2.1 Complex Numbers

Quantum mechanics is full of complex numbers, numbers involving

\begin{displaymath}
{\rm i}=\sqrt{-1}.
\end{displaymath}

Note that $\sqrt{-1}$ is not an ordinary, real, number, since there is no real number whose square is $\vphantom0\raisebox{1.5pt}{$-$}$1; the square of a real number is always positive. This section summarizes the most important properties of complex numbers.

First, any complex number, call it $c$, can by definition always be written in the form

\begin{displaymath}
c = c_r+{\rm i}c_i
\end{displaymath} (2.1)

where both $c_r$ and $c_i$ are ordinary real numbers, not involving $\sqrt{-1}$. The number $c_r$ is called the real part of $c$ and $c_i$ the imaginary part.

You can think of the real and imaginary parts of a complex number as the components of a two-di­men­sion­al vector:

\begin{displaymath}
\begin{picture}(200,70)(-100,-20)
\thinlines
\put(-5...
...(2,1){70}}
\put(0,23){\makebox(0,0)[tl]{$c$}}
\end{picture}
\end{displaymath}

The length of that vector is called the “magnitude,” or “absolute value” $\vert c\vert$ of the complex number. It equals

\begin{displaymath}
\vert c\vert = \sqrt{c_r^2+c_i^2}.
\end{displaymath}

Complex numbers can be manipulated pretty much in the same way as ordinary numbers can. A relation to remember is:

\begin{displaymath}
\frac{1}{{\rm i}} = -{\rm i}
\end{displaymath} (2.2)

which can be verified by multiplying the top and bottom of the fraction by ${\rm i}$ and noting that by definition ${\rm i}^2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1 in the bottom.

The complex conjugate of a complex number $c$, denoted by $c^*$, is found by replacing ${\rm i}$ everywhere by $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}$. In particular, if $c$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_r+{{\rm i}}c_i$, where $c_r$ and $c_i$ are real numbers, the complex conjugate is

\begin{displaymath}
c^* = c_r - {\rm i}c_i
\end{displaymath} (2.3)

The following picture shows that graphically, you get the complex conjugate of a complex number by flipping it over around the horizontal axis:

\begin{displaymath}
\begin{picture}(200,100)(-100,-50)
\thinlines
\put(-...
...1){70}}
\put(0,-23){\makebox(0,0)[bl]{$c^*$}}
\end{picture}
\end{displaymath}

You can get the magnitude of a complex number $c$ by multiplying $c$ with its complex conjugate $c^*$ and taking a square root:

\begin{displaymath}
\vert c\vert = \sqrt{c^* c}
\end{displaymath} (2.4)

If $c$ $\vphantom0\raisebox{1.5pt}{$=$}$ $c_r+{\rm i}{c}_i$, where $c_r$ and $c_i$ are real numbers, multiplying out $c^*c$ shows the magnitude of $c$ to be

\begin{displaymath}
\vert c\vert = \sqrt{c_r^2+c_i^2}
\end{displaymath}

which is indeed the same as before.

From the above graph of the vector representing a complex number $c$, the real part is $c_r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert c\vert\cos\alpha$ where $\alpha$ is the angle that the vector makes with the horizontal axis, and the imaginary part is $c_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert c\vert\sin\alpha$. So you can write any complex number in the form

\begin{displaymath}
c = \vert c\vert\left(\cos\alpha+{\rm i}\sin\alpha\right)
\end{displaymath}

The critically important Euler formula says that:
\begin{displaymath}
\cos\alpha + {\rm i}\sin\alpha = e^{{\rm i}\alpha} %
\end{displaymath} (2.5)

So, any complex number can be written in polar form as
\begin{displaymath}
c = \vert c\vert e^{{\rm i}\alpha} %
\end{displaymath} (2.6)

where both the magnitude $\vert c\vert$ and the phase angle (or argument) $\alpha$ are real numbers.

Any complex number of magnitude one can therefore be written as $e^{{\rm i}\alpha}$. Note that the only two real numbers of magnitude one, 1 and $\vphantom0\raisebox{1.5pt}{$-$}$1, are included for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, respectively $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$. The number ${\rm i}$ is obtained for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pi$$\raisebox{.5pt}{$/$}$​2 and $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}$ for $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$\pi$$\raisebox{.5pt}{$/$}$​2.

(See derivation {D.7} if you want to know where the Euler formula comes from.)


Key Points
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...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Complex numbers include the square root of minus one, ${\rm i}$, as a valid number.

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\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
All complex numbers can be written as a real part plus ${\rm i}$ times an imaginary part, where both parts are normal real numbers.

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\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
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The complex conjugate of a complex number is obtained by replacing ${\rm i}$ everywhere by $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}$.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The magnitude of a complex number is obtained by multiplying the number by its complex conjugate and then taking a square root.

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\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The Euler formula relates exponentials to sines and cosines.

2.1 Review Questions
1.

Multiply out $(2+3{\rm i})^2$ and then find its real and imaginary part.

Solution mathcplx-a

2.

Show more directly that 1$\raisebox{.5pt}{$/$}$${\rm i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}$.

Solution mathcplx-b

3.

Multiply out $(2+3{\rm i})(2-3{\rm i})$ and then find its real and imaginary part.

Solution mathcplx-c

4.

Find the magnitude or absolute value of $2+3{\rm i}$.

Solution mathcplx-d

5.

Verify that $(2-3{\rm i})^2$ is still the complex conjugate of $(2+3{\rm i})^2$ if both are multiplied out.

Solution mathcplx-e

6.

Verify that $e^{-2{\rm i}}$ is still the complex conjugate of $e^{2{\rm i}}$ after both are rewritten using the Euler formula.

Solution mathcplx-f

7.

Verify that $\left(e^{{\rm i}\alpha}+e^{-{\rm i}\alpha}\right)$$\raisebox{.5pt}{$/$}$​2 $\vphantom0\raisebox{1.5pt}{$=$}$ $\cos\alpha$.

Solution mathcplx-g

8.

Verify that $\left(e^{{\rm i}\alpha}-e^{-{\rm i}\alpha}\right)$$\raisebox{.5pt}{$/$}$$2{\rm i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sin\alpha$.

Solution mathcplx-h