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In quantum mechanics, we are generally interested in complex numbers. Let ${\bf C}^n$ denote the set of all $n$-tuples of complex numbers (a complex $n$-space). The elements of ${\bf C}^n$ may be called ``points'' or ``vectors'' in complex $n$-space. The elements of the complex numbers ${\bf C}$ are scalars.

A vector might be denoted by listing all its elements as $ (c_1 c_2
c_3 \cdots c_n)$. This format is called a row vector. Alternatively, the elements could be arranged as

$\displaystyle \left(
\begin{array}{c} c_1 \\  c_2 \\  c_3 \\  \vdots \\  c_n

which is a column vector. Our default choice will be to write vectors as column vectors, and row vectors will result from taking (complex conjugate) transposes of vectors.

When a vector is multiplied by a scalar (a number), this is the same as multiplying each component of the vector by that scalar, e.g.,

$\displaystyle 2
\left( \begin{array}{c} 1 \\  0 \\  3 \end{array}\right)$ $\textstyle =$ $\displaystyle \left( \begin{array}{c} 2 \\  0 \\  6 \end{array}\right).$ (1)

For real numbers, the dot (or inner) product between two vectors ${\bf a}$ and ${\bf b}$ is given by

$\displaystyle {\bf a} \cdot {\bf b}$ $\textstyle =$ $\displaystyle ( a_1 a_2 \cdots a_n) \cdot
\left( \begin{array}{c} b_1 \\  b_2 \\  \vdots \\  b_n \end{array} \right)$  
  $\textstyle =$ $\displaystyle a_1 b_1 + a_2 b_2 + \cdots + a_n b_n$ (2)

If ${\bf a}$ and ${\bf b}$ are complex, then one small complication arises. If ${\bf a}$ and ${\bf b}$ are given by

$\displaystyle {\bf a} =
\left( \begin{array}{c} a_1 \\  a_2 \\  \vdots \\  a_n ...
...} =
\left( \begin{array}{c} b_1 \\  b_2 \\  \vdots \\  b_n \end{array} \right),$     (3)

then to take the dot product of ${\bf a}$ and ${\bf b}$, we must first take the complex transpose of ${\bf a}$,
$\displaystyle {\bf a} \cdot {\bf b}$ $\textstyle =$ $\displaystyle ( a_1^* a_2^* \cdots a_n^*) \cdot
\left( \begin{array}{c} b_1 \\  b_2 \\  \vdots \\  b_n \end{array} \right)$  
  $\textstyle =$ $\displaystyle a_1^* b_1 + a_2^* b_2 + \cdots + a_n^* b_n.$ (4)

Recall that the complex conjugate of a complex number like $c + i d$ is $c - i d$ (the imaginary part has its sign reversed). Any time a row vector is converted into a column vector (or vice versa), if it is complex, then one needs to take complex conjugates of each element.
( a_1^* a_2^* \cdots a_n^*) \leftrightarrow
\left( \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array} \right).
\end{displaymath} (5)

A vector ${\bf a}$ is said to be normalized if its dot product with itself, ${\bf a} \cdot {\bf a}$, is 1. If this is not the case, it is always possible to normalize the vector to force the dot product to come out to 1; one merely needs to divide each component of the vector by the square root of the dot product that results before normalization. For example, the vector

{\bf a} = \left( \begin{array}{c} 1 \\ 1 \end{array} \right)
\end{displaymath} (6)

is not normalized, because its dot product with itself is $1 \cdot 1 + 1
\cdot 1 = 2$. However, if we divide the vector by $\sqrt{2}$, then the resulting vector
{\bf a}^{\prime} =
\frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 1 \end{array} \right)
\end{displaymath} (7)

is normalized:
\frac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \end{array...
...ft( 1 \cdot 1 + 1 \cdot 1
= \frac{1}{2} \cdot 2
= 1.
\end{displaymath} (8)

Two vectors ${\bf a}$ and ${\bf b}$ are said to be orthogonal if their dot product ${\bf a} \cdot {\bf b} = 0$. Two vectors are orthonormal if they are orthogonal and each one is normalized.

next up previous
Next: Matrices Up: linear_algebra Previous: linear_algebra
David Sherrill 2006-08-15