Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

Physics for IIT JAM, UGC - NET, CSIR NET

Created by: Akhilesh Thakur

Physics : Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

The document Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev is a part of the Physics Course Physics for IIT JAM, UGC - NET, CSIR NET.
All you need of Physics at this link: Physics

Schr¨odinger equation in three dimensions, central potential 

The knowledge we gained on angular momentum is particularly useful when treating real life problems. As our world is three dimensional, we need to generalize the treatment of Schro¨dinger equation to 3-d.
The time-independent Schro¨dinger equation becomes

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(56)

where, in 3-d,

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(57)

In many problems in physics, the potential is central, namely, V = V (r); this means that the potential is spherically symmetric, and is not a function of θ or φ. In this type of systems - the best representative may be the hydrogen atom to be discussed shortly, it is best to work in spherical coordinates, r, θ, φ.

In spherical coordinates, the laplacian becomes

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

Comparing to Equation 17, we see that the last two terms of the laplacian are equal toOrbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRevThus, we can write the Hamiltonian as

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

and the time-independent Schro¨dinger equation is

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

In order to proceed, we note that all of the angular momentum operators: Lx, Ly , Lz and Ldo not operate on the radial variable, r; this can be seen directly by their description in spherical coordinates, equation 17. This means that all these operators commute with V (r): [Lz , V (r)] = 0, etc. Furthermore, since Lx, Ly and Lz commute with L2, we conclude that all of the angular momentum operators commute with the Hamiltonian,

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

This means that it is possible to obtain solutions to Schro¨dinger equation (Equation 60) which are common eigenfunctions of Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev , L2 and Lz .

We already know simultaneous eigenfunctions of L2 and Lz : these are of course the spherical harmonics, Ylm(θ, φ). Thus, a full solution to Schro¨dinger equation can be written as
Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(62)

REl (r) is a radial function of r, which we need to find. The subscripts E and l mark the fact that in general, we obtain different functions for different values of the energy(E ) and the orbital angular momentum quantum number l. It is independent, though, on the magnetic quantum number m, as can be seen by inserting this solution into Schro¨dinger equation (in which the operator L2 appears explicitly, but not Lz ).

We may put the solution in Equation (62) in Schro¨dinger equation (60), and use the fact that  Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(Equation 25), to obtain an equation for REl (r),

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

To be physically acceptable, the wave functions must be square integrable, and normalized to 1:

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(64)

.We already know that the spherical part, Ylm(θ, φ) is normalized; see Equation 42. Thus, the radial part of the eigenfunctions must satisfy the normalization condition

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(65)

We may further simplify Equation 63 by changing a variable,

uEl (r) = rREl (r) (66)

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

where

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(68)
is an effective potential; in addition to the interaction potential, V (r) it contains a repulsive centrifugal barrier,  Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

With the inclusion of this potential, Equation 67 has an identical form to the 1-d (timeindependent) Schro¨dinger equation. The only difference is that it is physically meaningful only for r > 0, and we must provide the boundary condition at r = 0. The boundary conditions are provided by the physical requirement that the function REl (r) remains finite at the origin, r = 0. Since REl (r) = uEl (r)/r, this implies

uEl (0) = 0. (69)

 

 The hydrogen atom

Perhaps the most important demonstration of the above analysis (and of quantum mechanics in general) is the ability to predict the energy levels and wave functions of the hydrogen atom. This is the simplest atom, that contains one proton and one electron. The proton is heavy (mp/me = 1836) and is essentially motionless - we can assume it being at the origin, r = 0). The proton has a positive charge +q, and the electron a negative charge, −q. We can therefore use Coulumb’s law to calculate the potential energy:

 Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(70)

(in SI units).
The radial equation (67) becomes:

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(71)

Before proceeding to solve this equation, we note the following. At r → ∞, Vef f (r) → 0.
This means that for any value of positive energy (E > 0), one could find an acceptable eigenfunction uEl (r). Therefore, there is a continuous spectrum for E > 0, describing scattering between electron and proton (this will be dealt with in next year’s QM...). We focus here on solutions for which E < 0. These are called bound states.
We proceed by some change of variables: We write

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(72)

(note that E < 0, and so κ is real). Equation 71 becomes

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(73)

We next introduce

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(74)
and write Equation 73 as

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(75)

We proceed along lines which are somewhat similar to those taken in deriving the SHO.
We begin by examining the asymptotic behavior. We note that when ρ → ∞, Equation 75 is approximately d2uEl /dρ2 ≈ uEl , which admits the general solution

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

 However, in order for uEl to remain finite as ρ → ∞, we must demand B = 0, implying Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

 On the other hand, as ρ → 0, the centrifugal term dominates (apart when l = 0, but, as will be seen, the result is valid there too), and one can write

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

with the general solution

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

Again, the second term, ρ−l diverges as ρ → 0, implying that we must demand D = 0. Thus,

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

for small ρ.
This discussion motivates another change of variables, writing

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev          (76)

where v(ρ) ≡ vEl (ρ), and the subscripts E l are omitted for clarity.
With this change of variables, we have

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

and

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

 and Equation 75 becomes 

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(77)

We search for solution v(ρ) is terms of power series:

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(78)

We can thus write:

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(79)

and

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(80)

 similar to the analysis of the SHO, we insert these results into Equation 77, and equate the coefficients of each individual power law of ρ to write:

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

Similar to the SHO case, we note that for Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev we have

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

which gives Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRevthis of course is unacceptable, as it diverges at large ρ.

This means that the series must terminate, namely there is a maximum integer, jmax for which Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev. From Equation 81 this gives

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(82)

We can now define the principle quantum number, n via n ≡ jmax + l + 1. (83)

Thus, ρ0 = 2n. But ρdetermines the energy via equations 72 and 74:

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(84)

We therefore conclude that the allowed energies are

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(85)

This is the well-known Bohr’s formula.
Using again equation 74, one finds

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(86)

where

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(87)

is known as the Bohr’s radius.
The ground state (namely, the state of lowest energy) is obtained by putting n = 1 in Equation 85. Putting the values of the physical constants, one finds that

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(88)

This is the binding energy of the hydrogen atom - the amount of energy one needs to give to the electron in the hydrogen atom that is in its ground state to ionize the atom (=release the electron).
Furthermore, E2 = E1/22 = E1/4 = −3.4 eV, etc.

 

The wavefunctions 

Returning to Equation 62, the wavefunctions are given by

ψ(r, θ, φ) = REl (r)Ylm(θ, φ).

where (using Equations 66 and 76)

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(89)

Here, v(ρ) is given by the polynomial of degree jmax = n − l − 1 (see Equation 83).

In the ground state, n = 1; Equation 83 forces l = 0 and jmax = 0. Since l = 0, we known that m = 0 as well (see Equation 35). This means that the wave function is given by

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(90)

Using the recursion formula (Equation 81), with j = 0, leads to c1 = 0; that is, v(ρ) = 0 is simply a constant. This implies that

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(91)

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

The next energy level is n = 2, which represents the first excited state. There are, in fact four different states with this same energy: one state with l = 0, in which case also m = 0; and l = 1, in which case m = −1, 0, +1. For l = 0, Equation 81 gives c1 = −c0, c2 = 0, namely v(ρ) = c0(1 − ρ). This implies

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(93)

For l = 1, the recursion formula terminates the series after a single term, v(ρ) is constant, and one finds

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(94)
and so on.
In fact, one can write

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(95)
where

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(96)

is an associated Laguerre polynomial, and

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(97)

is known as the qth Laguerre polynomial. I list in table 5.1 the first few radial eigenfunctions of the hydrogen.

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

 Table 2: The first few radial wavefunctions for the hydrogen, REl (r).
According to the standard interpretation of the wavefunction, the quantity

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

represents the probability of finding the electron in the volume element Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev , when the system is in the stationary state specified by the quantum numbers (n, l, m).

 Since ψ(r, θ, φ) = REl (r)Ylm(θ, φ) (see Equation 62), the position probability density |ψnlm(r, θ, φ)|2 is composed of a radial part that depends only on r, and an angular part that depends only on θ (recall that the dependence on φ disappears).

We can write the radial part as

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(99)

which is known as the radial distribution function. In figure 2 I plot the first few radial functions REl and the radial distribution function.

Finally, in figures 3 – 5 I give a few examples of the full probability density of finding the electron in (r, θ) for a hydrogen atom, namely

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev(100)

for few values of the quantum numbers n, l, and m. As is obvious from the discussion above, this probability is independent on φ, but only on r and θ.

As a final remark, I would add that in the usual spectroscopic notation the quantum number l is replaced by a letter, according to the following table: Thus, the energy levels

 

Orbital Angular Momentum, Hydrogen Atom (Part - 2) - Angular Momentum, CSIR-NET Physical Sciences Physics Notes | EduRev

are denoted by two symbols: the first is the principal quantum number n, and the second is a letter corresponding to l.
The ground state (n = 1) is denoted by 1s; The first excited state (n = 2) contains one 2s state, and three 2p state, corresponding to m = −1, 0, +1 - so total 4 states; the second excited state contains one 3s state, three 3p states and five 3d states, with m = −2, −1, 0, +1, +2, so total of 9 states; etc.

Dynamic Test

Content Category

Related Searches

MCQs

,

Free

,

Sample Paper

,

Objective type Questions

,

Orbital Angular Momentum

,

CSIR-NET Physical Sciences Physics Notes | EduRev

,

CSIR-NET Physical Sciences Physics Notes | EduRev

,

Important questions

,

practice quizzes

,

Summary

,

Viva Questions

,

Previous Year Questions with Solutions

,

study material

,

Orbital Angular Momentum

,

Exam

,

Extra Questions

,

past year papers

,

Hydrogen Atom (Part - 2) - Angular Momentum

,

shortcuts and tricks

,

mock tests for examination

,

CSIR-NET Physical Sciences Physics Notes | EduRev

,

Orbital Angular Momentum

,

ppt

,

video lectures

,

Hydrogen Atom (Part - 2) - Angular Momentum

,

pdf

,

Hydrogen Atom (Part - 2) - Angular Momentum

,

Semester Notes

;