2.3 The Dot, oops, INNER Product

The dot product of vectors is an important tool. It makes it possible to find the length of a vector, by multiplying the vector by itself and taking the square root. It is also used to check if two vectors are orthogonal: if their dot product is zero, they are. In this subsection, the dot product is defined for complex vectors and functions.

The usual dot product of two vectors $\vec{f}$ and $\vec{g}$ can be found by multiplying components with the same index $i$ together and summing that:

\begin{displaymath}
\vec f \cdot \vec g \equiv f_1 g_1 + f_2 g_2 + f_3 g_3
\end{displaymath}

(The emphatic equal, $\vphantom0\raisebox{1.5pt}{$\equiv$}$, is commonly used to indicate “is by definition equal” or is always equal.) Figure 2.6 shows multiplied components using equal colors.

Figure 2.6: Forming the dot product of two vectors.
\begin{figure}
\centering
% \htmlimage{extrascale=3,notransparent}{}
\se...
...)[b]{3}}
\put(143,1.2){\makebox(0,0)[b]{$i$}}
\end{picture}
\end{figure}

Note the use of numeric subscripts, $f_1$, $f_2$, and $f_3$ rather than $f_x$, $f_y$, and $f_z$; it means the same thing. Numeric subscripts allow the three term sum above to be written more compactly as:

\begin{displaymath}
\vec f \cdot \vec g \equiv \sum_{\mbox{\scriptsize all }i} f_i g_i
\end{displaymath}

The $\Sigma$ is called the summation symbol.

The length of a vector $\vec{f}$, indicated by $\vert\vec{f}\vert$ or simply by $f$, is normally computed as

\begin{displaymath}
\vert\vec f\vert = \sqrt{\vec f \cdot \vec f}
= \sqrt{\sum_{\mbox{\scriptsize all }i} f_i^2}
\end{displaymath}

However, this does not work correctly for complex vectors. The difficulty is that terms of the form $f_i^2$ are no longer necessarily positive numbers. For example, ${\rm i}^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1.

Therefore, it is necessary to use a generalized “inner product” for complex vectors, which puts a complex conjugate on the first vector:

\begin{displaymath}
\fbox{$\displaystyle
\big\langle\vec f\big\vert\vec g\bi...
...le
\equiv
\sum_{\mbox{\scriptsize all }i} f^*_i g_i
$}
\end{displaymath} (2.7)

If the vector $\vec{f}$ is real, the complex conjugate does nothing, and the inner product $\big\langle\vec{f}\big\vert\vec{g}\big\rangle $ is the same as the dot product $\vec{f}\cdot\vec{g}$. Otherwise, in the inner product $\vec{f}$ and $\vec{g}$ are no longer interchangeable; the conjugates are only on the first factor, $\vec{f}$. Interchanging $\vec{f}$ and $\vec{g}$ changes the inner product’s value into its complex conjugate.

The length of a nonzero vector is now always a positive number:

\begin{displaymath}
\fbox{$\displaystyle
\vert\vec f\vert = \sqrt{\big\langl...
... \sqrt{\sum_{\mbox{\scriptsize all }i} \vert f_i\vert^2}
$}
\end{displaymath} (2.8)

Physicists take the inner product bracket verbally apart as

\begin{displaymath}
\begin{array}{ccc}
\big\langle \vec f\big\vert & & \big\...
...bra} & \makebox[0pt][l]{/}\mbox{c} & \mbox{ket}
\end{array}
\end{displaymath}

and refer to vectors as bras and kets.

The inner product of functions is defined in exactly the same way as for vectors, by multiplying values at the same $x$-​position together and summing. But since there are infinitely many $x$ values, the sum becomes an integral:

\begin{displaymath}
\fbox{$\displaystyle
\big\langle f \big\vert g\big\rangle = \int_{\mbox{\scriptsize all }x} f^*(x) g(x) {\,\rm d}x
$}
\end{displaymath} (2.9)

Figure 2.7 shows multiplied function values using equal colors:

Figure 2.7: Forming the inner product of two functions.
\begin{figure}
\centering
% \htmlimage{extrascale=3,notransparent}{}
\se...
...$g(x)$}}
\put(143,1.2){\makebox(0,0)[b]{$x$}}
\end{picture}
\end{figure}

The equivalent of the length of a vector is in the case of a function called its “norm:”

\begin{displaymath}
\fbox{$\displaystyle
\vert\vert f\vert\vert \equiv \sqrt...
...{\mbox{\scriptsize all }x} \vert f(x)\vert^2 {\,\rm d}x}
$}
\end{displaymath} (2.10)

The double bars are used to avoid confusion with the absolute value of the function.

A vector or function is called “normalized” if its length or norm is one:

\begin{displaymath}
\fbox{$\displaystyle
\big\langle f\big\vert f\big\rangle =1 \mbox{ iff $f$\ is normalized.}
$}
\end{displaymath} (2.11)

(“iff” should really be read as if and only if.)

Two vectors, or two functions, $f$ and $g$, are by definition orthogonal if their inner product is zero:

\begin{displaymath}
\fbox{$\displaystyle
\big\langle f\big\vert g\big\rangle =0 \mbox{ iff $f$\ and $g$\ are orthogonal.}
$}
\end{displaymath} (2.12)

Sets of vectors or functions that are all

occur a lot in quantum mechanics. Such sets should be called “orthonormal”, though the less precise term orthogonal is often used instead. This document will refer to them correctly as being orthonormal.

So, a set of functions or vectors $f_1,f_2,f_3,\ldots$ is orthonormal if

\begin{displaymath}
0=
\langle f_1\vert f_2\rangle=\langle f_2\vert f_1\rang...
...gle f_2\vert f_3\rangle=\langle f_3\vert f_2\rangle=
\ldots
\end{displaymath}

and

\begin{displaymath}
1=\langle f_1\vert f_1\rangle=\langle f_2\vert f_2\rangle=\langle f_3\vert f_3\rangle=
\ldots
\end{displaymath}


Key Points
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\end{picture}$
For complex vectors and functions, the normal dot product becomes the inner product.

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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
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To take an inner product of vectors,
  • take complex conjugates of the components of the first vector;
  • multiply corresponding components of the two vectors together;
  • sum these products.

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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
To take an inner product of functions,
  • take the complex conjugate of the first function;
  • multiply the two functions;
  • integrate the product function.

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To find the length of a vector, take the inner product of the vector with itself, and then a square root.

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To find the norm of a function, take the inner product of the function with itself, and then a square root.

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\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
A pair of vectors, or a pair of functions, is orthogonal if their inner product is zero.

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A set of vectors forms an orthonormal set if every one is orthogonal to all the rest, and every one is of unit length.

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A set of functions forms an orthonormal set if every one is orthogonal to all the rest, and every one is of unit norm.

2.3 Review Questions
1.

Find the following inner product of the two vectors:

\begin{displaymath}
\left\langle\left(
\begin{array}{c} 1+{\rm i}\\ 2-{\rm i}
\e...
...\begin{array}{c} 2{\rm i}\\ 3
\end{array}\right) \right\rangle
\end{displaymath}

Solution dot-a

2.

Find the length of the vector

\begin{displaymath}
\left(
\begin{array}{c} 1+{\rm i}\\ 3
\end{array}\right)
\end{displaymath}

Solution dot-b

3.

Find the inner product of the functions $\sin(x)$ and $\cos(x)$ on the interval 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ 1.

Solution dot-c

4.

Show that the functions $\sin(x)$ and $\cos(x)$ are orthogonal on the interval 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $2\pi$.

Solution dot-d

5.

Verify that $\sin(x)$ is not a normalized function on the interval 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $2\pi$, and normalize it by dividing by its norm.

Solution dot-e

6.

Verify that the most general multiple of $\sin(x)$ that is normalized on the interval 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $2\pi$ is $e^{{\rm i}\alpha}\sin(x)$$\raisebox{.5pt}{$/$}$$\sqrt{\pi}$ where $\alpha$ is any arbitrary real number. So, using the Euler formula, the following multiples of $\sin(x)$ are all normalized: $\sin(x)$$\raisebox{.5pt}{$/$}$$\sqrt{\pi}$, (for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0), $-\sin(x)$$\raisebox{.5pt}{$/$}$$\sqrt{\pi}$, (for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$), and ${\rm i}\sin(x)$$\raisebox{.5pt}{$/$}$$\sqrt{\pi}$, (for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$$\raisebox{.5pt}{$/$}$​2).

Solution dot-f

7.

Show that the functions $e^{4{\rm i}{\pi}x}$ and $e^{6{\rm i}{\pi}x}$ are an orthonormal set on the interval 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ 1.

Solution dot-g