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CRC Press eBooks

Particle in a Box: Quantisation of Translational Energy

Amita Dua

Published Nov 6, 2025• 0 citations
Publisher

Abstract

The motion and all the properties of a particle of mass ‘ m ’ which moves with potential energy, V ( x, y, z ) can be obtained from solution of well behaved fundamental Schrödinger wave equation https://www.w3.org/1998/Math/MathML" display="block"> − ℏ 2 2 m ∇ 2 ψ + V ( x , y , z ) ψ = E ψ

DOI
10.1201/9781003728634-4

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Examples

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Although one-dimensional problems are only an idealisation of naturally occurring physical situations, and hence may seem somewhat artificial, motion of a single particle in one dimension yields exact solutions of the Schrödinger equation which highlight most of the features of the three-dimensional quantum approach, avoiding, however, its complexity. Because the time-dependent Schrödinger equation in one dimension: 6.1 i ℏ ∂ Ψ ( x , t ) ∂ t = − ℏ 2 2 m ∂ 2 Ψ ( x , t ) ∂ x 2 + U ( x ) Ψ ( x , t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_1_B.tif"/> has the general solution, Eq.(5.62) , which is expressed in terms of the eigenvalues and eigenfunctions of the Hamiltonian operator only: 6.2 Ψ ( x , t ) = ∑ n a n e − i E n t / ℏ ψ n ( x ) + ∫ a ( E ) e − i E t / ℏ ψ ( x , E ) d E https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_2_B.tif"/> the problem is always reduced to solving the Schrödinger wave equation: 6.3 − ℏ 2 2 m ∂ 2 ψ ( x ) ∂ x 2 + U ( x ) ψ ( x ) = E ψ ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_3_B.tif"/> This is the energy eigenvalue equation, which can also be written as: 6.4 d 2 ψ ( x ) d x 2 + 2 m ℏ 2 [ E − U ( x ) ] ψ ( x ) = d 2 ψ ( x ) d x 2 + k 2 ( x ) ψ ( x ) = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_4_B.tif"/> by setting: 6.5 k 2 ( x ) = 2 m ℏ 2 [ E − U ( x ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_5_B.tif"/> 203 It follows that the energy eigenvalue problem is similar to the calculation of monochromatic waves in a medium with non-uniform refractive index. If E < U ( x ), it is clear that k ( x ) becomes imaginary, namely k { x ) = iγ (x), where: 6.6 γ 2 ( x ) = 2 m ℏ 2 [ U ( x ) − E ] , E < U ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_6_B.tif"/> For any given eigenvalue E, the second-order differential equation (6.4) yields two linearly independent solutions ψ 1 ( x ) and ψ 2 ( x ), suchthat: 6.7 ψ ( x ) = a 1 ψ 1 ( x ) + a 2 ψ 2 ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_7_B.tif"/> It follows that any eigenvalue might be at most two-fold degenerate. The solutions of interest are those which satisfy certain regularity conditions which make them admissible eigenfunctions: i ψ ( x ) and its first derivative should be continuous functions of x, for all x. This implies that we allow for finite discontinuities of the potential energy U ( x ) at some points, where d 2 ψ/dx 2 should exhibit an opposite finite discontinuity, so preventing the departure from continuity of the right-hand side of Eq.(6.1) , whose left-hand side ∂ Ψ /∂t is a continuous function. A step-like behaviour of d 2 ψ/dx 2 requires the continuity of dψ/dx at the same points, otherwise the discontinuity becomes infinite. ii ψ ( x ) should be finite or single-valued for all x. This ensures the possibility of normalizing the wave functions, according to Eq.(1.37) , and having a uniquely defined probability density || ψ ( x ) | 2 . The eigenfunctions ψ ( x ) can always be chosen to be real, and this becomes apparent by conjugation of Eq.(6.4) , which gives: d 2 ψ * ( x ) d x 2 + k 2 ( x ) ψ * ( x ) = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_8_B.tif"/> where k 2 ( x ) is always real. Because ψ ( x ) and ψ* ( x ) are eigenfunctions with the same eigenvalue E, their real and imaginary parts, namely: 6.8 ψ r ( x ) = ψ ( x ) + ψ * ( x ) 2 , ψ i ( x ) = ψ ( x ) − ψ * ( x ) 2 i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_9_B.tif"/> are real eigenfunctions corresponding to E. Hence, we will limit ourselves to the real solutions ψ ( x ) to Eq.(6.4) .

CRC Press eBooks•Nov 5, 2020•0 citations
The motion and all the properties of a particle of mass ‘ m ’ which moves with potential energy, V ( x, y, z ) can be obtained from solution of well behaved fundamental Schrödinger wave equation https://www.w3.org/1998/Math/MathML" display="block"> − ℏ 2 2 m ∇ 2 ψ + V ( x , y , z ) ψ = E ψ
All the properties and motion of a particle of mass ‘ m ’ which moves with potential energy, V ( x, y, z ) can be obtained from solution of well behaved Schrodinger equation https://www.w3.org/1998/Math/MathML" display="block"> − ħ 2 2 m ∇ 2 ψ + V ( x , y , z ) ψ = E ψ https://www.w3.org/1999/xlink" content-type="black-white" xlink:href=" https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003490135/247a5333-15ea-46ba-8a55-ecb37a472373/content/math4-1_B.tif "/>
There is more information in a quantum state than in the classical initial values https://www.w3.org/1998/Math/MathML" display="inline"> ( x 0 , p 0 ) that determine the complete 1-dimensional motion of a point particle for a given force. The initial wave function https://www.w3.org/1998/Math/MathML" display="inline"> Ψ ( x , 0 ) requires an infinite number of parameters, which can be expressed in observational terms as the two basic expectation values https://www.w3.org/1998/Math/MathML" display="inline"> ⟨ x ^ ⟩ and https://www.w3.org/1998/Math/MathML" display="inline"> ⟨ p ^ ⟩ together with the two variances https://www.w3.org/1998/Math/MathML" display="inline"> ( σ x 2 , σ p 2 ) and the covariance 5.1 https://www.w3.org/1998/Math/MathML" display="block"> C x p = 1 2 ⟨ x ^ p ^ + p ^ x ^ ⟩ − ⟨ x ^ ⟩ ⟨ p ^ ⟩ ,
This 1-D Schrodinger equation describes the non relativistic motion of a single charged particle (mass m = 1, = 1) with a potential V in a uniform electric field t → u(t). With v = −u, (1) represents also the dynamics of a particle in a non Galilean frame of absolute position v (see, e.g., [9]). A change of independent variables (t, q) → (t, z) and dependent variable ψ → φ, transform (1) into (2) where the control appears as a shift on the space variable. These classical transformations are as follows (see, e.g., [2]). Instead of considering u as control, take v defined by v = −u as control. Then
Although one-dimensional problems are only an idealisation of naturally occurring physical situations, and hence may seem somewhat artificial, motion of a single particle in one dimension yields exact solutions of the Schrödinger equation which highlight most of the features of the three-dimensional quantum approach, avoiding, however, its complexity. Because the time-dependent Schrödinger equation in one dimension: 6.1 i ℏ ∂ Ψ ( x , t ) ∂ t = − ℏ 2 2 m ∂ 2 Ψ ( x , t ) ∂ x 2 + U ( x ) Ψ ( x , t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_1_B.tif"/> has the general solution, Eq.(5.62) , which is expressed in terms of the eigenvalues and eigenfunctions of the Hamiltonian operator only: 6.2 Ψ ( x , t ) = ∑ n a n e − i E n t / ℏ ψ n ( x ) + ∫ a ( E ) e − i E t / ℏ ψ ( x , E ) d E https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_2_B.tif"/> the problem is always reduced to solving the Schrödinger wave equation: 6.3 − ℏ 2 2 m ∂ 2 ψ ( x ) ∂ x 2 + U ( x ) ψ ( x ) = E ψ ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_3_B.tif"/> This is the energy eigenvalue equation, which can also be written as: 6.4 d 2 ψ ( x ) d x 2 + 2 m ℏ 2 [ E − U ( x ) ] ψ ( x ) = d 2 ψ ( x ) d x 2 + k 2 ( x ) ψ ( x ) = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_4_B.tif"/> by setting: 6.5 k 2 ( x ) = 2 m ℏ 2 [ E − U ( x ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_5_B.tif"/> 203 It follows that the energy eigenvalue problem is similar to the calculation of monochromatic waves in a medium with non-uniform refractive index. If E < U ( x ), it is clear that k ( x ) becomes imaginary, namely k { x ) = iγ (x), where: 6.6 γ 2 ( x ) = 2 m ℏ 2 [ U ( x ) − E ] , E < U ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_6_B.tif"/> For any given eigenvalue E, the second-order differential equation (6.4) yields two linearly independent solutions ψ 1 ( x ) and ψ 2 ( x ), suchthat: 6.7 ψ ( x ) = a 1 ψ 1 ( x ) + a 2 ψ 2 ( x ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_7_B.tif"/> It follows that any eigenvalue might be at most two-fold degenerate. The solutions of interest are those which satisfy certain regularity conditions which make them admissible eigenfunctions: i ψ ( x ) and its first derivative should be continuous functions of x, for all x. This implies that we allow for finite discontinuities of the potential energy U ( x ) at some points, where d 2 ψ/dx 2 should exhibit an opposite finite discontinuity, so preventing the departure from continuity of the right-hand side of Eq.(6.1) , whose left-hand side ∂ Ψ /∂t is a continuous function. A step-like behaviour of d 2 ψ/dx 2 requires the continuity of dψ/dx at the same points, otherwise the discontinuity becomes infinite. ii ψ ( x ) should be finite or single-valued for all x. This ensures the possibility of normalizing the wave functions, according to Eq.(1.37) , and having a uniquely defined probability density || ψ ( x ) | 2 . The eigenfunctions ψ ( x ) can always be chosen to be real, and this becomes apparent by conjugation of Eq.(6.4) , which gives: d 2 ψ * ( x ) d x 2 + k 2 ( x ) ψ * ( x ) = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_8_B.tif"/> where k 2 ( x ) is always real. Because ψ ( x ) and ψ* ( x ) are eigenfunctions with the same eigenvalue E, their real and imaginary parts, namely: 6.8 ψ r ( x ) = ψ ( x ) + ψ * ( x ) 2 , ψ i ( x ) = ψ ( x ) − ψ * ( x ) 2 i https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn6_9_B.tif"/> are real eigenfunctions corresponding to E. Hence, we will limit ourselves to the real solutions ψ ( x ) to Eq.(6.4) .