Abstract
In this chapter, the machinery to deal with single particles is worked out, culminating in the vital solutions for the hydrogen atom and hydrogen molecular ion.
Before the hydrogen atom can be discussed however, first the quantum mechanics of angular momentum needs to be covered. Just like you need angular momentum to solve the motion of a planet around the sun in classical physics, so do you need angular momentum for the motion of an electron around a nucleus in quantum mechanics. The eigenvalues of angular momentum and their quantum numbers are critically important for many other reasons besides the hydrogen atom.
After angular momentum, the hydrogen atom can be discussed. The solution is messy, but fundamentally not much different from that of the particle in the pipe or the harmonic oscillator of the previous chapter.
The hydrogen atom is the major step towards explaining heavier atoms and then chemical bonds. One rather unusual chemical bond can already be discussed in this chapter: that of a ionized hydrogen molecule. A hydrogen molecular ion has only one electron.
But the hydrogen molecular ion cannot readily be solved exactly, even if the motion of the nuclei is ignored. So an approximate method will be used. Before this can be done, however, a problem must be addressed.
The hydrogen molecular ion ground state is defined to be the state of lowest energy. But an approximate ground state is not an exact energy eigenfunction and has uncertain energy. So how should the term “lowest energy” be defined for the approximation?
To answer that, before tackling the molecular ion, first systems with uncertainty in a variable of interest are discussed. The “expectation value” of a variable will be defined to be the average of the eigenvalues,
91
weighted by their probability. The “standard deviation” will be defined as a measure of how much uncertainty there is to that expectation value.
With a precise mathematical definition of uncertainty, the obvious next question is whether two different variables can be certain at the same time. The “commutator” of the two operators will be introduced to answer it. That then allows the Heisenberg uncertainty relationship to be formulated. Not only can position and linear momentum not be certain at the same time; a specific equation can be written down for how big the uncertainty must be, at the very least.
With the mathematical machinery of uncertainty defined, the hydrogen molecular ion is solved last.
4.1 Angular Momentum
Before a solution can be found for the important electronic structure of the hy-drogen atom, the basis for the description of all the other elements and chemical bonds, first angular momentum must be discussed. Like in the classical New-tonian case, angular momentum is essential for the analysis, and in quantum mechanics, angular momentum is also essential for describing the final solution.
Moreover, the quantum properties of angular momentum turn out to be quite unexpected and important for practical applications.
4.1.1 Definition of angular momentum
The old Newtonian physics defines angular momentum ~L as the vectorial prod-uct ~r× ~p, where ~r is the position of the particle in question and ~p is its linear momentum.
Following the Newtonian analogy, quantum mechanics substitutes the gradi-ent operator ¯h∇/i for the linear momentum, so the angular momentum operator becomes:
~b
L = h¯
i~rb× ∇ ~rb≡ (x,b y,b z)b ∇ ≡ ∂
∂x, ∂
∂y, ∂
∂z
!
(4.1) Unlike the Hamiltonian, the angular momentum operator is not specific to a given system. All observations about angular momentum will apply regardless of the physical system being studied.
Key Points
◦ The angular momentum operator (4.1) has been identified.
4.1. ANGULAR MOMENTUM 93
4.1.2 Angular momentum in an arbitrary direction
The intent in this subsection is to find the operator for the angular momentum in an arbitrary direction and its eigenfunctions and eigenvalues.
For convenience, the direction in which the angular momentum is desired will be taken as the z-axis of the coordinate system. In fact, much of the mathematics that you do in quantum mechanics requires you to select some arbitrary direction as your z-axis, even if the physics itself does not have any preferred direction. It is further conventional in the quantum mechanics of atoms and molecules to draw the chosen z-axis horizontal, (though not in [21]
or [43]), and that is what will be done here.
x
y
z r
θ φ
P
Figure 4.1: Spherical coordinates of an arbitrary point P.
Things further simplify greatly if you switch from Cartesian coordinates x, y, and z to “spherical coordinates” r, θ, and φ, as shown in figure 4.1. The coordinate r is the distance from the chosen origin, θ is the angular position away from the chosen z-axis, and φ is the angular position around the z-axis, measured from the chosen x-axis.
In terms of these spherical coordinates, the z-component of angular momen-tum simplifies to:
Lbz ≡ ¯h i
∂
∂φ (4.2)
This can be verified by looking up the gradient operator ∇ in spherical coor-dinates in [35, pp. 124-126] and then taking the component of ~r × ∇ in the z-direction.
In any case, with a bit of thought, it clearly makes sense: the z-component of linear momentum classically describes the motion in the direction of the
z-axis, while the z-component of angular momentum describes the motion around the z-axis. So if in quantum mechanics the z-linear momentum is ¯h/i times the derivative with respect the coordinate z along the z-axis, then surely the logical equivalent for z-angular momentum is ¯h/i times the derivative with respect to the angle φ around the z-axis?
Anyway, the eigenfunctions of the operator Lbz above turn out to be expo-nentials in φ. More precisely, the eigenfunctions are of the form
C(r, θ)eimφ (4.3)
where m is a constant and C(r, θ) can be any arbitrary function of r and θ.
For historical reasons, the number m is called the “magnetic quantum num-ber”. Historically, physicists have never seen a need to get rid of obsolete and confusing terms. The magnetic quantum number must be an integer, one of . . . ,−2, −1, 0, 1, 2, 3, . . . The reason is that if you increase the angle φ by 2π, you make a complete circle around the z-axis and return to the same point.
Then the eigenfunction (4.3) must again be the same, but that is only the case if m is an integer, as can be verified from the Euler formula (2.5).
The above solution is easily verified directly, and the eigenvalue Lz identi-fied, by substitution into the eigenvalue problemLbzCeimφ = LzCeimφ using the expression forLbz above:
¯ h i
∂Ceimφ
∂φ = LzCeimφ =⇒ h¯
iimCeimφ = LzCeimφ It follows that every eigenvalue is of the form:
Lz = m¯h for m an integer (4.4)
So the angular momentum in a given direction cannot just take on any value:
it must be a whole multiple m, (possibly negative), of Planck’s constant ¯h.
Compare that with the linear momentum component pz which can take on any value, within the accuracy that the uncertainty principle allows. Lz can only take discrete values, but they will be precise. And since the z-axis was arbitrary, this is true in any direction you choose.
It is important to keep in mind that if the surroundings of the particle has no preferred direction, the angular momentum in the arbitrarily chosen z-direction is physically irrelevant. For example, for the motion of the electron in an isolated hydrogen atom, no preferred direction of space can be identified. Therefore, the energy of the electron will only depend on its total angular momentum, not on the angular momentum in whatever is completely arbitrarily chosen to be the z-direction. In terms of quantum mechanics, that means that the value of m does not affect the energy. (Actually, this is not exactly true, although it is
4.1. ANGULAR MOMENTUM 95 true to very high accuracy. The electron and nucleus have magnetic fields that give them inherent directionality. It remains true that the z-component of net angular momentum of the complete atom is not relevant. However, the space in which the electron moves has a preferred direction due to the magnetic field of the nucleus and vice-versa. It affects energy very slightly. Therefore the electron and nucleus must coordinate their angular momentum components, addendum {A.32}.)
Key Points
◦ Even if the physics that you want to describe has no preferred di-rection, you usually need to select some arbitrary z-axis to do the mathematics of quantum mechanics.
◦ Spherical coordinates based on the chosen z-axis are needed in this and subsequent analysis. They are defined in figure 4.1.
◦ The operator for the z-component of angular momentum is (4.2), where φ is the angle around the z-axis.
◦ The eigenvalues, or measurable values, of angular momentum in any arbitrary direction are whole multiples m, possibly negative, of ¯h.
◦ The whole multiple m is called the magnetic quantum number.
4.1.2 Review Questions
1 If the angular momentum in a given direction is a multiple of ¯h = 1.054,57 10−34 J s, then ¯h should have units of angular momentum. Verify that.
Solution angub-a
2 What is the magnetic quantum number of a macroscopic, 1 kg, particle that is encircling the z-axis at a distance of 1 m at a speed of 1 m/s? Write out as an integer, and show digits you are not sure about as a question mark.
Solution angub-b
3 Actually, based on the derived eigenfunction, C(r, θ)eimφ, would any macroscopic particle ever be at a single magnetic quantum number in the first place? In par-ticular, what can you say about where the particle can be found in an eigenstate?
Solution angub-c
4.1.3 Square angular momentum
Besides the angular momentum in an arbitrary direction, the other quantity of primary importance is the magnitude of the angular momentum. This is the
length of the angular momentum vector,
q~L· ~L. The square root is awkward, though; it is easier to work with the square angular momentum:
L2 ≡ ~L · ~L
This subsection discusses the Lb2 operator and its eigenvalues.
Like theLbz operator of the previous subsection,Lb2 can be written in terms of spherical coordinates. To do so, note first that, {D.11},
~b
L·L =~b h¯
i(~r× ∇) ·¯h
i(~r× ∇) = −¯h2~r· (∇ × (~r × ∇))
and then look up the gradient and the curl in [35, pp. 124-126]. The result is:
Lb2 ≡ − ¯h2
Obviously, this result is not as intuitive as the Lbz-operator of the previous subsection, but once again, it only involves the spherical coordinate angles.
The measurable values of square angular momentum will be the eigenvalues of this operator. However, that eigenvalue problem is not easy to solve. In fact the solution is not even unique.
Y00=
Table 4.1: The first few spherical harmonics.
The solution to the problem may be summarized as follows. First, the non uniqueness is removed by demanding that the eigenfunctions are also eigenfunc-tions of Lbz, the operator of angular momentum in the z-direction. This makes
4.1. ANGULAR MOMENTUM 97 the problem solvable, {D.12}, and the resulting eigenfunctions are called the
“spherical harmonics” Ylm(θ, φ). The first few are given explicitly in table 4.1.
In case you need more of them for some reason, there is a generic expression (D.3) in derivation {D.12}.
These eigenfunctions can additionally be multiplied by any arbitrary func-tion of the distance from the origin r. They are normalized to be orthonormal integrated over the surface of the unit sphere:
Z π
The spherical harmonics Ylm are sometimes symbolically written in “ket nota-tion” as |l mi.
What to say about them, except that they are in general a mess? Well, at least every one is proportional to eimφ, as an eigenfunction ofLbz should be. More importantly, the very first one, Y00 is independent of angular position compared to the origin (it is the same for all θ and φ angular positions.) This eigenfunction corresponds to the state in which there is no angular momentum around the origin at all. If a particle has no angular momentum around the origin, it can be found at all angular locations relative to it with equal probability.
Y00=
Table 4.2: The first few spherical harmonics rewritten.
There is a different way of looking at the angular momentum eigenfunctions.
It is shown in table 4.2. It shows that rlYlmis always a polynomial in the position component of degree l. Furthermore, you can check that ∇2rlYlm = 0: the Laplacian of rlYlmis always zero. This way of looking at the spherical harmonics
is often very helpful in understanding more advanced quantum topics. These solutions may be indicated as
Ylm ≡ rlYlm (4.7)
and referred to as the “harmonic polynomials.” In general the term “harmonic”
indicates a function whose Laplacian∇2 is zero.
Far more important than the details of the eigenfunctions themselves are the eigenvalues that come rolling out of the analysis. A spherical harmonic Ylm has an angular momentum in the z-direction
Lz = m¯h (4.8)
where the integer m is called the magnetic quantum number, as noted in the previous subsection. That is no surprise, because the analysis demanded that they take that form. The new result is that a spherical harmonic has a square angular momentum
L2 = l(l + 1)¯h2 (4.9)
where l is also an integer, and is called the “azimuthal quantum number” for reasons you do not want to know. It is maybe a weird result, (why not simply l2¯h2?) but that is what square angular momentum turns out to be.
The azimuthal quantum number is at least as large as the magnitude of the magnetic quantum number m:
l≥ |m| (4.10)
The reason is thatLb2 =Lb2x+Lb2y+Lb2z must be at least as large asLb2z; in terms of eigenvalues, l(l + 1)¯h2 must be at least as large as m2¯h2. As it is, with l ≥ |m|, either the angular momentum is completely zero, for l = m = 0, or L2 is always greater than L2z.
Key Points
◦ The operator for square angular momentum is (4.5).
◦ The eigenfunctions of both square angular momentum and angular momentum in the chosen z-direction are called the spherical harmon-ics Ylm.
◦ If a particle has no angular momentum around the origin, it can be found at all angular locations relative to it with equal probability.
◦ The eigenvalues for square angular momentum take the counter-in-tuitive form L2 = l(l + 1)¯h2 where l is a nonnegative integer, one of 0, 1, 2, 3, . . ., and is called the azimuthal quantum number.
4.1. ANGULAR MOMENTUM 99
◦ The azimuthal quantum number l is always at least as big as the absolute value of the magnetic quantum number m.
4.1.3 Review Questions
1 The general wave function of a state with azimuthal quantum number l and mag-netic quantum number m is Ψ = R(r)Ylm(θ, φ), where R(r) is some further ar-bitrary function of r. Show that the condition for this wave function to be nor-malized, so that the total probability of finding the particle integrated over all possible positions is one, is that
Z ∞
r=0R(r)∗R(r)r2dr = 1.
Solution anguc-a
2 Can you invert the statement about zero angular momentum and say: if a particle can be found at all angular positions compared to the origin with equal probability, it will have zero angular momentum?
Solution anguc-b
3 What is the minimum amount that the total square angular momentum is larger than just the square angular momentum in the z-direction for a given value of l?
Solution anguc-c
4.1.4 Angular momentum uncertainty
Rephrasing the final results of the previous subsection, if there is nonzero an-gular momentum, the anan-gular momentum in the z-direction is always less than the total angular momentum. There is something funny going on here. The z-direction can be chosen arbitrarily, and if you choose it in the same direction as the angular momentum vector, then the z-component should be the entire vector. So, how can it always be less?
The answer of quantum mechanics is that the looked-for angular momentum vector does not exist. No axis, however arbitrarily chosen, can align with a nonexisting vector.
There is an uncertainty principle here, similar to the one of Heisenberg for position and linear momentum. For angular momentum, it turns out that if the component of angular momentum in a given direction, here taken to be z, has a definite value, then the components in both the x and y directions will be uncertain. (Details will be given in chapter 12.2). The wave function will be in a state where Lx and Ly have a range of possible values m1¯h, m2¯h, . . ., each with some probability. Without definite x and y components, there simply is no angular momentum vector.
It is tempting to think of quantities that have not been measured, such as the angular momentum vector in this example, as being merely “hidden.” However, the impossibility for the z-axis to ever align with any angular momentum vector shows that there is a fundamental difference between “being hidden” and “not existing”.
Key Points
◦ According to quantum mechanics, an exact nonzero angular momen-tum vector will never exist. If one component of angular momenmomen-tum has a definite value, then the other two components will be uncertain.
4.2 The Hydrogen Atom
This section examines the critically important case of the hydrogen atom. The hydrogen atom consists of a nucleus which is just a single proton, and an electron encircling that nucleus. The nucleus, being much heavier than the electron, can be assumed to be at rest, and only the motion of the electron is of concern.
The energy levels of the electron determine the photons that the atom will absorb or emit, allowing the powerful scientific tool of spectral analysis. The electronic structure is also essential for understanding the properties of the other elements and of chemical bonds.
4.2.1 The Hamiltonian
The first step is to find the Hamiltonian of the electron. The electron experiences an electrostatic Coulomb attraction to the oppositely charged nucleus. The corresponding potential energy is
V =− e2
4πǫ0r (4.11)
with r the distance from the nucleus. The constant
e = 1.6 10−19 C (4.12)
is the magnitude of the electric charges of the electron and proton, and the constant
ǫ0 = 8.85 10−12 C2/J m (4.13) is called the “permittivity of space.”
Unlike for the harmonic oscillator discussed earlier, this potential energy cannot be split into separate parts for Cartesian coordinates x, y, and z. To
4.2. THE HYDROGEN ATOM 101 do the analysis for the hydrogen atom, you must put the nucleus at the origin of the coordinate system and use spherical coordinates r (the distance from the nucleus), θ (the angle from an arbitrarily chosen z-axis), and φ (the angle around the z-axis); see figure 4.1. In terms of spherical coordinates, the potential energy above depends on just the single coordinate r.
To get the Hamiltonian, you need to add to this potential energy the kinetic energy operator T of chapter 3.3, which involves the Laplacian. The Laplacianb in spherical coordinates is readily available in table books, [35, p. 126], and the Hamiltonian is thus found to be:
H =− ¯h2
is the mass of the electron.
It may be noted that the small proton motion can be corrected for by slightly adjusting the mass of the electron to be an effective 9.104,4 10−31kg,{A.4}. This makes the solution exact, except for extremely small errors due to relativistic effects. (These are discussed in addendum {A.32}.)
Key Points
◦ To analyze the hydrogen atom, you must use spherical coordinates.
◦ The Hamiltonian in spherical coordinates has been written down. It is (4.14).
4.2.2 Solution using separation of variables
This subsection describes in general lines how the eigenvalue problem for the electron of the hydrogen atom is solved. The basic ideas are like those used to solve the particle in a pipe and the harmonic oscillator, but in this case, they are used in spherical coordinates rather than Cartesian ones. Without getting too much caught up in the mathematical details, do not miss the opportunity of learning where the hydrogen energy eigenfunctions and eigenvalues come from. This is the crown jewel of quantum mechanics; brilliant, almost flawless, critically important; one of the greatest works of physical analysis ever.
The eigenvalue problem for the Hamiltonian, as formulated in the previous subsection, can be solved by searching for solutions ψ that take the form of
The eigenvalue problem for the Hamiltonian, as formulated in the previous subsection, can be solved by searching for solutions ψ that take the form of