Comprehensive Study Notes

In the next section we shall solve the eigenvalue problem for angular momentum, which is a vector property. We therefore first review vectors.

Physical properties (for example, mass, length, energy) that are completely specified by their magnitude are called scalars. Physical properties (for example, force, velocity, momentum) that require specification of both magnitude and direction are called vectors. A vector is represented by a directed line segment whose length and direction give the magnitude and direction of the property.

The sum of two vectors $\mathbf{A}$ and $\mathbf{B}$ is defined by the following procedure: Slide the first vector so that its tail touches the head of the second vector, keeping the direction of the first vector fixed. Then draw a new vector from the tail of the second vector to the head of the first vector. See Fig. 5.1. The product of a vector and a scalar, $c \mathbf{A}$, is defined as a vector of length $|c|$ times the length of $\mathbf{A}$ with the same direction as $\mathbf{A}$ if $c$ is positive, or the opposite direction to $\mathbf{A}$ if $c$ is negative.

FIGURE 5.1 Addition of two vectors.

(a)

(b) $\mathbf{C}=\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}$

FIGURE 5.2 Unit vectors $\mathbf{i}, \mathbf{j}$,
$\mathbf{k}$, and components of $\mathbf{A}$.

To obtain an algebraic (as well as geometric) way of representing vectors, we set up Cartesian coordinates in space. We draw a vector of unit length directed along the positive $x$ axis and call it i. (No connection with $i=\sqrt{-1}$.) Unit vectors in the positive $y$ and $z$ directions are called $\mathbf{j}$ and $\mathbf{k}$ (Fig. 5.2). To represent any vector $\mathbf{A}$ in terms of the three unit vectors, we first slide $\mathbf{A}$ so that its tail is at the origin, preserving its direction during this process. We then find the projections of $\mathbf{A}$ on the $x, y$, and $z$ axes: $A{x}, A{y}$, and $A_{z}$. From the definition of vector addition, it follows that (Fig. 5.2)

\(
\begin{equation}
\mathbf{A}=A{x} \mathbf{i}+A{y} \mathbf{j}+A_{z} \mathbf{k} \tag{5.17}
\end{equation}
\)

We can specify $\mathbf{A}$ by specifying its three components: $\left(A{x}, A{y}, A_{z}\right)$. A vector in threedimensional space can therefore be defined as an ordered set of three numbers.

Two vectors $\mathbf{A}$ and $\mathbf{B}$ are equal if and only if all their corresponding components are equal: $A{x}=B{x}, A{y}=B{y}, A{z}=B{z}$. Therefore a vector equation is equivalent to three scalar equations.

To add two vectors, we add corresponding components:

\(
\mathbf{A}+\mathbf{B}=A{x} \mathbf{i}+A{y} \mathbf{j}+A{z} \mathbf{k}+B{x} \mathbf{i}+B{y} \mathbf{j}+B{z} \mathbf{k}
\)

\(
\begin{equation}
\mathbf{A}+\mathbf{B}=\left(A{x}+B{x}\right) \mathbf{i}+\left(A{y}+B{y}\right) \mathbf{j}+\left(A{z}+B{z}\right) \mathbf{k} \tag{5.18}
\end{equation}
\)

Also, if $c$ is a scalar, then

\(
\begin{equation}
c \mathbf{A}=c A{x} \mathbf{i}+c A{y} \mathbf{j}+c A_{z} \mathbf{k} \tag{5.19}
\end{equation}
\)

The magnitude of a vector $\mathbf{A}$ is its length and is denoted by $A$ or $|\mathbf{A}|$. The magnitude $A$ is a scalar.

The dot product or scalar product $\mathbf{A} \cdot \mathbf{B}$ of two vectors is defined by

\(
\begin{equation}
\mathbf{A} \cdot \mathbf{B}=|\mathbf{A}||\mathbf{B}| \cos \theta=\mathbf{B} \cdot \mathbf{A} \tag{5.20}
\end{equation}
\)

where $\theta$ is the angle between the vectors. The dot product, being the product of three scalars, is a scalar. Note that $|\mathbf{A}| \cos \theta$ is the projection of $\mathbf{A}$ on $\mathbf{B}$. From the definition of vector addition, it follows that the projection of the vector $\mathbf{A}+\mathbf{B}$ on some vector $\mathbf{C}$ is the sum of the projections of $\mathbf{A}$ and of $\mathbf{B}$ on $\mathbf{C}$. Therefore

\(
\begin{equation}
(\mathbf{A}+\mathbf{B}) \cdot \mathbf{C}=\mathbf{A} \cdot \mathbf{C}+\mathbf{B} \cdot \mathbf{C} \tag{5.21}
\end{equation}
\)

Since the three unit vectors $\mathbf{i}, \mathbf{j}$, and $\mathbf{k}$ are each of unit length and are mutually perpendicular, we have

\(
\begin{equation}
\mathbf{i} \cdot \mathbf{i}=\mathbf{j} \cdot \mathbf{j}=\mathbf{k} \cdot \mathbf{k}=\cos 0=1, \quad \mathbf{i} \cdot \mathbf{j}=\mathbf{j} \cdot \mathbf{k}=\mathbf{k} \cdot \mathbf{i}=\cos (\pi / 2)=0 \tag{5.22}
\end{equation}
\)

Using (5.22) and the distributive law (5.21), we have

\(
\begin{gather}
\mathbf{A} \cdot \mathbf{B}=\left(A{x} \mathbf{i}+A{y} \mathbf{j}+A{z} \mathbf{k}\right) \cdot\left(B{x} \mathbf{i}+B{y} \mathbf{j}+B{z} \mathbf{k}\right) \
\mathbf{A} \cdot \mathbf{B}=A{x} B{x}+A{y} B{y}+A{z} B{z} \tag{5.23}
\end{gather}
\)

where six of the nine terms in the dot product are zero.
Equation (5.20) gives

\(
\begin{equation}
\mathbf{A} \cdot \mathbf{A}=|\mathbf{A}|^{2} \tag{5.24}
\end{equation}
\)

Using (5.23), we therefore have

\(
\begin{equation}
|\mathbf{A}|=\left(A{x}^{2}+A{y}^{2}+A_{z}^{2}\right)^{1 / 2} \tag{5.25}
\end{equation}
\)

For three-dimensional vectors, there is another type of product. The cross product or vector product $\mathbf{A} \times \mathbf{B}$ is a vector whose magnitude is

\(
\begin{equation}
|\mathbf{A} \times \mathbf{B}|=|\mathbf{A}||\mathbf{B}| \sin \theta \tag{5.26}
\end{equation}
\)

whose line segment is perpendicular to the plane defined by $\mathbf{A}$ and $\mathbf{B}$, and whose direction is such that $\mathbf{A}, \mathbf{B}$, and $\mathbf{A} \times \mathbf{B}$ form a right-handed system (just as the $x, y$, and $z$ axes form a right-handed system). See Fig. 5.3. From the definition it follows that

\(
\mathbf{B} \times \mathbf{A}=-\mathbf{A} \times \mathbf{B}
\)

Also, it can be shown that (Taylor and Mann, Section 10.2)

\(
\begin{equation}
\mathbf{A} \times(\mathbf{B}+\mathbf{C})=\mathbf{A} \times \mathbf{B}+\mathbf{A} \times \mathbf{C} \tag{5.27}
\end{equation}
\)

For the three unit vectors, we have

\(
\begin{gathered}
\mathbf{i} \times \mathbf{i}=\mathbf{j} \times \mathbf{j}=\mathbf{k} \times \mathbf{k}=\sin 0=0 \
\mathbf{i} \times \mathbf{j}=\mathbf{k}, \quad \mathbf{j} \times \mathbf{i}=-\mathbf{k}, \quad \mathbf{j} \times \mathbf{k}=\mathbf{i}, \quad \mathbf{k} \times \mathbf{j}=-\mathbf{i}, \quad \mathbf{k} \times \mathbf{i}=\mathbf{j}, \quad \mathbf{i} \times \mathbf{k}=-\mathbf{j}
\end{gathered}
\)

Using these equations and the distributive property (5.27), we find

\(
\begin{gathered}
\mathbf{A} \times \mathbf{B}=\left(A{x} \mathbf{i}+A{y} \mathbf{j}+A{z} \mathbf{k}\right) \times\left(B{x} \mathbf{i}+B{y} \mathbf{j}+B{z} \mathbf{k}\right) \
\mathbf{A} \times \mathbf{B}=\left(A{y} B{z}-A{z} B{y}\right) \mathbf{i}+\left(A{z} B{x}-A{x} B{z}\right) \mathbf{j}+\left(A{x} B{y}-A{y} B{x}\right) \mathbf{k}
\end{gathered}
\)

FIGURE 5.3 Cross product of two vectors.

As a memory aid, we can express the cross product as a determinant (see Section 8.3):

\(
\mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \tag{5.28}\
A{x} & A{y} & A{z} \
B{x} & B{y} & B{z}
\end{array}\right|=\mathbf{i}\left|\begin{array}{cc}
A{y} & A{z} \
B{y} & B{z}
\end{array}\right|-\mathbf{j}\left|\begin{array}{cc}
A{x} & A{z} \
B{x} & B{z}
\end{array}\right|+\mathbf{k}\left|\begin{array}{cc}
A{x} & A{y} \
B{x} & B{y}
\end{array}\right|
\)

We define the vector operator del as

\(
\begin{equation}
\nabla \equiv \mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z} \tag{5.29}
\end{equation}
\)

From Eq. (3.23), the operator for the linear-momentum vector is $\hat{\mathbf{p}}=-i \hbar \nabla$.
The gradient of a function $g(x, y, z)$ is defined as the result of operating on $g$ with del:

\(
\begin{equation}
\operatorname{grad} g(x, y, z) \equiv \nabla g(x, y, z) \equiv \mathbf{i} \frac{\partial g}{\partial x}+\mathbf{j} \frac{\partial g}{\partial y}+\mathbf{k} \frac{\partial g}{\partial z} \tag{5.30}
\end{equation}
\)

The gradient of a scalar function is a vector function. The vector $\nabla g(x, y, z)$ represents the spatial rate of change of the function $g$ : The $x$ component of $\nabla g$ is the rate of change of $g$ with respect to $x$, and so on. It can be shown that the vector $\nabla g$ points in the direction in which the rate of change of $g$ is greatest. From Eq. (4.24), the relation between force and potential energy is

\(
\begin{equation}
\mathbf{F}=-\nabla V(x, y, z)=-\mathbf{i} \frac{\partial V}{\partial x}-\mathbf{j} \frac{\partial V}{\partial y}-\mathbf{k} \frac{\partial V}{\partial z} \tag{5.31}
\end{equation}
\)

Suppose that the components of the vector $\mathbf{A}$ are each functions of some parameter $t$; $A{x}=A{x}(t), A{y}=A{y}(t), A{z}=A{z}(t)$. The derivative of $\mathbf{A}$ with respect to $t$ is defined as

\(
\begin{equation}
\frac{d \mathbf{A}}{d t}=\mathbf{i} \frac{d A{x}}{d t}+\mathbf{j} \frac{d A{y}}{d t}+\mathbf{k} \frac{d A_{z}}{d t} \tag{5.32}
\end{equation}
\)

Vector notation is a convenient way to represent the variables of a function. The wave function of a two-particle system can be written as $\psi\left(x{1}, y{1}, z{1}, x{2}, y{2}, z{2}\right)$. If $\mathbf{r}{1}$ is the vector from the origin to particle 1 , then $\mathbf{r}{1}$ has components $x{1}, y{1}, z{1}$ and specification of $\mathbf{r}{1}$ is equivalent to specification of the three coordinates $x{1}, y{1}, z{1}$. The same is true for the vector $\mathbf{r}{2}$ from the origin to particle 2 . Therefore, we can write the wave function as $\psi\left(\mathbf{r}{1}, \mathbf{r}{2}\right)$. Vector notation can be used in integrals. For example, the integral over all space in Eq. (3.57) is often written as $\int \cdots \int\left|\Psi\left(\mathbf{r}{1}, \ldots, \mathbf{r}{n}, t\right)\right|^{2} d \mathbf{r}{1} \cdots d \mathbf{r}{n}$.

Vectors in $n$-Dimensional Space

The definition of a vector can be generalized to more than three dimensions. A vector $\mathbf{A}$ in three-dimensional space can be defined by its magnitude $|\mathbf{A}|$ and its direction, or it can be defined by its three components $\left(A{x}, A{y}, A{z}\right)$ in a Cartesian coordinate system. Therefore, we can define a three-dimensional vector as a set of three real numbers $\left(A{x}, A{y}, A{z}\right)$ in a particular order. A vector $\mathbf{B}$ in an $n$-dimensional real vector "space" (sometimes called a hyperspace) is defined as an ordered set of $n$ real numbers $\left(B{1}, B{2}, \ldots, B{n}\right)$, where $B{1}, B{2}, \ldots, B{n}$ are the components of $\mathbf{B}$. Don't be concerned that you can't visualize vectors in an $n$-dimensional space.

The variables of a function are often denoted using $n$-dimensional vector notation. For example, instead of writing the wave function of a two-particle system as $\psi\left(\mathbf{r}{1}, \mathbf{r}{2}\right)$, we can define a six-dimensional vector $\mathbf{q}$ whose components are $q{1}=x{1}, q{2}=y{1}$,
$q{3}=z{1}, q{4}=x{2}, q{5}=y{2}, q{6}=z{2}$ and write the wave function as $\psi(\mathbf{q})$. For an $n$-particle system, we can define $\mathbf{q}$ to have $3 n$ components and write the wave function as $\psi(\mathbf{q})$ and the normalization integral over all space as $\int|\psi(\mathbf{q})|^{2} d \mathbf{q}$.

The theory of searching for the equilibrium geometry of a molecule uses $n$-dimensional vectors (Section 15.10). The rest of Section 5.2 is relevant to Section 15.10 and need not be read until you study Section 15.10.

Two $n$-dimensional vectors are equal if all their corresponding components are equal; $\mathbf{B}=\mathbf{C}$ if and only if $B{1}=C{1}, B{2}=C{2}, \ldots, B{n}=C{n}$. Therefore, in $n$-dimensional space, a vector equation is equivalent to $n$ scalar equations. The sum of two $n$-dimensional vectors $\mathbf{B}$ and $\mathbf{D}$ is defined as the vector $\left(B{1}+D{1}, B{2}+D{2}, \ldots, B{n}+D{n}\right)$. The difference is defined similarly. The vector $k \mathbf{B}$ is defined as the vector $\left(k B{1}, k B{2}, \ldots, k B{n}\right)$, where $k$ is a scalar. In three-dimensional space, the vectors $k \mathbf{A}$, where $k>0$, all lie in the same direction. In $n$-dimensional space the vectors $k \mathbf{B}$ all lie in the same direction. Just as the numbers $\left(A{x}, A{y}, A{z}\right)$ define a point in three-dimensional space, the numbers $\left(B{1}, B{2}, \ldots, B_{n}\right)$ define a point in $n$-dimensional space.

The length (or magnitude or Euclidean norm) $|\mathbf{B}|$ (sometimes denoted $|\mathbf{B}|$ ) of an $n$-dimensional real vector is defined as

\(
|\mathbf{B}| \equiv(\mathbf{B} \cdot \mathbf{B})^{1 / 2}=\left(B{1}^{2}+B{2}^{2}+\cdots+B_{n}^{2}\right)^{1 / 2}
\)

A vector whose length is 1 is said to be normalized.
The inner product (or scalar product) $\mathbf{B} \cdot \mathbf{G}$ of two real $n$-dimensional vectors $\mathbf{B}$ and $\mathbf{G}$ is defined as the scalar

\(
\mathbf{B} \cdot \mathbf{G} \equiv B{1} G{1}+B{2} G{2}+\cdots+B{n} G{n}
\)

If $\mathbf{B} \cdot \mathbf{G}=0$, the vectors $\mathbf{B}$ and $\mathbf{G}$ are said to be orthogonal. The cosine of the angle $\theta$ between two $n$-dimensional vectors $\mathbf{B}$ and $\mathbf{C}$ is defined by analogy to (5.20) as $\cos \theta \equiv \mathbf{B} \cdot \mathbf{C} /|\mathbf{B} | \mathbf{C}|$. One can show that this definition makes $\cos \theta$ lie in the range -1 to 1 .

In three-dimensional space, the unit vectors $\mathbf{i}=(1,0,0), \mathbf{j}=(0,1,0), \mathbf{k}=(0,0,1)$ are mutually perpendicular. Also, any vector can be written as a linear combination of these three vectors [Eq. (5.17)]. In an $n$-dimensional real vector space, the unit vectors $\mathbf{e}{1} \equiv(1,0,0, \ldots, 0), \mathbf{e}{2} \equiv(0,1,0, \ldots, 0), \ldots, \mathbf{e}{n} \equiv(0,0,0, \ldots, 1)$ are mutually orthogonal. Since the $n$-dimensional vector $\mathbf{B}$ equals $B{1} \mathbf{e}{1}+B{2} \mathbf{e}{2}+\cdots+B{n} \mathbf{e}{n}$, any $n$-dimensional real vector can be written as a linear combination of the $n$ unit vectors $\mathbf{e}{1}, \mathbf{e}{2}, \ldots, \mathbf{e}{n}$. This set of $n$ vectors is therefore said to be a basis for the $n$-dimensional real vector space. Since the vectors $\mathbf{e}{1}, \mathbf{e}{2}, \ldots, \mathbf{e}{n}$ are orthogonal and normalized, they are an orthonormal basis for the real vector space. The scalar product $\mathbf{B} \cdot \mathbf{e}{i}$ gives the component of $\mathbf{B}$ in the direction of the basis vector $\mathbf{e}_{i}$. A vector space has many possible basis sets. Any set of $n$ linearly independent real vectors can serve as a basis for the $n$-dimensional real vector space.

A three-dimensional vector can be specified by its three components or by its length and its direction. The direction can be specified by giving the three angles that the vector makes with the positive halves of the $x, y$, and $z$ axes. These angles are the direction angles of the vector and lie in the range 0 to $180^{\circ}$. However, the direction angle with the $z$ axis is fixed once the other two direction angles have been given, so only two direction angles are independent. Thus a three-dimensional vector can be specified by its length and two direction angles. Similarly, in $n$-dimensional space, the direction angles between a vector and each unit vector $\mathbf{e}{1}, \mathbf{e}{2}, \ldots, \mathbf{e}_{n}$ can be found from the above formula for the cosine of the angle between two vectors. An $n$-dimensional vector can thus be specified by its length and $n-1$ direction angles.

The gradient of a function of three variables is defined by (5.30). The gradient $\nabla f$ of a function $f\left(q{1}, q{2}, \ldots, q_{n}\right)$ of $n$ variables is defined as the $n$-dimensional vector whose components are the first partial derivatives of $f$ :

\(
\nabla f \equiv\left(\partial f / \partial q{1}\right) \mathbf{e}{1}+\left(\partial f / \partial q{2}\right) \mathbf{e}{2}+\cdots+\left(\partial f / \partial q{n}\right) \mathbf{e}{n}
\)

We have considered real, $n$-dimensional vector spaces. Dirac's formulation of quantum mechanics uses a complex, infinite-dimensional vector space, discussion of which is omitted.