Complex numbers: a review
Useful facts and identities:
- It is well worth memorizing a few commonly seen
\(\theta\)'s (with \(r=1\)):
$$\eqb{
e^{i\pi} & \eql & \cos\pi + i \sin\pi \eql -1 \\
e^{i2\pi} & \eql & \cos 2\pi + i \sin 2\pi \eql 1\\
e^{i\frac{\pi}{2}} & \eql & \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \eql i\\
e^{i\frac{\pi}{4}} & \eql & \cos \frac{\pi}{4} + i \sin
\frac{\pi}{4} \eql \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i \\
e^{i\frac{\pi}{3}} & \eql & \cos \frac{\pi}{3} + i \sin
\frac{\pi}{3} \eql \frac{1}{2} + \frac{\sqrt{3}}{2} i \\
e^{i\frac{\pi}{6}} & \eql & \cos \frac{\pi}{6} + i \sin
\frac{\pi}{6} \eql \frac{\sqrt{3}}{2} + \frac{1}{2} i \\
}$$
- Useful trig identities for quantum computing:
$$\eqb{
\sin(-\theta) & \eql & - \sin\theta \\
\cos(-\theta) & \eql & \cos\theta \\
\sin^2 \theta + \cos^2 \theta & \eql & 1 \\
\sin(\theta_1 + \theta_2) & \eql &
\sin\theta_1 \cos\theta_2 + \cos\theta_1 \sin\theta_2\\
\cos(\theta_1 + \theta_2) & \eql &
\cos\theta_1 \cos\theta_2 - \sin\theta_1 \sin\theta_2 \\
}$$
- From which
$$\eqb{
\cos(2\theta) & \eql & 2\cos^2\theta - 1 \\
\sin(2\theta) & \eql & 2\sin\theta \cos\theta \\
}$$
Problem:
Simplify \(i^{15}\)
Solution:
$$
i^{15} \eql i \cdot i^{14} \eql i \cdot (i^2)^7 \eql i (-1)^7 \eql -i
$$
Problem:
Simplify
\(
\frac{i-1}{3-2i}
\)
into \(a+ib\) form.
Solution:
Use the division trick of multiplying and dividing by the
conjugate of the denominator:
$$\eqb{
\frac{i-1}{3-2i}
& \eql &
\frac{i-1}{3-2i} \; \frac{3+2i}{3+2i} \\
& \eql &
\frac{3i + 2i^2 -3 - 2i}{3^2 + 2^2}\\
& \eql &
\frac{-5 + i}{13}\\
& \eql &
-\frac{5}{13} + \frac{1}{13} i
}$$
Problem:
Calculate \(\re( (3+2i)(1-i) )\)
Solution:
$$
(3+2i)(1-i) \eql 3 - 3i + 2i - 2i^2 \eql 5 - i
$$
Thus, \(\re( (3+2i)(1-i) ) = 5\)
Problem:
Compute \((1+i)^8\)
Solution:
Instead of laboriously multiplying, we'll write \(1+i\) in polar
form, where powers are straightforward:
$$\eqb{
1 + i
& \eql &
\sqrt{2} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i) \\
& \eql &
\sqrt{2} e^{i\frac{\pi}{4}}
}$$
Thus,
$$\eqb{
(1+i)^8
& \eql &
(\sqrt{2} e^{i\frac{\pi}{4}})^8 \\
& \eql &
(\sqrt{2})^8 e^{2\pi i} \\
& \eql & 2^4 \\
& \eql & 16 \\
}$$
Problem:
Use the polar representation to calculate all three
cube roots of unity. Express the angles (phases) in degrees as well.
Solution:
The first step is to express the complex number 1 in polar form:
$$
1 \eql \cos2\pi + i\sin2\pi \eql e^{i2\pi}
$$
At first, it may seem that it's enough to write
$$
1^{\frac{1}{3}} \eql e^{i\frac{2\pi}{3}}
\;\;\;\;\;\; (\mbox{angle = }120^\circ)
$$
This is one of the three cube roots of 1. To find the other two,
we must use angle redundancy and instead write
$$
1 \eql e^{i(2\pi + 2\pi k)}
$$
Then,
$$
1^{\frac{1}{3}} \eql e^{i\frac{(2\pi + 2\pi k)}{3}}
$$
With \(k=0\), we get the earlier solution. With \(k=1\):
$$
e^{i\frac{(2\pi + 2\pi k)}{3}} \eql e^{i\frac{4\pi}{3}}
\;\;\;\;\;\; (\mbox{angle = }240^\circ)
$$
Next, with \(k=2\):
$$
e^{i\frac{(2\pi + 2\pi k)}{3}} \eql e^{i\frac{6\pi}{3}}
\eql 1
\;\;\;\;\;\; (\mbox{angle = }360^\circ = 0^\circ)
$$
With \(k=3\):
$$
e^{i\frac{(2\pi + 2\pi k)}{3}} \eql e^{i\frac{8\pi}{3}}
\eql e^{i\frac{2\pi}{3}}
\;\;\;\;\;\; (\mbox{angle = }120^\circ)
$$
Thus, the \(k=3\) number is the same as the \(k=0\) number,
and the cycle repeats.
Hence, the three cube roots are \(1, e^{i\frac{2\pi}{3}}, e^{i\frac{4\pi}{3}}\).
We'll later see that \(e^{i\frac{2\pi}{3}}\) is called the
principal cube root of 1.
Problem:
Prove, for complex numbers \(z_1, z_2\):
- \( (z_1 + z_2)^* = z_1^* + z_2^* \)
- \( (z_1 z_2)^* = z_1^* z_2^* \)
- \( |z_1 z_2| = |z_1| |z_2| \)
Solution:
Let \(z_1 = a_1 + ib_1\) and \(z_2 = a_2 + ib_2\).
-
$$
(z_1 + z_2)^*
\eql (a_1 + ib_1 + a_2 + ib_2)^*
\eql a_1+a_2 - i(b_1 + b_2)
\eql (a_1 - ib_1) + (a_2 - ib_2)
\eql z_1^* + z_2^*
$$
-
$$
(z_1 z_2)^*
\eql ( (a_1 + ib_1)(a_2 + ib_2) )^*
\eql ( a_1 a_2 - b_1 b_2 + (a_1 b_1 + a_2 b_1) i )^*
\eql a_1 a_2 - b_1 b_2 - (a_1 b_1 + a_2 b_1) i )
\eql a_1 a_2 - b_1 b_2 - (a_1 b_1 + a_2 b_1) i )
\eql (a_1 - ib_1) (a_2 - ib_2)
\eql z_1^* z_2^*
$$
-
$$
|z_1 z_2|^2
\eql (z_1 z_2) (z_1 z_2)^*
\eql (z_1 z_2) (z_1^* z_2^*)
\eql (z_1 z_z^*) (z_2 z_2^*)
\eql |z_1|^2 |z_2|^2
$$
Since magnitudes are positive real numbers,
\(|z_1 z_2| = |z_1| |z_2|\).
Complex vectors with Dirac notation
Problem:
With \(\kt{u}=(1, 2-i, 3i),\; \kt{v}=(2,1,i), \; \kt{w}=(1,-2,3)\),
and \(\alpha=(2-3i), \beta=-1\), calculate
- the vectors \(\alpha\kt{v}, \kt{\alpha v}, \br{\alpha u}, \; \alpha^*\br{u}\)
- the numbers \(\inr{u}{v}, \inr{v}{u}, \inr{v}{u}^*\), and compare the first
and third.
- the matrix \(\otr{u}{v}\), the matrix-vector product
\(\otr{u}{v}\kt{w}\), and the scalar-vector product \(\inr{v}{w}\kt{u}\),
and compare the latter two results.
- \(\magsq{\kt{u}}, \inr{u}{u}\)
Solution:
- Since \(\kt{v}\) is already a column vector
$$
\kt{\alpha v} \eql \alpha\kt{v}
\eql (2-3i)\vecthree{2}{1}{i}
\eql \vecthree{4-6i}{2-3i}{3+2i}
$$
Next,
$$
\kt{\alpha u} \eql (2-3i)\vecthree{1}{2-i}{3i}
\eql \vecthree{2-3i}{1-8i}{9+6i}
$$
and so
$$
\br{\alpha u} \eql \mat{2+3i & 1+8i & 9-6i}
$$
If, on the other hand, we conjugate and then scalar-multiply:
$$
\alpha^*\br{u} \eql
(2+3i) \mat{1 & 2+i & -3i}
\eql \mat{2+3i & 1+8i & 9-6i}
$$
as expected.
- First,
$$
\inr{u}{v} \eql \mat{1 & 2+i & -3i} \vecthree{2}{1}{i}
\eql 2 + (2+i) - 3i^2
\eql 7+i
$$
Next
$$
\inr{v}{u} \eql \mat{2 & 1 & -i} \vecthree{1}{2-i}{3i}
\eql 7-i
$$
and so
$$
\inr{v}{u}^* \eql 7 + i \eql \inr{u}{v}
$$
- First
$$
\otr{u}{v} \eql
\mat{2 & 1 & -i\\
4-2i & 2-i & -1-2i\\
6i & 3i & 3}
$$
Multiplying this matrix into \(\kt{w}\) gives
$$
\otr{u}{v} \; \kt{w} \eql
\mat{2 & 1 & -i\\
4-2i & 2-i & -1-2i\\
6i & 3i & 3}
\vecthree{1}{-2}{3}
\eql
\vecthree{-3i}{-3-6i}{9}
$$
Next,
$$
\inr{v}{w} \eql
\mat{2 & 1 & -i} \vecthree{1}{-2}{3}
\eql -3i
$$
Multiplying into \(\kt{u}\) gives
$$
\inr{v}{w} \; \kt{u}
\eql -3i \vecthree{1}{2-i}{3i}
\eql \vecthree{-3i}{-3-6i}{9}
$$
Thus, we see that
$$
(\otr{u}{v}) \kt{w}
\eql \kt{u}\; (\inr{v}{w})
\eql (\inr{v}{w}) \kt{u}
$$
since \(\inr{v}{w}\) is a number.
- Clearly
$$
\magsq{\kt{u}} \eql |1|^2 + |2-i|^2 + |-3i|^2 \eql 15
$$
and
$$
\inr{u}{u}
\eql \mat{1 & 2+i & -3i}\vecthree{1}{2-i}{3i}
\eql 15
$$
As expected, they should be the same.
Inner products, projectors, Hermitians and unitaries
Problem:
Let $$\kt{u} \eql \vectwo{i}{0}$$
What is \(\magsq{u}\)?
Solution:
Note that this is vector with two complex numbers:
the complex number \(i\) as the first element, and \(0\) as the
second. Then note that
$$
\br{u} \eql \mat{ -i & 0 } \mbx{Conjugated row}
$$
Thus,
$$\eqb{
\magsq{u} & \eql & \inr{u}{u} \\
& \eql & \mat{ -i & 0 } \mat{ i \\ 0 } \\
& \eql & -i^2 + 0 \\
& \eql & 1
}$$
Note: one might mistakenly think that \(\magsq{u} = i^2 + 0^2\).
Instead, \(\magsq{u} = \magsq{i} + \magsq{0} = 1 + 0 = 1\).
Recall that if \(z=i\), then \(\magsq{z} = zz^* = i(-i) = -i^2 = 1\).
Problem:
Thus far we've seen projectors defined with unit-length
vectors, as \(\kt{v}{v}\), and the completeness
relation for an orthonormal basis as \(\sum_i \kt{v_i}{v_i} = I\).
Does the completeness relation work for non-unit length
vectors? If not, what is the tweak needed to make it work?
Solution:
Consider, for example, \(\kt{v_1}=(1,1), \kt{v_2}=(-1,1)\).
Both are not unit-length since
$$\eqb{
\inr{v_1}{v_1} & \eql & 2\\
\inr{v_2}{v_2} & \eql & 2\\
}$$
The sum
$$
\otr{v_1}{v_1} + \otr{v_2}{v_2}
\eql
\vectwo{1}{1} \mat{1 & 1} + \vectwo{-1}{1} \mat{-1 & 1}
\eql
\mat{1 & 1\\ 1 & 1} + \mat{1 & -1\\ -1 & 1}
\eql
\mat{2 & 0\\ 0 & 2}
\; \neq \; I
$$
But it appears that scaling by the length (in this case, 2)
will result in \(I\). We can formalize this as follows.
Let \(\kt{v_1},\ldots,\kt{v_n}\) be an non-unit length
but orthogonal basis and \(\kt{u}\) be a vector.
that we express as:
$$
\kt{u} \eql
\sum_i \alpha_i \kt{v_i} \\
$$
where, as usual, each \(\alpha_i\) is obtained by multiplying
each side by \(\br{v_i}\):
$$
\inr{v_i}{u} \eql \alpha_i \inr{v_i}{v_i}
$$
(all other inner products are zero). And so,
$$
\alpha_i \eql \frac{ \inr{v_i}{u} }{ \inr{v_i}{v_i} }
\eql \frac{ \inr{v_i}{u} }{ |{v_i}|^2 }
$$
Now return to the expression of \(\kt{u}\):
$$\eqb{
\kt{u} & \eql &
\sum_i \alpha_i \kt{v_i} \\
& \eql &
\sum_i \frac{ \inr{v_i}{u} }{ |{v_i}|^2 } \; \kt{v_i} \\
& \eql &
\sum_i \frac{1}{ |{v_i}|^2 } \otr{v_i}{v_i} \; \kt{u} \\
& \eql &
\left( \sum_i \frac{1}{ |{v_i}|^2 } \otr{v_i}{v_i} \right) \; \kt{u} \\
}$$
Thus,
$$
\sum_i \frac{1}{ |{v_i}|^2 } \otr{v_i}{v_i} \eql I
$$
Which is the desired completeness relation. The real goal
of this exercise is to get more comfortable with Dirac
notation and "moving" the scalar to reformulate in terms
of the outerproduct.
Problem:
This problem is intended to increase familiarity with Dirac symbols.
Suppose \(\kt{0}, \kt{1}, \ldots, \kt{N-1}\) are \(N\) basis
vectors and that \(\kt{u}, \kt{v}\) are two vectors
in the same space expressed as
$$\eqb{
\kt{u} & \eql & \alpha_0 \kt{0} + \ldots + \alpha_{N-1} \kt{N-1}\\
\kt{v} & \eql & \beta_0 \kt{0} + \ldots + \beta_{N-1} \kt{N-1}\\
}$$
Show that
$$
\inr{u}{v} \eql \sum_{i=0}^{N-1} \alpha_i^* \beta_i
$$
Solution:
$$\eqb{
& \; & \inrh{ \alpha_0 \kt{0} + \ldots + \alpha_{N-1} \kt{N-1} }{\beta_0 \kt{0} + \ldots + \beta_{N-1} \kt{N-1} } \\
& \; & \eql
\beta_0 \inrh{ \alpha_0 \kt{0} + \ldots + \alpha_{N-1} \kt{N-1} }{0}
+ \ldots +
\beta_{N-1} \inrh{ \alpha_0 \kt{0} + \ldots + \alpha_{N-1} \kt{N-1} }{N-1} \\
& \; & \eql
\beta_0 \parenh{ \alpha_0^* \br{0} + \ldots + \alpha_{N-1}^*
\br{N-1} } \kt{0}
+ \ldots +
\beta_{N-1} \parenh{ \alpha_0^* \br{0} + \ldots + \alpha_{N-1}^* \br{N-1} } \kt{N-1} \\
& \; & \eql
\beta_0 \alpha_0^* \inr{0}{0} + \ldots +
\beta_{N-1} \alpha_{N-1}^* \inr{N-1}{N-1} \\
& \; & \eql
\alpha_0^* \beta_0 + \ldots + \alpha_{N-1}^* \beta_{N-1}
}$$
Notice that, in the first line, for additional clarity, we have left the vector
form in the left (conjugated) side of an inner-product:
$$
\inrh{ \mathbf{\alpha_0 \kt{0} + \ldots + \alpha_{N-1} \kt{N-1}} }{\beta_0 \kt{0} + \ldots + \beta_{N-1} \kt{N-1} } \\
$$
Here, the intention is to allow anything to sit inside the
inner-product, and to simplify it at the right time.
Problem:
Since inner products are computed often, let's do another
example problem. Show that
$$\eqb{
\kt{u} & \eql &
\frac{i}{\sqrt{3}} \kt{0} - \frac{\sqrt{2}}{\sqrt{3}} \kt{1} \\
\kt{u^\perp} & \eql &
- \frac{\sqrt{2}}{\sqrt{3}} \kt{0} -\frac{i}{\sqrt{3}} \kt{1} \\
}$$
are a 2D basis, and then compute the coefficients
when expressing \(\kt{+}\) in this basis.
Solution:
First, let's show they are orthogonal and of unit-length.
$$\eqb{
\inr{u}{u^\perp}
& \eql &
\inrh{ -\frac{i}{\sqrt{3}} \br{0} - \frac{\sqrt{2}}{\sqrt{3}}
\br{1} }{ - \frac{\sqrt{2}}{\sqrt{3}} \kt{0} + \frac{i}{\sqrt{3}} \kt{1} }\\
& \eql &
\frac{i\sqrt{2}}{3} \inr{0}{0}
-
\frac{i\sqrt{2}}{3} \inr{1}{1} \\
& \eql & 0
}$$
Unit-length:
$$\eqb{
\inr{u}{u} & \eql &
\inrh{ -\frac{i}{\sqrt{3}} \br{0} - \frac{\sqrt{2}}{\sqrt{3}}
\br{1} }{ \frac{i}{\sqrt{3}} \kt{0} - \frac{\sqrt{2}}{\sqrt{3}} \kt{1} }\\
& \eql &
-\frac{i^2}{3} + \frac{2}{3}\\
& \eql & 1
}$$
Similarly, \(\inr{u^\perp}{u^\perp} = 1\).
Next, the coefficients for \(\kt{+}\)
$$
\inr{u}{+}
\eql
\inrh{ -\frac{i}{\sqrt{3}} \br{0} - \frac{\sqrt{2}}{\sqrt{3}}
\br{1} }{ \isqt{1} \kt{0} + \isqt{1} \kt{1} }
\eql
- \frac{\sqrt{2}}{\sqrt{6}} - \frac{i}{\sqrt{6}}
$$
and
$$
\inr{u^\perp}{+}
\eql
\inrh{ -\frac{\sqrt{2}}{\sqrt{3}} \br{0} - \frac{i}{\sqrt{3}}
\br{1} }{ \isqt{1} \kt{0} + \isqt{1} \kt{1} }
\eql
- \frac{\sqrt{2}}{\sqrt{6}} - \frac{i}{\sqrt{6}}
$$
Check:
$$
\magsq{\inr{u}{+}} + \magsq{\inr{u^\perp}{+}}
\eql
\frac{1+2}{6} + \frac{1+2}{6}
\eql 1
$$
(See Proposition 2.9). Thus,
$$
\kt{+} \eql
\inr{u}{+} \kt{u} + \inr{u^\perp}{+} \kt{u^\perp}
\eql
\left(- \frac{\sqrt{2}}{\sqrt{6}} - \frac{i}{\sqrt{6}} \right) \kt{u}
\; + \;
\left(- \frac{\sqrt{2}}{\sqrt{6}} - \frac{i}{\sqrt{6}} \right) \kt{u^\perp}
$$
© 2022, Rahul Simha