A hydrogenic system is a two-particle system, where the two particles in question are the nucleus and an electron
The nucleus contains two types of particles, protons and neutrons
Electron is much lighter than either proton or neutron
Both proton and electron carries an electric charge
Protons are positively charged and electrons are negatively charged
Their electric charge are of the same magnitude but of different sign
Proton:+qeElectron:−qe
We shall approximate the nucleus to be fixed in space, with the electron moving around it. This approximation is called the Born-Oppenheimer Approximation
The nucleus is a lot more massive than the electrons, so despite having similar kinetic energy, the nucleus moves a lot slower than the electrons. This implies that electrons appear to adjust their position instantaneously to follow the nucleus
The consequence of fixing the nucleus is that the nucleus has no kinetic energy and the Hamiltonian can be simplified to one that only contains the kinetic energy and potential energy operator of the electrons
In other words, we have approximated the wavefunction of the system to be the electronic wavefunction
When we are dealing with the case where the particle is rotating on a sphere, we assumed that the radius is fixed. The same assumption cannot be made since we have no evidence to show that it is true
We shall construct the potential energy operator of the hydrogenic system
There exists a Columbic interaction between the negatively charged electron and the positively charged nucleus. Moreover, the nucleus and electron can be treated as point charges
V(r)=−4πϵ0rZqe2⟹V=−4πϵ0rZqe2×
Z refers to the atomic number, which is the number of positively charged proton
The potential energy depends only on the distance, r, separating the nucleus and the electron, not the angles. The potential energy is therefore centrosymmetric
We can therefore construct the Schrödinger equation for the hydrogenic system
HΨ(r,θ,ϕ)=E×Ψ(r,θ,ϕ)
The Hamiltonian is the sum of the kinetic energy operator and potential energy operator
(KE+V)Ψ(r,θ,ϕ)=E×Ψ(r,θ,ϕ)
Substituting in the expressions of those two operators in the spherical position basis
Since R(r) and A(θ,ϕ) do not vary with the angles and the radius respectively, their partial derivatives with those variables will be zero. Using the product rule will give us
Since there is always an electrostatic force between the nucleus and the electron, the eigenenergy of the system is always negative unless they are infinitely far apart ( r=∞ )
We can see that the equation involves the second derivative of the product of the radial wavefunction, R(r), and its independent variable, r. This is not ideal, so we should define their product as an intermediate function R( r )
R(r)=R(r)×r⟹R(r)=rR(r)
Now, when we substitute this back to the differential equation, we can cancel out the independent variable, r
Unlike the angular wavefunction, the radial wavefunction is a lot trickier to solve
We need to consider how the wavefunction behaves when the electron is far away or close to the nucleus
Knowing how wavefunctions should behave at large and small r values allow us to find asymptotic solutions at both limits
The radial wavefunction is heavily influenced by the potential energy experienced by the electron. >- Hence, if we know how the potential energy changes with the radius, we can have a crude prediction of what the wavefunction looks like
The effective potential energy of the system is given by the sum of the actual potential energy and the rotational kinetic energy
Veffective(r)=V(r)+KErotation(r)
The potential energy term is simply the electrostatic potential energy from classical physics, while the rotational kinetic energy is given by the energy of a particle on a sphere
V(r)=4πϵ0rZqe2KErotation(r)=2mer2ℓ(ℓ+1)ℏ2
To predict what the radial wavefunctions will look like, we shall examine how the effective potential energy behave at different distance from the nucleus
The electrostatic potential energy does not depend on the quantum number ℓ
The rotational kinetic energy goes to infinity as r approaches zero, but its slope becomes less steep as the value of ℓ increases
The effective potential is the sum of the two functions
The effective potential gives us a vague idea of what the radial wavefunction should look like
From the effective potential, we know that the radial wavefunctions for ℓ=0 and ℓ=0 must be very different at r=0. Wavefunctions with ℓ=0 should peak at r=0, while wavefunctions with ℓ=0 should have a value of zero at r=0
The radial wavefunctions for both ℓ=0 and ℓ=0 must both decay to zero, which makes them both square-integrable
We shall start by finding the asymptotic solution when the electron is far away from the nucleus
When the value of r is large, the terms with r as the denominator will approach zero such that
We now have a second-order homogeneous linear differential equation with constant coefficients, so the radial wavefunction at large value of r must have the form eλr , where λ is a suitable constant
r→∞limR(r)=eλr
The characteristic equation of the differential equation is
λ2+(0)λ−ℏ22me∣E∣
Hence, λ must be
λ=±ℏ22me∣E∣=±ℏ2me∣E∣
To save time writing, we can define a quantity, κ, which has a unit of \text{length}^
κ≡ℏ2me∣E∣
The radial wavefunction at large values of r is therefore
r→∞limR(r)=Aeκr+Be−κr
In order for R(r) to vanish at the origin, we need to set A=0
r→∞limR(r)=Be−κr
Physically, such wavefunctions correspond to states in which the electron is not bound to the nucleus. Hence, the energy in the wavefunction only refers to non-negative energy (E≥0)
Just like a free particle, all non-negative energies are allowed, so the energy is continuous instead of quantized. The positive-energy eigenfunctions are therefore called the continuum eigenfunctions
We can now find the asymptotic solution when the electron is close to the nucleus, or the bound state of the electron
When the value of r is small, the terms with r as the denominator will approach infinity. Moreover, the terms with r2 as the denominator will go to infinity more rapidly than the terms with r or r0 as denominator, and will therefore dominate
We now have a second-order homogeneous linear differential equation with a non-constant variable. We can use the power series method to solve it by setting the ordinary point at r = 0
r→0limR(r)=t=0∑∞ξtrt
Plugging it into the equation, we will have
t=2∑∞t(t−1)ξtrt−2−r2ℓ(ℓ+1)t=0∑∞ξtrt=0
Multiplying the coefficient into the series will leave us with
t=2∑∞t(t−1)ξtrt−2−t=0∑∞−ℓ(ℓ+1)ξtrt−2=0
In order to make the index of the two power series the same, we shall first recognize that the term corresponding to t=0 and t=1 in the first series amounts to zero and we can therefore shift the index to t=0 without affecting the value
t=0∑∞t(t−1)ξtrt−2−t=0∑∞−ℓ(ℓ+1)ξtrt−2=0
With both series starting from the same point, we can combine them and perform some simple factorization
t=2∑∞ξtrt−2[t(t−1)−ℓ(ℓ+1)]=0
Since the exponential cannot be zero, this gives us the relation
t(t−1)−ℓ(ℓ+1)=0⟹t2−t−ℓ(ℓ+1)=0
After expanding the expression, we will notice that it is a quadratic equation. Solving for t
The overall function R(r) should be written as a product of the two asymptotic solutions and an unknown function that bridges the two extremes, B(r)
R(r)=r→0rℓ+1×B(r)BridgingFunction×r→∞e−κr
The two asymptotic solutions dictates the behavior of the radial wavefunction at large and small values of r
B(r) will complete the picture by bridging the two ends of the function
To determine what B(ρ) is, we need to construct a differential equation for it. This can be done by substituting this newly defined R(r) into the original differential equation
The only way to avoid this infinity catastrophe is to ensure the series terminate after a finite number of terms such that e–κr can ensure the radial wavefunction goes to zero eventually
B(r) overpowers the asymptotic part of the solution unless its power series representation is truncated
Since we need to terminate the series, this is the same as saying that B(r) should be a finite polynomial of degree N. In other words
μN=0μN+1=0
Because we have a one-step recurrence relation, saying μN+1=0 is the same as saying all the subsequent coefficients are also 0
To see how we can make sure the series terminate when v=N+1, we should revisit the recurrence relation
μNμN+1=(N+1)(2ℓ+2+N)2κ(N+ℓ+1−n)
In order for μN+1 to be zero, the numerator on the right hand side must also become zero
μN+1=2κ(N+ℓ+1−n)=0⟹n=(N+ℓ+1)
Since both N and ℓ are integers, the only way for the equality to hold true is for n to only take on integer values
We have created a new quantum number, n
The smallest value of N and ℓ can get is 0, so the smallest value n can take on is 1
Now that we have both the asymptotic solutions and the bridging solution, we can finally find the radial wavefunction of the system
The radial function, R(r) is related to the intermediate function by
R(r)=rR(r)
R(r) can be written as a product
R(r)=rrℓ+1B(r)e−κr
Finally, we obtain the unnormalized radial wavefunction
R(r)=rℓB(r)e−κr
We can now plot out the graphs of the radial wavefunctions of the hydrogenic system
Since the radial wavefunctions contain the rℓ terms, they all vanishes at the origin except for the ones with ℓ=0. The credit goes to the centrifugal potential, which repels the electrons from coming too close to the nuclei
The number of nodes in the radial wavefunction increases with n and decreases with ℓ. This is because that the truncated bridging function is a polynomial of degree N and a polynomial passes through the x-axis N times. From the recurrence relation we derived above, we know that N=n–ℓ–1, and so the number of nodes in the radial wavefunction is also n–ℓ–1
