Subsections


4.2 Angular Momentum

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.


4.2.1 Definition of angular momentum

The old Newtonian physics defines angular momentum $\vec{L}$ as the vectorial product ${\skew0\vec r}$ $\times$ ${\skew0\vec p}$, where ${\skew0\vec r}$ is the position of the particle in question and ${\skew0\vec p}$ is its linear momentum.

Following the Newtonian analogy, quantum mechanics substitutes the gradient operator $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$ for the linear momentum, so the angular momentum operator becomes:

\begin{displaymath}
{\skew 4\widehat{\vec L}}= \frac{\hbar}{{\rm i}} {\skew 2\...
...rtial}{\partial y},
\frac{\partial}{\partial z}
\right) %
\end{displaymath} (4.18)

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
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The angular momentum operator (4.18) has been identified.


4.2.2 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 [25] or [51]), and that is what will be done here.

Figure 4.7: Spherical coordinates of an arbitrary point P.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...(0,0){$\phi$}}
\put(40,120){\makebox(0,0){P}}
\end{picture}
\end{figure}

Things further simplify greatly if you switch from Cartesian coordinates $x$, $y$, and $z$ to “spherical coordinates” $r$, $\theta$, and $\phi$, as shown in figure 4.7. The coordinate $r$ is the distance from the chosen origin, $\theta$ is the angular position away from the chosen $z$-​axis, and $\phi$ 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 momentum simplifies to:

\begin{displaymath}
\fbox{$\displaystyle
\L_z \equiv \frac{\hbar}{{\rm i}}
\frac{\partial}{\partial\phi}
$} %
\end{displaymath} (4.19)

This can be verified by looking up the gradient operator $\nabla$ in spherical coordinates in [40, pp. 124-126] and then taking the component of ${\skew0\vec r}\times\nabla$ 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 $\hbar$$\raisebox{.5pt}{$/$}$${\rm i}$ times the derivative with respect the coordinate $z$ along the $z$-​axis, then surely the logical equivalent for $z$ angular momentum is $\hbar$$\raisebox{.5pt}{$/$}$${\rm i}$ times the derivative with respect to the angle $\phi$ around the $z$-​axis?

Anyway, the eigenfunctions of the operator $\L _z$ above turn out to be exponentials in $\phi$. More precisely, the eigenfunctions are of the form

\begin{displaymath}
C(r,\theta) e^{{\rm i}m \phi} %
\end{displaymath} (4.20)

Here $m$ is a constant and $C(r,\theta)$ can be any arbitrary function of $r$ and $\theta$. For historical reasons, the number $m$ is called the “magnetic quantum number”. 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 $\ldots,-2,-1,0,1,2,3,\ldots$ The reason is that if you increase the angle $\phi$ by $2\pi$, you make a complete circle around the $z$-​axis and return to the same point. Then the eigenfunction (4.20) must again be the same, but that is only the case if $m$ is an integer. To verify this, use the Euler formula (2.5).

Note further that the orbital momentum is associated with a particle whose mass is also indicated by $m$. This book will more specifically indicate the magnetic quantum number as $m_l$ if confusion between the two is likely.

The above solution is easily verified directly, and the eigenvalue $L_z$ identified, by substitution into the eigenvalue problem $\L _zCe^{{{\rm i}}m\phi}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $L_zCe^{{{\rm i}}m\phi}$ using the expression for $\L _z$ above:

\begin{displaymath}
\frac{\hbar}{{\rm i}} \frac{\partial C e^{{\rm i}m\phi}}{\...
...rm i}} {\rm i}m C e^{{\rm i}m\phi}
= L_z C e^{{\rm i}m\phi}
\end{displaymath}

It follows that every eigenvalue is of the form:

\begin{displaymath}
\fbox{$\displaystyle
L_z = m_l\hbar \mbox{ for $m_l$\ an integer}
$} %
\end{displaymath} (4.21)

So the angular momentum in a given direction cannot just take on any value: it must be a whole multiple $m_l$, (possibly negative), of Planck's constant $\hbar$.

Compare that with the linear momentum component $p_z$ which can take on any value, within the accuracy that the uncertainty principle allows. $L_z$ 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 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.38}.)


Key Points
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Even if the physics that you want to describe has no preferred direction, you usually need to select some arbitrary $z$-​axis to do the mathematics of quantum mechanics.

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Spherical coordinates based on the chosen $z$-​axis are needed in this and subsequent analysis. They are defined in figure 4.7.

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The operator for the $z$-​component of angular momentum is (4.19), where $\phi$ is the angle around the $z$-​axis.

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The eigenvalues, or measurable values, of angular momentum in any arbitrary direction are whole multiples $m$, possibly negative, of $\hbar$.

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The whole multiple $m$ is called the magnetic quantum number.

4.2.2 Review Questions
1.

If the angular momentum in a given direction is a multiple of $\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.054,57 10$\POW9,{-34}$ J s, then $\hbar$ 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, $C(r,\theta)e^{{\rm i}{m}\phi}$, 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?

Solution angub-c


4.2.3 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, $\sqrt{\vec{L}\cdot\vec{L}}$. The square root is awkward, though; it is easier to work with the square angular momentum:

\begin{displaymath}
L^2 \equiv \vec L \cdot \vec L
\end{displaymath}

This subsection discusses the $\L ^2$ operator and its eigenvalues.

Like the $\L _z$ operator of the previous subsection, $\L ^2$ can be written in terms of spherical coordinates. To do so, note first that

\begin{displaymath}
{\skew 4\widehat{\vec L}}\cdot {\skew 4\widehat{\vec L}}=
...
...ew0\vec r}\cdot (\nabla \times ({\skew0\vec r}\times \nabla))
\end{displaymath}

(That is the basic vector identity (D.2) with vectors ${\skew0\vec r}$, $\nabla$, and ${\skew0\vec r}\times\nabla$.) Next look up the gradient and the curl in [40, pp. 124-126]. The result is:
\begin{displaymath}
\L ^2 \equiv
-\frac{\hbar^2}{\sin\theta}
\frac{\partia...
...\hbar^2}{\sin^2\theta}
\frac{\partial^2}{\partial \phi^2} %
\end{displaymath} (4.22)

Obviously, this result is not as intuitive as the $\L _z$ 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.


Table 4.2: The first few spherical harmonics.
\begin{table}\begin{displaymath}
\renewedcommand{arraystretch}{2.8}
\begin{a...
...\phi}
\\ [5pt]\hline\hline
\end{array}
\end{displaymath}
\end{table}


The solution to the problem may be summarized as follows. First, the nonuniqueness is removed by demanding that the eigenfunctions are also eigenfunctions of $\L _z$, the operator of angular momentum in the $z$-​direction. This makes the problem solvable, {D.14}, and the resulting eigenfunctions are called the spherical harmonics $Y_l^m(\theta,\phi)$. 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 $r$. They are normalized to be orthonormal integrated over the surface of the unit sphere:

\begin{displaymath}
\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}
Y_l^m(\theta,\...
...line m}$} \\
0 & \mbox{otherwise}
\end{array}
\right. %
\end{displaymath} (4.23)

The spherical harmonics $Y_l^m$ are sometimes symbolically written in “ket notation” as $\big\vert l\:m\big\rangle $.

What to say about them, except that they are in general a mess? Well, at least every one is proportional to $e^{{{\rm i}}m\phi}$, as an eigenfunction of $\L _z$ should be. More importantly, the very first one, $Y^0_0$ is independent of angular position compared to the origin (it is the same for all $\theta$ and $\phi$ 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.


Table 4.3: The first few spherical harmonics rewritten.
\begin{table}\begin{displaymath}
\renewedcommand{arraystretch}{2.8}
\begin{a...
...}y)^2
\\ [5pt]\hline\hline
\end{array}
\end{displaymath}
\end{table}


There is a different way of looking at the angular momentum eigenfunctions. It is shown in table 4.3. It shows that $r^lY_l^m$ is always a polynomial in the position component of degree $l$. Furthermore, you can check that $\nabla^2r^lY_l^m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0: the Laplacian of $r^lY_l^m$ 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

\begin{displaymath}
{\cal Y}_l^m \equiv r^l Y_l^m %
\end{displaymath} (4.24)

and referred to as the “harmonic polynomials.” In general the term harmonic indicates a function whose Laplacian $\nabla^2$ 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 $Y_l^m$ has an angular momentum in the $z$-​direction

\begin{displaymath}
L_z = m\hbar
\end{displaymath} (4.25)

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
\begin{displaymath}
\fbox{$\displaystyle L^2 = l (l+1) \hbar^2 $}
\end{displaymath} (4.26)

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 $l^2\hbar^2$?) 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 $m$:

\begin{displaymath}
\fbox{$\displaystyle l \mathrel{\raisebox{-1pt}{$\geqslant$}}\vert m\vert $}
\end{displaymath} (4.27)

The reason is that $\L ^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\L _x^2+\L _y^2+\L _z^2$ must be at least as large as $\L _z^2$; in terms of eigenvalues, $l(l+1)\hbar^2$ must be at least as large as $m^2\hbar^2$. As it is, with $l$ $\raisebox{-.5pt}{$\geqslant$}$ $\vert m\vert$, either the angular momentum is completely zero, for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, or $L^2$ is always greater than $L_z^2$.


Key Points
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The operator for square angular momentum is (4.22).

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The eigenfunctions of both square angular momentum and angular momentum in the chosen $z$-​direction are called the spherical harmonics $Y_l^m$.

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If a particle has no angular momentum around the origin, it can be found at all angular locations relative to it with equal probability.

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The eigenvalues for square angular momentum take the counter-​intuitive form $L^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l(l+1)\hbar^2$ where $l$ is a nonnegative integer, one of 0, 1, 2, 3, ..., and is called the azimuthal quantum number.

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The azimuthal quantum number $l$ is always at least as big as the absolute value of the magnetic quantum number $m$.

4.2.3 Review Questions
1.

The general wave function of a state with azimuthal quantum number $l$ and magnetic quantum number $m$ is $\Psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R(r)Y_l^m(\theta ,\phi)$, where $R(r)$ is some further arbitrary function of $r$. 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

\begin{displaymath}
\int_{r=0}^\infty R(r)^* R(r) r^2 {\,\rm d}r = 1.
\end{displaymath}

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


4.2.4 Angular momentum uncertainty

Rephrasing the final results of the previous subsection, if there is nonzero angular momentum, the angular 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 $L_x$ and $L_y$ have a range of possible values $m_1\hbar$, $m_2\hbar$, ..., 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
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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.