issue-3, January 2019 Eigen Energy Values and Eigen Functions of a Particle in an Infinite Square Well Potential by Laplace Transforms Rohit Gupta, Rahul Gupta, Dinesh Verma Abstract- Quantum mechanics is one of the branches of physics and is a fundamental theory which explains the nature of atoms II. DEFINITION OF LAPLACE and subatomic particles at the smallest scale of energy. In this TRANSFORMATIONS mechanics, physical problems are solved by algebraic and analytic methods. By applying Laplace Transforms we can find Let F (t) is a well-defined function of real numbers t ≥ 0. the general solutions of one dimensional Schrodinger’s time- The Laplace transformations of F (t), denoted by f (푝) or L independent wave equation for a particle in an infinite square {F (t)}, is defined as well potential. In this paper, we will discuss the Eigen energy ∞ −푝푡 L {F (t)} = 푒 퐹 푡 푑푡 = 푓(푝), provided that the values and Eigen functions of a particle in an infinite square well 0 potential. We will find that general solutions (Eigen functions) of integral exists, i.e. convergent. If the integral is convergent one dimensional Schrodinger’s time- independent wave equation for some value of 푝 , then the Laplace transformations of F for a particle in an infinite square well potential have very (t) exists otherwise not. Where 푝 푠 the parameter which interesting characteristics that each Eigen function is associated may be real or complex number and L isthe Laplace with a particular value of energy (Eigen energy value) of the particle confined inside an infinite square well potential. transformations operator [1-3]. Keywords: Eigen functions, Eigen values, Infinite square A. Laplace Transformations of Elementary Functions well, Laplace Transforms. 1 1. 퐿 1 = , 푝 > 0 I. INTRODUCTION 푝 푛! 2. 퐿 푡푛 = , In Quantum mechanics, physical problems are solved by 푝푛+1 algebraic and analytic methods. By applying Laplace 푤ℎ푒푟푒 푛 = 0,1,2,3 … Transforms to the one dimensional Schrodinger’s time- 1 independent wave equation, we can obtain the Eigenenergy 3. 퐿 푒푏푡 = , 푝 > 푏 values and Eigenfunctions of a particle in a square well 푝 − 푏 푏 potential of infinite height. To define Eigenvalues and 4. 퐿 푠푛푏푡 = , 푝 > 0 Eigenfunctions, suppose that an operator Ô is operated on 푝2 + 푏2 the function g and resultsthe same function g multiplied by 푏 5. 퐿 푠푛ℎ푏푡 = , 푝 > 푏 some constant α i.e. Ô g = α g. In this equation, g is the 푝2 − 푏2 Eigenfunction of operator Ô and the constant α is the Eigen 푝 6. 퐿 푐표푠푏푡 = , 푝 > 0 value of the operator Ô related with the Eigenfunction g and 푝2 + 푏2 푝 the equation is known as Eigen value equation. The 7.퐿 푐표푠ℎ푏푡 = , 푝 > |푏| Eigenfunctions are selected from a special class of 푝2−푏2 functions. For example, in bound state problem, all wave B. Proof of Laplace Transformations of Some functions and their derivatives must be continuous, single Elementary Functions valued and finite everywhere. They must also vanish at infinity. Such functions are called as well-behaved According to the definition of Laplace transformations, ∞ −푝푡 functions[1-4]. As an example, to illustrate the Eigen value 1. L {F (t)} = 0 푒 퐹(푡)푑푡 , then d of an operator, consider the operator operating on a ∞ dz L {1} = 푒−푝푡 1 푑푡 −6z 0 well – behaved function e . The result is d −6z −6z 1 (e ) = −6e . = − ( 푒−∞ − 푒−0) dz 푝 Comparing this equation with standard Eigen value equation 1 Ô g = α g, we find that (-6) is the Eigen value of = − 0 − 1 d 푝 operator( ) associated with the Eigen functione−6z. dz 1 = = 푓(푝) , 푝 > 1 푝 ∞ −푝푡 2. L {F (t)} = 0 푒 퐹(푡)푑푡 , then Revised Manuscript Received on 18 December 2018. Rohit Gupta, Lecturer, Yogananda College of Engineering and ∞ −푝푡 L { 푠푛ℎ푏푡 } = 푒 푠푛ℎ푏푡 푑푡 Technology (YCET), Jammu, India. 0 Rahul Gupta, Lecturer, Yogananda College of Engineering and 푏푡 −푏푡 Technology (YCET), Jammu, India. ∞ −푝푡 푒 −푒 = 푒 푑푡 Dr. Dinesh Verma, Associate Professor, Yogananda College of 0 2 Engineering and Technology (YCET), Jammu, India. Published By: Blue Eyes Intelligence Engineering Retrieval Number: C2558018319 6 & Sciences Publication Eigen Energy Values and Eigen Functions of a Particle in an Infinite Square Well Potential by Laplace Transforms ∞ 푒− 푝−푏 푡−푒 − 푝+푏 푡 III. FORMULATION = 푑푡 0 2 A. Infinite Square Well Potential 1 = − ( 푒−∞ − 푒−0) A potential well is a potential energy function with a 2 푝−푏 minimum. If a particle is left in the well and the total energy 1 + ( 푒−∞ − 푒−0) of the particle is less than the height of the potential well, we 2(푝+푏) can say that the particle is trapped in the well. In classical 1 1 mechanics, a particle trapped in the potential well can = − 2(푝−푏) 2(푝+푏) vibrate back and forth but cannot leave the well. In quantum mechanics, such a trapped particle is called bound state [4- 1 2푏 = . 6]. Considering a particle confined to the region 0 < z < a. it 2 푝2−푏2 can move freely with in the region 0 < z < a, but subject to 푏 Therefore, 퐿 푠푛ℎ푏푡 = , 푝 > 푏 strong forces at z = 0 and z = a. Therefore, the particle never 푝2−푏2 cross to the right to the regionz > a, or to the left of z < 0. It means V = 0 in the region 0 < z < a, and rises to infinity at z C. Laplace Transformations of derivatives = 0 and z = a. This situation is called one dimensional Let the function F is having exponential order, that is F is potential box. Therefore, an infinite square well potential V continuous function and is piecewise continuous function on (z) (shown in figure 1) is defined as any interval, then the Laplace transforms of derivative of F V (z) = 0 for 0 < z <a (t), denoted by 퐿 퐹′ 푡 , is given by = ∞ for z ≤ 0 and z ≥ a. ∞ 퐿 퐹′ 푡 = 푒−푝푡 퐹′ 푡 푑푡 0 Integrating by parts and using the condition that,푒−푝푡 퐹 푡 = 0 푤ℎ푒푛 푡 = ∞, we get ′ ∞ −푝푡 퐿 퐹 푡 = 0 − 퐹 0 − 0 −푝푒 퐹 푡 푑푡 ′ ∞ −푝푡 퐿 퐹 푡 = −퐹 0 + 푝 0 푒 퐹 푡 푑푡 퐿 퐹′ 푡 = 푝퐿 퐹 푡 − 퐹(0) 퐿 퐹′ 푡 = 푝푓 푝 − 퐹 0 Now, since 퐿 퐹′ 푡 = 푝퐿 퐹 푡 − 퐹 0 , Therefore, 퐿 퐹′′ 푡 = 푝퐿 퐹′ 푡 − 퐹′ 0 The time - independent Schrodinger’sequation in one dimension is written as[7-8]: ′′ ′ 2m 퐿 퐹 푡 = 푝 푝퐿 퐹 푡 − 퐹(0) − 퐹 0 퐷2휓(z) + {E–V (z) }휓(z) = 0 (1) 푧 ℏ2 퐿 퐹′′ 푡 = 푝2퐿 퐹 푡 − p퐹(0) − 퐹′ 0 ∂ 퐷푧 ≡ ′′ 2 ′ ∂z 퐿 퐹 푡 = 푝 푓(푝) − p퐹(0) − 퐹 0 This equation is second-order linear differential equation. In Similarly, 퐿 퐹′′′ 푡 = equation (1), 휓(z) is probability wave function of the 푝3푓 푝 − 푝2퐹 0 − 푝퐹′ 0 − 퐹′ ′ 0 and so on. particle and V (z) is the potential energy. Now we will solve equation (1) to obtain Eigenenergy values and D. Inverse Laplace Transformations Eigenfunctions of a particle in an infinite square well The inverse Laplace transformsof the function f(p) is L-1[f potential. (p)] =F (t). If we write L [F(t)] = f(p), then L-1[f(p)] = F (t), For a particle inside an infinite square well potential, where L-1 is called the inverse Laplace transforms operator V(z) = 0. [1-3]. Substitute this value of potential in equation (1), we get 2 2m 퐷푧 휓 z + E 휓(z) = 0 (2) E. Inverse Laplace Transformations of Some Functions ℏ2 1 Where z belongs to [0, a] with boundary conditions 휓(0) = 1. L-1{ } = 1 푝 휓(a) = 0. 1 2. L-1{ } = 푒푏푡 For convenience, let us substitute (푝−푏) 2m 2 1 1 2 E = 푘 (3) 3. L-1{ }= sinbt ℏ 푝2+푏2 푏 Therefore, equation (2) becomes 푝 4. L-1{ } = cosbt 퐷2휓(z) + 푘2휓(z) = 0 (4) 푝2+푏2 푧 푝 Taking Laplace Transform of equation (4), we get 5. L-1{ } = coshbt 푝2−푏2 2 2 퐿[퐷푧 휓(z) ] + 푘 퐿[휓(z)] = 0 1 1 6. L-1{ } = sinhbt 푝2−푏2 푏 1 푡 푛−1 7. L-1{ } = 푝푛 (푛−1)! Published By: Blue Eyes Intelligence Engineering Retrieval Number: C2558018319 7 & Sciences Publication International Journal of Innovative Technology and Exploring Engineering (IJITEE) ISSN: 2278-3075, Volume-8 Issue-3, January 2019 This equation gives C. Determination of constant A 2 2 푝 휓(p) – 푝휓(0) -퐷푧 ψ 0 + 푘 휓(p) = 0 (5) Since the probability density between z = 0 and z = a, is one, Applying boundary condition: 휓(0) = 0, equation (5) because the particle is somewhere within this boundary, that becomes, is inside the infinite square well potential. Hence applying 2 2 푝 휓 (p) –퐷푧 ψ 0 + 푘 휓 (p) = 0 normalization condition, we can write 2 2 푧=푎 ∗ Or 푝 휓(p) + 푘 휓(p) = 퐷푧 ψ 0 (6) 푧=0 휓(z) 휓(z) 푑푧 = 1 (12) ∗ In this equation, 퐷푧 ψ 0 is some constant. Here휓(z) is the complex conjugate of 휓(z). Let us substitute퐷푧 ψ 0 = 퐴, Using equation (10) in equation (12), we can write Equation (6) becomes 2 퐴 푧=푎 2 n휋 2 2 n휋 sin ( z) 푑푧 = 1 푝 휓 (p) + 푘 휓 (p) = A 푧=0 푎 푎 퐴 2 Or 휓(p) = (7) (푝2 + 푘2) 퐴 푧=푎 1 2n휋 Or n휋 [1 − cos z ]dz = 1 푧=0 2 푎 Taking inverse Laplace transforms of equation (7), we get 푎 퐴 Solving the integration and arranging, we get 휓(z) = sin (k z) (8) 푘 1 퐴 n휋 2 2 Applying boundary condition: 휓(a) = 0, equation (8) gives A = (13) 푘 푎 푎 sin (k a) = 0 D.