Let us consider a particle constrained to move in a circular path with a fixed radius
By convention, the particle is rotating in the xy-plane
The situation is similar to that of a particle in a box. The particle is free to move around the ring at the fixed radius ( r ) since the potential energy is 0 along the path, but it cannot escape the ring as the potential energy outside is infinite
We shall construct the Schrödinger equation for this motion in spherical coordinates
The radius of rotation must be constant, so the two partial derivatives with respect to radius must disappear. Moreover, the system is on the x-y plane, θ is always at 90° and unchanging, the derivatives with respect to θ will become 0 and the sin(θ) will become 1
The ring ensures that the particle experience zero potential energy along the circular path, but infinite potential energy outside it
V(r,θ,ϕ)={0∞along the ringoutside of the ring⟹V={0×∞×along the ringoutside of the ring
Hence, we will have two Schrödinger equations, one for the path along the ring and one for the space outside the ring
along the ring[−2mℏ2(r21∂ϕ2∂2)+0]Ψ(ϕ)=E×Ψ(ϕ)outside of the ring[−2mℏ2(r21∂ϕ2∂2)+∞]Ψ(ϕ)=E×Ψ(ϕ)
Just like the particle in a box situation, to satisfy the Schrödinger equation outside the ring, the value of the wavefunction outside the ring must be equal to zero
Hence, we will only be focusing on solving the Schrödinger equation inside the ring
After write it in its normal form simplification, it should be apparent that the Schrödinger equation here is nothing but a second-order linear homogeneous equation with constant coefficients
dϕ2d2Ψ(ϕ)+ℏ22mr2E
We know that the unknown function must have the form eλϕ , where λ is a suitable constant
Ψ(ϕ)=eλϕ
The characteristic equation of the differential equation is
λ2+(0)λ+ℏ22mr2E=0
Solving the characteristic equation will give us a pair of complex conjugate as root
λ=±−ℏ22mr2E=±iℏ2mr2E
Hence, the general solution is given by the linear combination
Ψ(ϕ)=Aeiℏ2mr2Eϕ+Be−iℏ2mr2Eϕ
To save time writing, we shall define the constant term as m_
mℓ=ℏ2mr2E⟹Ψ(ϕ)=Aeimℓϕ+Be−imℓϕ
The values of the constant A and B in a particular solution depend on how the state of motion of the particle is prepared
If the particle is rotating in the clockwise direction, B = 0 and A is a normalization factor
If the particle is rotating in the counter-clockwise direction, A = 0 and B is a normalization factor
Although the wavefunction appears to be acceptable in the eyes of the Schrödinger equation, we have yet considered the restrictions on the system
Since the particle is in circular motion, it retraces its path after 2π. The continuity and single-valued requirement forces the value of the wavefunction to match at points separated by a complete revolution
Ψ(ϕ)=Ψ(ϕ+2π)
Plugging our newly derived wavefunction into the cyclic boundary condition gives us
Aeimℓϕ+Be−imℓϕ=Aeimℓϕ+2π+Be−imℓϕ+2π
Simplifying the expression and using some exponential properties, we will arrive at
(Aeimℓϕ+Be−imℓϕ)=(Aeimℓϕ+Be−imℓϕ)×eimℓ2π
If we cancel out the terms, we will be left with
1=eimℓ2π
Using Euler's identity, we can express the exponential in terms of sine and cosine
1=cos(2πmℓ)+isin(2πmℓ)
Since 1 has no imaginary part to it, the sine term must be equal to zero and the cosine term must be equal to one. This will occur only when 2πmℓ is an integer multiple of 2π
mℓ×2π=0(2π),±1(2π),±2(2π),±3(2π),⋯
Hence, mℓ can only adopt integer values, making it a quantum number
mℓ=0,±1,±2,±3,⋯
Positive values of mℓ corresponds to rotation in a clockwise sense around the z-axis, while negative values of mℓ corresponds to rotation in a counter-clockwise sense around the z-axis
The wavefunctions can therefore be written in the form eimℓϕ, where mℓ is allowed to take either positive or negative values (or 0)
Ψ(ϕ)=Neimℓϕ
N is the normalization constant
The determination of the wavefunction is not yet complete as it still needs to be normalized
Due to the cyclic nature of this situation, the interval of the inner product integral will be between 0 and 2π
Both the cosine and sine part are acceptable solution to the Schrödinger equation. We can therefore interpret the cosine and sine part are wavefunctions that describe rotation in the positive and negative direction respectively
Cosine function as an acceptable wavefunction
Ψ+(ϕ)=π1cos(∣mℓ∣ϕ)
Sine function as an acceptable wavefunction
Ψ−(ϕ)=π1sin(∣mℓ∣ϕ)
The wavefunction shows that the particle will have a standing wave-like behavior when confined within the ring. But note that the case in which mℓ=0 is not allowed for the sine function since the particle vanishes everywhere and is untainable
The linear representation will be as such
The cyclic representation will be as such
If we are only concern with the relative sign of the wavefunction, we can represent it as such
Since the potential energy along the ring is zero, the particle's rotational kinetic energy is equal to the total energy
We can determine the eigenvalue energy through some simple substitution
mℓ=ℏ2mr2E⟹E=2mr2mℓ2ℏ2
mr2 is nothing but the moment of inertia of the system, I
E=2Imℓ2ℏ2
The expression of the eigenvalue energy has important implicantions on the meaning of the quantum number m_
The occurence of mℓ as its square implies that the energy of rotation is independent of the direction of rotation ( sing of mℓ ). Hence, states with a non-zero value of mℓ are two-fold degenerate
The state described by mℓ=0 is non-degenerate and has an energy equal to zero, which means that there is no zero-point energy in this situation
This can easily be rationalized by the fact that all the sine and cosine wavefunctions of the same mℓ for all mℓ>0 have identical form. Such symmetry results in the two-fold degeneracy
The angular momentum operator can be determined by revisiting classical physics
Since we are working with a particle rotating on the x-y plane, its total angular angular momentum is given by the angular momentum along the z-axis, which can be defined as such in classical physics
Lz=xpy−ypx
Substituting the linear momentum and coordinate with their respective operators, and we will obtain the expression for the angular momentum operator along the z-axis
Now that we have the angular momentum operator along the z-axis, we can find the angular momentum with ease
Due to the spherical symmetry of the system, the system, and thus the Hamiltonian operator, is invariant under rotation. In other words, the Hamiltonian commutes with the angular momentum operator, [H,Lz]=0, and so the wavefunction is an eigenfunction of the angular momentum operator
LzΨ(ϕ)=LzΨ(ϕ)
Inserting the wavefunction and angular momentum operator along the z-axis we derived into the equation will give us
−iℏ∂ϕ∂[2π1eimℓϕ]=Lz[2π1eimℓϕ]
After performing the differentiation, we are left with
−i2mℓℏ[2π1eimℓϕ]=Lz[2π1eimℓϕ]
After some simplification, we will get the expression
Lz=mℓℏ
The fact that the angular momentum is directly proportional to mℓ reinforces a few key concept
The sign of mℓ determines the direction of rotation as it directly influences the sign of the angular momentum
The speed of rotation depends on the magnitude of m_
There are some new implication now that we related angular momentum and m_
Since mℓ is quantized, the angular momentum must also be quantized
Unlike in classical physics, the quantized angular momentum does not depend on the the mass or moment of inertia
Since the quantized values of angular momentum depend on mℓ , but not mℓ^2, every wavefunction has its own characteristic value of angular momentum. Degenerates states will have the same magnitude of angular momentum, but opposite sign
Since the number of nodes increases with mℓ, an increase in the number of nodes is related to an increase in magnitude of the angular momentum
Let us consider a particle constrained to move anywhere on the surface of a sphere with a fixed radius
The sphere can be thought of as a three-dimensional stacking of rings
The situation is similar to that of a particle in a box. The particle is free to move on the surface of the sphere since the potential energy is 0 along the path, but it cannot escape the sphere as the potential energy outside is infinite
The sphere ensures that the particle experience zero potential energy along its surface, but infinite potential energy outside of it
V(r,θ,ϕ)={0∞on the surface of the sphereoutside of the surface of the sphere⟹V={0×∞×on the surface of the sphereoutside of the surface of the sphere
Hence, we will have two Schrödinger equations, one for the area on the surface of the sphere and one for the space outside the surface of the sphere
on the surface of the sphere[2mr2−ℏ2(sin2(θ)1∂ϕ2∂2+sin(θ)1∂θ∂sin(θ)∂θ∂)+0]Ψ(θ,ϕ)=E×Ψ(θ,ϕ)outside of the surface of the sphere[2mr2−ℏ2(sin2(θ)1∂ϕ2∂2+sin(θ)1∂θ∂sin(θ)∂θ∂)+∞]Ψ(θ,ϕ)=E×Ψ(θ,ϕ)
Just like before, to satisfy the Schrödinger equation outside the ring, the value of the wavefunction outside the ring must be equal to zero. Hence, we will only be focusing on solving the Schrödinger equation inside the ring
The wavefunction only has two independent variables, θ and ϕ. These two angles do not depend on each other, so we can separate the wavefunction
Ψ(θ,ϕ)=Θ(θ)×Φ(ϕ)
Θ(θ) is a function that varies with θ only, and is independent of ϕ
Φ(ϕ) is a function that varies with ϕ only, and is independent of θ
We can now solve the Schrödinger equation in terms of those two newly defined functions
Since Θ(θ) and Φ(ϕ) do not vary with ϕ and θ respectively, their partial derivatives with those variables will be zero. Using the product rule will give us
Our task is to find physically acceptable solutions for this equation over the range of 0≤θ≤π. For this purpose, it is mathematically convenient to change the independent variable from θ to sin(θ) or cos(θ)
We shall first derive the differential equations with sin(θ) as the independent variable
Using the chain rule, we can change the independent variable of the differential operator from θ to sin(θ)
dθd=dθdsin(θ)×dsin(θ)d=cos(θ)dsin(θ)d
Rewritting the differential equation using the new differential operator
The series we obtained is divergent and thus does not yield an acceptable wavefunction. To avoid this infinity catastrophe is to ensure the series terminate after a finite number of terms. In other words, we are assuming that there is a maximum value of w–2 for which aw–2=0 , we call this value ℓ
The series terminates at ςℓ, so ςℓ+2 must be equal to zero. This is only true if
ςℓ+2=0⟹ℓ(ℓ+1)=ℏ22IE
Since ℓ represents the degree of the polynomial, it can be any non-negative integers
ℓ=0,1,2,3,⋯
We have effectively introduced a new quantum number, ℓ. Its emergence originates from the additional cyclic boundary condition that requires the wavefunction to match itself across the pole of the sphere
Now that we have obtained the new quantum number, ℓ, we are in a position to actually solve for Θ(θ). This can be done by deriving the other variant of the differential equation with cos(θ) as the independent variable
Before we change the variable, we should update our differential equation using our newly found relation
ℓ indicates the maximum power of θ terms and the total order of combination of cos(θ) and sin(θ) terms in the Legendre polynomials
The fact that the associated Lengendre polynomials are the solutions to the equation, it introduces a relation between the two quantum number, mℓ and ℓ
We note that the associated Lengendre polynomials are polynomials of degree "ℓ–mℓ". Therefore, we have the relation
∣mℓ≤ℓ∣
The absolute value of mℓ is always less than or equal to the value of ℓ of the wavefunction
Because mℓ can have integer values from –ℓ to +ℓ ( including 0 ), there are a total of 2ℓ+1 possible values of mℓ for each value of ℓ
mℓ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧ℓℓ−1ℓ−2:−ℓ+2−ℓ+1−ℓmaximum value ofmℓminimum value ofmℓ
Now that we have the solution to both Φ(ϕ) and Θ(θ), we can multiply them together and normalize the wavefunction to get
These wavefunctions does not depend on either the mass of the particle or the radius of the sphere that defines the system
These wavefunctions are so important in quantum mechanics that they are given the special name, spherical harmonics, and are denoted as Υℓ,mℓ(θ,ϕ)
The reason why they are so important is due to the fact that they are fundamental description of waves on spherical surfaces
We should be familiar with the notion of harmonics, even though we have not explicitly stated it in this manner
When we considered the case where the particle is travelling in one-dimensional, we determined that the eigenfunctions are simple trigonometric or exponential functions of x. The trigonometric form is identical to the harmonic amplitude of a standing wave and we can refer to these functions as "linear harmonics"
When we considered the case where the particle is rotating in a ring, we determined that the eigenfunctions can be expressed as simple trigonometric or exponential functions of ϕ. By analogy with the linear motion, we can refer to these functions as "circular harmonics"
Now, when we are considering the case where the particle is free to move on the surface of the sphere, we determined that the eigenfunctions can also be expressed in terms of trigonometric and exponential functions of ϕ and θ. Hence, we refer to these type of functions as spherical harmonics
The spherical harmonics shows that the particle will have a standing wave-like behavior when confined on the surface of the sphere
If we only focus on the relative position of the nodes, we will notice that the number of nodes of the wavefunction increases with the value of ℓ
Wavefunctions with the same absolute value of mℓ yields wavefunction of the same shape, but they are different because they are "pointing" in different directions
Although the spherical harmonics seem to have wildly different shapes, it is important to bear in mind that they are really nothing but spherical wavefunctions
All the perculier geometries are just the result of plotting the wavefunctions on the surface of the sphere
The lobes of the spherical harmonics are identical to the crests and troughs of a sinusodial function
Just like any functions, spherical harmonics can have different signs or phases at different region and in between those regions are the nodes
Apart from the spherical harmonic that corresponds to ℓ=0, all the spherical harmonics are highly antisotropic
Antisotropic means that, for a given radial dependence, the probability distribution is concentrated in the vicinity of certain preferred directions and vanishes along other directions
We shall see later that the antisotropy plays an important role in chemistry
Since the potential energy on the surface of the sphere is zero, the particle's rotational kinetic energy is equal to the total energy
We can determine the eigenvalue energy of the particle by using one of the relation we derived earlier
ℏ22IE=ℓ(ℓ+1)
After some simple rearrangements, we will obtain the expression for the eigenenergy of a particle on a sphere
E=2Iℓ(ℓ+1)ℏ2
The expression of the eigenvalue energy tells us that the energy depends on ℓ, but not mℓ
This is consistent with the fact that the kinetic energy increases with the curvature of the wavefunction
Since it is ℓ, not mℓ ,that determines the number of nodes, and thus the curvature of the wavefunction, it is ℓ that determines the eigenenergy, not mℓ
We can see that each value of mℓ corresponds to one particular wavefunction, but the eigenenergy is independent of mℓ
We can therefore conclude that the number of degeneracy depends on the number of different values of mℓ for a given value of ℓ
Since there are 2ℓ+1 different value of mℓ, and thus 2ℓ+1 different wavefunctions, for each value of ℓ, each energy level is 2ℓ+1-fold degenerate
The degeneracy with respect to mℓ is due to the symmetry of the spherical harmonics
