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In quantum mechanics, we are generally interested in complex numbers. Let
denote the set of all
-tuples of complex numbers (a
complex
-space). The elements of
may be called ``points''
or ``vectors'' in complex
-space. The elements of the complex
numbers
are scalars.
A vector might be denoted by listing all its elements as
. This format is called a row vector. Alternatively,
the elements could be arranged as
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.,
For real numbers, the dot (or inner) product between two vectors
and
is given by
If
and
are complex, then one small complication arises.
If
and
are given by
![$\displaystyle {\bf a} =
\left( \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n ...
...} =
\left( \begin{array}{c} b_1 \\ b_2 \\ \vdots \\ b_n \end{array} \right),$](img14.png) |
|
|
(3) |
then to take the dot product of
and
, we must first
take the complex transpose of
,
Recall that the complex conjugate of a complex number like
is
(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.
![\begin{displaymath}
( a_1^* a_2^* \cdots a_n^*) \leftrightarrow
\left( \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array} \right).
\end{displaymath}](img19.png) |
(5) |
A vector
is said to be normalized if its dot product
with itself,
, 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
![\begin{displaymath}
{\bf a} = \left( \begin{array}{c} 1 \\ 1 \end{array} \right)
\end{displaymath}](img21.png) |
(6) |
is not normalized, because its dot product with itself is
. However, if we divide the vector by
, then the
resulting vector
![\begin{displaymath}
{\bf a}^{\prime} =
\frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 1 \end{array} \right)
\end{displaymath}](img24.png) |
(7) |
is normalized:
![\begin{displaymath}
\frac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \end{array...
...ft( 1 \cdot 1 + 1 \cdot 1
\right)
= \frac{1}{2} \cdot 2
= 1.
\end{displaymath}](img25.png) |
(8) |
Two vectors
and
are said to be orthogonal if
their dot product
. Two vectors are
orthonormal if they are orthogonal and each one is normalized.
Next: Matrices
Up: linear_algebra
Previous: linear_algebra
David Sherrill
2006-08-15