Before a solution can be found for the important electronic structure of the hydrogen atom, the basis for the description of all the other elements and chemical bonds, first angular momentum must be discussed. Like in the classical Newtonian 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.
The old Newtonian physics defines angular momentum as the vectorial product , where is the position of the particle in question and is its linear momentum.
Following the Newtonian analogy, quantum mechanics substitutes
the gradient operator for the linear momentum,
so the angular momentum operator becomes:
Key Points
- The angular momentum operator (4.18) has been identified.
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 -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 -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 -axis horizontal, (though not in [25] or [51]), and that is what will be done here.
Things further simplify greatly if you switch from Cartesian coordinates , , and to “spherical coordinates” , , and , as shown in figure 4.7. The coordinate is the distance from the chosen origin, is the angular position away from the chosen -axis, and is the angular position around the -axis, measured from the chosen -axis.
In terms of these spherical coordinates, the -component of angular
momentum simplifies to:
In any case, with a bit of thought, it clearly makes sense: the -component of linear momentum classically describes the motion in the direction of the -axis, while the -component of angular momentum describes the motion around the -axis. So if in quantum mechanics the linear momentum is times the derivative with respect the coordinate along the -axis, then surely the logical equivalent for angular momentum is times the derivative with respect to the angle around the -axis?
Anyway, the eigenfunctions of the operator above turn out to be
exponentials in . More precisely, the eigenfunctions are
of the form
Note further that the orbital momentum is associated with a particle whose mass is also indicated by . This book will more specifically indicate the magnetic quantum number as if confusion between the two is likely.
The above solution is easily verified directly, and the eigenvalue
identified, by substitution into the eigenvalue
problem using the
expression for above:
It follows that every eigenvalue is of the form:
Compare that with the linear momentum component which can take on any value, within the accuracy that the uncertainty principle allows. can only take discrete values, but they will be precise. And since the -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 -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 -direction. In terms of quantum mechanics, that means that the value of does not affect the energy. (Actually, this is not exactly true, although it is true to very high accuracy. The electron and nucleus have magnetic fields that give them inherent directionality. It remains true that the -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.38}.)
Key Points
- Even if the physics that you want to describe has no preferred direction, you usually need to select some arbitrary -axis to do the mathematics of quantum mechanics.
- Spherical coordinates based on the chosen -axis are needed in this and subsequent analysis. They are defined in figure 4.7.
- The operator for the -component of angular momentum is (4.19), where is the angle around the -axis.
- The eigenvalues, or measurable values, of angular momentum in any arbitrary direction are whole multiples , possibly negative, of .
- The whole multiple is called the magnetic quantum number.
If the angular momentum in a given direction is a multiple of 1.054,57 10 J s, then should have units of angular momentum. Verify that.
What is the magnetic quantum number of a macroscopic, 1 kg, particle that is encircling the -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.
Actually, based on the derived eigenfunction, , would any macroscopic particle ever be at a single magnetic quantum number in the first place? In particular, what can you say about where the particle can be found in an eigenstate?
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,
. The square root is awkward,
though; it is easier to work with the square angular momentum:
Like the operator of the previous subsection, can be
written in terms of spherical coordinates. To do so, note first that
The solution to the problem may be summarized as follows. First, the
nonuniqueness is removed by demanding that the eigenfunctions are
also eigenfunctions of , the operator of angular
momentum in the -direction. This makes the problem solvable,
{D.14}, and the resulting eigenfunctions are called
the spherical harmonics
.
The first few are given explicitly in table 4.2. In case you
need more of them for some reason, there is a generic expression
(D.5) in derivation {D.14}.
These eigenfunctions can additionally be multiplied by any arbitrary
function of the distance from the origin . They are
normalized to be orthonormal integrated over the surface of the unit
sphere:
What to say about them, except that they are in general a mess? Well, at least every one is proportional to , as an eigenfunction of should be. More importantly, the very first one, is independent of angular position compared to the origin (it is the same for all and 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.
There is a different way of looking at the angular momentum
eigenfunctions. It is shown in table 4.3. It shows that
is always a polynomial in the position component of degree
. Furthermore, you can check that
0: the Laplacian of is 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
harmonicindicates a function whose Laplacian 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 has an angular momentum in the
-direction
(4.25) |
(4.26) |
The azimuthal quantum number is at least as large as the magnitude of
the magnetic quantum number :
(4.27) |
Key Points
- The operator for square angular momentum is (4.22).
- The eigenfunctions of both square angular momentum and angular momentum in the chosen -direction are called the spherical harmonics .
- 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-intuitive form where is a nonnegative integer, one of 0, 1, 2, 3, ..., and is called the azimuthal quantum number.
- The azimuthal quantum number is always at least as big as the absolute value of the magnetic quantum number .
The general wave function of a state with azimuthal quantum number and magnetic quantum number is , where is some further arbitrary function of . Show that the condition for this wave function to be normalized, so that the total probability of finding the particle integrated over all possible positions is one, is that
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?
What is the minimum amount that the total square angular momentum is larger than just the square angular momentum in the -direction for a given value of ?
Rephrasing the final results of the previous subsection, if there is nonzero angular momentum, the angular momentum in the -direction is always less than the total angular momentum. There is something funny going on here. The -direction can be chosen arbitrarily, and if you choose it in the same direction as the angular momentum vector, then the -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 , has a definite value, then the components in both the and directions will be uncertain. (Details will be given in chapter 12.2). The wave function will be in a state where and have a range of possible values , , ..., each with some probability. Without definite and 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 -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 momentum vector will never exist. If one component of angular momentum has a definite value, then the other two components will be uncertain.