The solutions to the Time‐Independent Schrödinger Equation (TISE) can be found as long as we can define the Hamiltonian, which in the simplest cases, means defining the potential in which the particle of interest is moving. In the following sections, we will consider a single particle of mass, m p , moving in a potential that is time‐independent, but can vary in space. We will follow four steps: Define the Hamiltonian by defining the Potential Energy Operator since H ^ = - ħ 2 2 m p ∇ ^ 2 + V ^ ( r ) $ \hat{H} = - \frac{{{\hbar}^{2} }}{{2m_{p} }}\hat{\nabla }^{2} + \hat{V}(r) $ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315381381/5a219413-b541-4543-a0c3-fab4f5358e87/content/inline-math4_1.tif"/> . Solve the TISE to get the wave functions ψ ( x , y , z ) $ \psi (x, y, z) $ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315381381/5a219413-b541-4543-a0c3-fab4f5358e87/content/inline-math4_2.tif"/> and then the total stationary state wave functions Ψ ( x , y , z ; t ) = ψ ( x , y , z ) e - i ω t . $ \iPsi (x, y, z;t) = \psi (x, y, z)e^{ - i\omega t} . $ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781315381381/5a219413-b541-4543-a0c3-fab4f5358e87/content/inline-math4_3.tif"/> Determine the energies of the various states of the system. Examine the implications of the solutions to the Schrödinger equations.
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) .
In the previous chapter, we did not examine the change of a state with time https://www.w3.org/1998/Math/MathML"> t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math2_1.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> . It has been assumed that we have been examining a state at a given moment https://www.w3.org/1998/Math/MathML"> t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math2_2.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> . In the general case, the state of a system changes with time https://www.w3.org/1998/Math/MathML"> t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math2_3.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> . For example, say the motion of an electron in a potential field, its wave function depends on time https://www.w3.org/1998/Math/MathML"> t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math2_4.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> : https://www.w3.org/1998/Math/MathML"> Ψ = Ψ q , t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math2_5.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> . Such a dependence has to be described by a differential equation in which we may have https://www.w3.org/1998/Math/MathML"> ∂ Ψ ∂ t , ∂ 2 Ψ ∂ t 2 , ⋯ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math2_6.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> .
Spherically symmetric systems, where the potential energy U(r) is independent of the direction of r → https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/inline-eqn7_1_B.tif"/> , are usually treated using spherical polar coordinates (r,θ,φ), related to the Cartesian coordinates as shown by Eqs.(2.20) and ( 2.21) . Therefore, in view of the appropriate form ( 2.36) of the Hamiltonian, the Schrodinger wave equation for Ψ (r.θ,φ) reads: 7.1 H ^ Ψ = − h 2 m [ 1 r 2 ∂ ∂ r ( r 2 ∂ Ψ ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ Ψ ∂ θ ) + 1 r 2 sin 2 θ ∂ 2 Ψ ∂ φ 2 ] + U ( r ) Ψ = E Ψ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_1_B.tif"/> In contrast to the motion of a particle whose potential energy depends on Cartesian coordinates, where the Schrödinger equation can be separated into three equations pertaining to motions along the axes x, y, z, which are independent of one another (see Problem 5.4.2 ), Eq.(7.1) also separates into equations pertaining to motions along the coordinates r, θ and φ, but the three types of motion are no longer independent. The separation of variables is made in two stages, first substituting into the energy eigenvalue equation the trial solution: 7.2 Ψ ( r , θ , φ ) = R ( r ) Y ( θ , φ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_2_B.tif"/> This transforms Eq.(7.1) in the form: 7.3 h 2 2 m [ 1 r ∂ ∂ r ( r 2 ∂ R ∂ r ) ] + r 2 [ E − U ( r ) ] = − h 2 2 m 1 Y [ 1 sin θ ∂ ∂ θ ( sin θ ∂ Y ∂ θ ) + 1 sin 2 θ ∂ 2 Y ∂ φ 2 ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_3_B.tif"/> where one side is independent of θ and φ , and the other side is independent of r , so that they must be separately equal to some constant, which is conveniently chosen as L 2 / 2 m. In this way we separate the radial motion, described by an ordinary differential equation where ∂ /∂r is replaced by d/dr: 7.4 [ − h 2 2 m 1 r 2 d d r ( r 2 d d r ) + U ( r ) + L 2 2 m r 2 ] R ( r ) = E R ( r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_4_B.tif"/> 243 from the rotational motion, described by the angular equation: 7.5 − h 2 [ 1 sin θ ∂ ∂ θ ( sin θ ∂ ∂ θ ) + 1 sin 2 θ ∂ 2 ∂ φ 2 ] Y= L 2 Y or L ^ 2 Y= L 2 Y https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_5_B.tif"/> which is the eigenvalue problem for orbital angular momentum, represented by the operator ( 2.33) . This separation corresponds to the classical splitting of kinetic energy among the two types of motion. The second stage of the separation process assumes that: 7.6 Y( θ , φ )= P ( θ ) Φ ( φ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_6_B.tif"/> such that Eq.(7.5) becomes: 7.7 h 2 [ sin θ P ( θ ) ∂ ∂ θ ( sin θ ∂ P ( θ ) ∂ θ ) + L 2 h 2 sin 2 θ ] = − h 2 1 Φ ( φ ) ∂ 2 Φ ( φ ) ∂ φ 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_7_B.tif"/> Since one side is independent of φ , while the other is independent of θ , they must each be equal to a constant, which we call L z 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/inline-eqn7_2_B.tif"/> , and this yields two other ordinary differential equations: 7.8 − h 2 d 2 d φ 2 Φ ( φ ) = L z 2 Φ ( φ ) or L ^ z 2 Φ ( φ ) = L z 2 Φ ( φ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_8_B.tif"/> where L ^ z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/inline-eqn7_3_B.tif"/> is defined by Eq.(2.25) , and: 7.9 1 sin θ d d θ ( sin θ d P ( θ ) d θ ) ( L 2 h 2 − L z 2 h 2 sin 2 θ ) P ( θ ) = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003078265/e77b8ce0-ccb3-48ed-9f6d-a84c8cb529c5/content/eqn7_9_B.tif"/>
We look for a Hamiltonian to use in the time-dependent Schrödinger equation, and because this is first order in time we stay first order in space trying to treat all coordinates on a similar basis. Write 4.1 https://www.w3.org/1998/Math/MathML"> i ∂ ψ ∂ t = H ψ = − i α i ∂ ψ ∂ x i + m β ψ i = 1 , 2 , 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429184550/8ac61231-bcf3-4e9e-a16f-9d6f78005bbb/content/math4_1_B.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> where ψ is now a column and the α i and β are square matrices.
To obtain a Lorentz invariant Schrödinger equation, we considered the square root of the Klein–Gordon equation. This had the disadvantage that the Hamiltonian https://www.w3.org/1998/Math/MathML"> H = p → 2 + m 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429073601/03ca065d-1bf0-4218-811d-2f95276252a3/content/equ_603.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> contains an infinite number of powers of https://www.w3.org/1998/Math/MathML"> p → 2 / m 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429073601/03ca065d-1bf0-4218-811d-2f95276252a3/content/equ_604.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> , the parameter in which the square root should be expanded. It would have been better to treat space and time on a more equal footing in the Schrödinger equation. This is what Dirac took as his starting point. As the Schrödinger equation is linear in https://www.w3.org/1998/Math/MathML"> p 0 = i ∂ / ∂ t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429073601/03ca065d-1bf0-4218-811d-2f95276252a3/content/equ_605.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> , one is looking for a Hamiltonian that is linear in the momenta https://www.w3.org/1998/Math/MathML"> p j = i ∂ / ∂ x j ( = − p j ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429073601/03ca065d-1bf0-4218-811d-2f95276252a3/content/equ_606.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> ). ( 12.1 ) https://www.w3.org/1998/Math/MathML"> i ∂ Ψ ∂ t = H Ψ = − i α k ∂ Ψ ∂ x k + β m Ψ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429073601/03ca065d-1bf0-4218-811d-2f95276252a3/content/equ_607.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/>
Abstract After revisiting the time-dependent Schrödinger equation and introducing the Hamiltonian operator, the time-independent Schrödinger equation (TISE) is obtained. Conditions on acceptable wavefunctions are then stated. The method for solution of the TISE is laid out, and the infinite potential well is solved as an example. A discussion of the physical interpretation of the solutions — as energy eigenstates in position space — follows. This provides an opportunity to introduce overall and relative phases, and to point out that the latter are responsible for time evolution. The TISE is then applied to the important problems of scattering from a step potential and tunneling through a rectangular potential barrier. Use of the TISE for such problems greatly simplifies calculations, but it can also instill misconceptions; both aspects are discussed.