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 "/>
We re-examine the square of the Hamiltonian, https://www.w3.org/1998/Math/MathML"> Η https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math16_1.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> , in Equation 15.7 and also the Dirac approach in finding the relativistic Hamiltonian. The classical Hamiltonian for the free particle should be the positive square root of Equation 15.7 which is not a linear function of the momenta. Applying this to the problem, say for example in Equation 15.10, we have the resultant wave equation that is asymmetrical with respect to space and time derivatives and so the necessity to linearize the equation in the momentum operators. So for Equation 15.10, it is impossible to include the Pauli spin matrices without destroying the invariance of the theory since the spin matrices transform the components of a three-dimensional rather than a four-dimensional vector. In addition, the function, https://www.w3.org/1998/Math/MathML"> Ψ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math16_2.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> , has one component rather than two components like spinors and so the Schrödinger relativistic equation represents a particle with no spin. So for a potential to be added in Equation 15.10, the Lorentz-transformation properties of such an addition should be guaranteed by matrix algebra which is the requirement for the Hamiltonian https://www.w3.org/1998/Math/MathML"> Η ^ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003273073/d236e21f-f9f6-4e42-b186-ef1c50f0a68b/content/math16_3.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> to be linear in the momentum operators and guarantees relativistic invariance. This matrix algebra which permits the linearization of the Hamiltonian in the momentum operator was introduced by Dirac with the help of the Pauli matrices.
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 "/> .
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) .