One of the most important problems in quantum mechanics concerns the motion of a particle in a potential of the form V ( r ) = V ( r ), where r ≡ | r | https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203741481/4ab22e1a-c1c5-4151-9df7-a6d7ea2b05e0/content/eq763.tif"/> . This system has the simplifying feature of spherical symmetry, which allows the angular dependence of the motion to be reduced to purely geometrical considerations and hence solved once and for all, while the physics enters only through the radial motion. Treating this problem in detail also provides a good opportunity to illustrate the quantum mechanical formalism.
One of the most important problems in quantum mechanics concerns the motion of a particle in a potential of the form V ( r ) = V ( r ), where r ≡ | r | https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203741481/4ab22e1a-c1c5-4151-9df7-a6d7ea2b05e0/content/eq763.tif"/> . This system has the simplifying feature of spherical symmetry, which allows the angular dependence of the motion to be reduced to purely geometrical considerations and hence solved once and for all, while the physics enters only through the radial motion. Treating this problem in detail also provides a good opportunity to illustrate the quantum mechanical formalism.
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"/>
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) .
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.
The problem of interacting particles can usually be reduced in quantum mechanics to that of one particle, as can be done in classical mechanics. Normally, this problem is one of an electron orbiting around, or being affected by, a positive atomic core or a scattering center with the interaction governed by the Coulomb potential. This simple problem of a particle in a quadratic potential energy is one of the typical problems of quantization. In two dimensions, the problem is more interesting, but not particularly more complicated. In three dimensions, people could extend the treatment to a three-dimensional harmonic oscillator, but a more important problem has spherical symmetry in the potential, and the latter is not the quadratic potential of the harmonic oscillator. It is of more interest to turn our attention to the motion of an electron about the nucleus of an atom.
Q36(1) The spherical harmonics Y 1,1 ( θ , φ ), Y 1,0 ( θ , φ ), Y 1, −1( θ , φ ) are given by Eqs. (16.65) to (16.67). In Cartesian coordinates, these functions are given by 1 Y 1 , − 1 ( x , y , z ) = 3 8 π x − i y r , Y 1 , 0 ( x , y , z ) = 3 4 π z r , Y 1 , 1 ( x , y , z ) = − 3 8 π x + i y r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0667.tif"/> (a) Using the expression in Cartesian coordinates for the quantised orbital angular momentum operator L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1694.tif"/> in Eq. (27.86), verify by explicit calculations that Y 1,1 ( x , y , z ), Y 1,0 ( x , y , z )and Y 1,-1 ( x , y , z ) are the eigenfunctions of L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1695.tif"/> corresponding to eigenvalues ħ , 0 and – ħ . (b) Consider the following coordinate transformations: x → z ′ , y → x ′ , z → y ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0668.tif"/> (1) 202 Show that the component of the orbital angular momentum operator along the z ′-direction is the same as that along the x -direction, i.e., show that L ^ c z ′ = L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1696.tif"/> . (2) Show that the simultaneous eigenfunctions of L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1697.tif"/> and L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1698.tif"/> corresponding to eigenvalues of L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1699.tif"/> equal to 2 ħ 2 are given in Cartesian coordinates by X 1 , − 1 ( x , y , z ) = 3 8 π y − i z r , X 1 , 0 ( x , y , z ) = 3 4 π x r , X 1 , 1 ( x , y , z ) = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0669.tif"/> SQ36(1)(a) In Cartesian coordinates we have 2 : p ^ x Y → 1 , − 1 : = − i ℏ 3 8 π ∂ ∂ x x − i y r = − i ℏ 3 8 π ( 1 r − x − i y r 2 ∂ ∂ x r ) = − i ℏ 3 8 π ( 1 r − x − i y r 2 x r ) = − i ℏ 3 8 π 1 r 3 ( r 2 − x 2 + i x y ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0670.tif"/> p ^ y Y → 1 , − 1 : = − i ℏ 3 8 π ∂ ∂ y x − i y r = − i ℏ 3 8 π ( − i r − x − i y r 2 ∂ ∂ y r ) = − i ℏ 3 8 π ( − i r − x − i y r 2 y r ) = − i ℏ 3 8 π 1 r 3 ( − i r 2 − x y + i y 2 ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0671.tif"/> L ^ z Y → 1 , − 1 : = ( x p ^ y − y p ^ x ) Y 1 , − 1 = − i ℏ 3 8 π 1 r 3 ( ( − i x r 2 − x 2 y + i x y 2 ) − ( y r 2 − y x 2 + i x y 2 ) ) = − i ℏ 3 8 π 1 r 3 ( − i x r 2 − y r 2 ) = − ℏ 3 8 π 1 r { x − i y } = − ℏ Y 1 , − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0672.tif"/> 203 Next we have p ^ x Y → 1 , 0 : = − i ℏ 3 4 π ∂ ∂ x z r = − i ℏ 3 4 π ( − z r 2 ∂ ∂ x r ) = i ℏ 3 4 π z x r 3 . p ^ y Y → 1 , 0 : = − i ℏ 3 4 π ∂ ∂ y z r = − i ℏ 3 4 π ( − z r 2 ∂ ∂ y r ) = i ℏ 3 4 π z y r 3 . L ^ z Y → 1 , 0 : = ( x p ^ y − y p ^ x ) Y 1 , 0 = x ( i ℏ 3 4 π z y r 3 ) − y ( i ℏ 3 4 π z x r 3 ) = 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0673.tif"/> Finally we have p ^ x Y → 1 , 1 : = i ℏ 3 8 π ∂ ∂ x x + i y r = − i ℏ 3 8 π ( 1 r − x + i y r 2 ∂ ∂ x r ) = i ℏ 3 8 π ( 1 r − x + i y r 2 x r ) = i ℏ 3 8 π 1 r 3 ( r 2 − x 2 − i x y ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0674.tif"/> p ^ y Y → 1 , 1 : = i ℏ 3 8 π ∂ ∂ y x + i y r = i ℏ 3 8 π ( i r − x + i y r 2 ∂ ∂ y r ) = i ℏ 3 8 π ( i r − x + i y r 2 y r ) = i ℏ 3 8 π 1 r 3 ( i r 2 − x y − i y 2 ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0675.tif"/> L ^ z Y → 1 , 1 : = ( x p ^ y − y p ^ x ) Y 1 , 1 = i ℏ 3 8 π 1 r 3 { ( i x r 2 − x 2 y − i x y 2 ) = − ( y r 2 − y x 2 − i x y 2 ) } = i ℏ 3 8 π 1 r 3 { i x r 2 − y r 2 } = − ℏ 3 8 π 1 r ( x + i y ) = ℏ Y 1 , 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0676.tif"/> 204 SQ36(1)(b) (1) Consider a transformation from the original coordinate system ( x , y , z ) to a new coordinate system ( x ′, y ′, z ′): x → z ′ , y → x ′ , z → y ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0677.tif"/> We can see that the new z ′-axis coincides with the original x -axis, and that r = x 2 + y 2 + z 2 = x ′ 2 + y ′ 2 + z ′ 2 = r ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0678.tif"/> Physically we expect the angular momentum about the x -axis to be the same as the angular momentum about the z′ -axis. We can confirm this mathematically by showing that L ^ x = y p ^ z − z p ^ y = − i ℏ ( y ∂ ∂ z − z ∂ ∂ y ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0679.tif"/> is equal to L ^ z ′ = x ′ p ^ y ′ − y ′ p ^ x ′ = − i ℏ ( x ′ ∂ ∂ y ′ − y ′ ∂ ∂ x ′ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0680.tif"/> This is obvious since x′ = y , y′ = z . It follows that the eigenfunctions and eigenvalues of L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1700.tif"/> are the same in terms of the dashed coordinates as that of L ^ z ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1701.tif"/> . (2) We know the eigenfunctions Y ℓ , m ℓ ′ ( x ′ , y ′ , z ′ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1702.tif"/> and eigenvalues of L ^ z ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1703.tif"/> in the transformed coordinates (x′, y′, z′) as they are of the same form as those of the eigenfunctions and eigenvalues of L ^ z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1704.tif"/> in the original coordinates, i.e., we have Y 1 , − 1 ′ ( x ′ , y ′ , z ′ ) = 3 8 π x ′ − i y ′ r ′ , Y 1 , 0 ′ ( x ′ , y ′ , z ′ ) = 3 4 π z ′ r ′ Y 1 , 1 ′ ( x ′ , y ′ , z ′ ) = − 3 8 π x ′ + i y ′ r ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0681.tif"/> 205 Since L ^ z ′ = L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1705.tif"/> these are also eigenfunctions of L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1706.tif"/> Written in terms of the original coordinates these functions become Y 1 , − 1 ′ = 3 8 π y − i z r , Y 1 , 0 ′ = 3 4 π x r , Y 1 , 1 ′ = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0682.tif"/> We can conclude that the required eigenfunctions L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1707.tif"/> denoted more conveniently by X 1,1 , X 1,0 , X 1, −1 are X 1 , − 1 = 3 8 π y − i z r , X 1 , 0 = 3 4 π x r , X 1 , 1 = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0683.tif"/> Q36(2) Suppose L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1708.tif"/> and L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1709.tif"/> are measured giving the eigenvalues 2 ħ 2 and – ħ , respectively. A measurement of L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1710.tif"/> is then made. What are the possible results of the measurement of L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1711.tif"/> ? Find the probability of each of these possible results.
The problem of interacting particles can usually be reduced in quantum mechanics to that of one particle, as can be done in classical mechanics. Normally, this problem is one of an electron orbiting around, or being affected by, a positive atomic core or a scattering center with the interaction governed by the Coulomb potential. In general, this is a multi-dimensional problem, and not one of the simpler one-dimensional problems with which we have been concerned in the previous chapters. Once we begin to treat multiple dimensions, then state degeneracy begins to arise more frequently, and the most common problem treated is that of the hydrogen atom vs. the helium atom. These extra dimensions provide more degrees of freedom and more complexity. In the hydrogen atom, there is only a single electron, and the problem is solvable. On the other hand, adding a second electron for the helium atom makes the problem extremely difficult to solve, as it has become a three-body problem in which the interactions are difficult to handle.