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5.2
The determinant \mathop{det}(A) of
a two–by–two matrix A
is defined as \mathop{det}\left (\array{
a&b\cr
c &d } \right ) ≡\left |\array{
a&b\cr
c &d } \right | := ad−cb.
5.2.1 This is the trivial mapping represented by the unit or identity matrix,
I ,
\begin{eqnarray}
I = \left (\array{
1&0\cr
0 &1 } \right ).& & %&(5.10)\\
\end{eqnarray}
We have \mathop{det}(I) = 1 .
Check that Ix = x for
any vector x .
5.2.2 These are linear mappings A represented by
the multiples of the unit matrix , where c
is a real number such that
\begin{eqnarray}
A = \left (\array{
c&0\cr
0 &c } \right ).& & %&(5.11)\\
\end{eqnarray}
We have \mathop{det}(A) = {c}^{2} > 1 . Check
that in this case Ax = cx
for any vector x .
5.2.3 These are linear mappings A
such as
\begin{eqnarray}
A = \left (\array{
1&0\cr
0 &0 } \right ).& & %&(5.12)\\
\end{eqnarray}
We have \mathop{det}(A) = 0 . Check that in
this case, for any vector x = (x,y) ,
Ax = (x,0) : the vector is projected
onto the x -axis.
5.2.4 These are mappings R(θ) that rotate
vectors around the origin by an angle θ ,
\begin{eqnarray}
R(θ) = \left (\array{
\mathop{cos}\nolimits θ& −\mathop{ sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & \mathop{ cos} \nolimits θ } \right ).& & %&(5.13)\\
\end{eqnarray}
In this case, \mathop{det}(R(θ)) ={\mathop{ cos}\nolimits }^{2}θ − (−{\mathop{sin}\nolimits }^{2}θ) = 1 .
A vector x = (x,y) is
rotated into
\begin{eqnarray}
R(θ)x = \left (\array{
\mathop{cos}\nolimits θ& −\mathop{ sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & \mathop{ cos} \nolimits θ } \right )\left (\array{
x\cr
y } \right ) = \left (\array{
x\mathop{cos}\nolimits θ − y\mathop{sin}\nolimits θ\cr
x\mathop{ sin} \nolimits θ + y\mathop{ cos} \nolimits θ } \right ).& & %&(5.14)\\
\end{eqnarray}
Examples for rotations are
\begin{eqnarray}
\left (\array{
\mathop{cos}\nolimits θ& −\mathop{ sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & \mathop{ cos} \nolimits θ } \right )\left (\array{
1\cr
0 } \right ) = \left (\array{
\mathop{cos}\nolimits θ\cr
\mathop{ sin} \nolimits θ } \right ),\quad \left (\array{
\mathop{cos}\nolimits θ& −\mathop{ sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & \mathop{ cos} \nolimits θ } \right )\left (\array{
0\cr
1 } \right ) = \left (\array{
−\mathop{ sin}\nolimits θ\cr
\mathop{ cos} \nolimits θ } \right ).& & %&(5.15)\\
\end{eqnarray}
θ = 0 In this case,
\begin{eqnarray}
R(θ = 0) = \left (\array{
1&0\cr
0 &1 } \right ) = I\quad \text{(unit matrix).}& & %&(5.16)\\
\end{eqnarray}
θ = {π\over
2} In this case,
\begin{eqnarray}
R\left (θ = {π\over
2} \right ) = \left (\array{
0& − 1\cr
1 & 0 } \right ) = −i{σ}_{y}\quad \text{($ − i$ times Pauli Matrix ${σ}_{y}$).}& & %&(5.17)\\
\end{eqnarray}
5.2.5 These are mappings S(θ)
that reflect a vectors at a fixed axis:
\begin{eqnarray}
S(θ) = \left (\array{
\mathop{cos}\nolimits θ& \mathop{sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & −\mathop{ cos} \nolimits θ } \right ).& & %&(5.18)\\
\end{eqnarray}
In this case, \mathop{det}(S(θ)) = −{\mathop{cos}\nolimits }^{2}θ −{\mathop{ sin}\nolimits }^{2}θ = −1 .
A vector x = (x,y) is
transformed into
\begin{eqnarray}
S(θ)x = \left (\array{
\mathop{cos}\nolimits θ& \mathop{sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & −\mathop{ cos} \nolimits θ } \right )\left (\array{
x\cr
y } \right ) = \left (\array{
x\mathop{cos}\nolimits θ + y\mathop{sin}\nolimits θ\cr
x\mathop{ sin} \nolimits θ − y\mathop{ cos} \nolimits θ } \right ).& & %&(5.19)\\
\end{eqnarray}
Examples:
\begin{eqnarray}
\left (\array{
\mathop{cos}\nolimits θ& \mathop{sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & −\mathop{ cos} \nolimits θ } \right )\left (\array{
\mathop{cos}\nolimits {1\over
2}θ
\cr
\mathop{sin}\nolimits {1\over
2}θ } \right ) = \left (\array{
\mathop{cos}\nolimits {1\over
2}θ\mathop{cos}\nolimits θ +\mathop{ sin}\nolimits {1\over
2}θ\mathop{sin}\nolimits θ
\cr
\mathop{cos}\nolimits {1\over
2}θ\mathop{sin}\nolimits θ −\mathop{ sin}\nolimits {1\over
2}θ\mathop{cos}\nolimits θ } \right ) = \left (\array{
\mathop{cos}\nolimits {1\over
2}θ
\cr
\mathop{sin}\nolimits {1\over
2}θ } \right ),& & %&(5.20)\\
\end{eqnarray}
where we have a formula for trigonometric functions (CHECK). Furthermore, we have
\begin{eqnarray}
\left (\array{
\mathop{cos}\nolimits θ& \mathop{sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & −\mathop{ cos} \nolimits θ } \right )\left (\array{
1\cr
0 } \right ) = \left (\array{
\mathop{cos}\nolimits θ\cr
\mathop{ sin} \nolimits θ } \right ),\quad \left (\array{
\mathop{cos}\nolimits θ& \mathop{sin}\nolimits θ\cr
\mathop{ sin} \nolimits θ & −\mathop{ cos} \nolimits θ } \right )\left (\array{
0\cr
1 } \right ) = \left (\array{
\mathop{sin}\nolimits θ\cr
−\mathop{ cos} \nolimits θ } \right ).& & %&(5.21)\\
\end{eqnarray}
Sketch this in the x -y -plane
(lecture). We recognise that S(θ)
defines a reflection at the axis defined by the direction of the vector
(\mathop{cos}\nolimits {1\over
2}θ,\mathop{sin}\nolimits {1\over
2}θ)
θ = 0 In this case,
\begin{eqnarray}
S(θ = 0) = \left (\array{
1& 0\cr
0 & −1 } \right ) = {σ}_{z}\quad \text{(Pauli Matrix ${σ}_{z}$).}& & %&(5.22)\\
\end{eqnarray}
θ = {π\over
2} In this case,
\begin{eqnarray}
S\left (θ = {π\over
2} \right ) = \left (\array{
0&1\cr
1 &0 } \right ) = {σ}_{x}\quad \text{(Pauli Matrix ${σ}_{x}$).}& & %&(5.23)\\
\end{eqnarray}