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src/Advanced/Introduction to Quantum/main.typ Normal file
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#import "@local/handout:0.1.0": *
#show: handout.with(
title: [Intro to Quantum Computing],
by: "Mark",
)
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
#include "src/parts/01 bits.typ"
#pagebreak()
#include "src/parts/02 qubit.typ"
#pagebreak()
#include "src/parts/03 two qubits.typ"
#pagebreak()
// DONE UNTIL HERE
#include "src/parts/04 logic gates.typ"
#pagebreak()
#include "src/parts/05 quantum gates.typ"
#pagebreak()
#include "src/parts/06 hxh.typ"
#pagebreak()
#include "src/parts/07 superdense.typ"
#pagebreak()
#include "src/parts/08 teleport.typ"

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src/Advanced/Introduction to Quantum/src/parts/01 bits.typ Normal file
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#import "@local/handout:0.1.0": *
#import "@preview/cetz:0.4.2"
= Probabilistic Bits
#definition()
As we already know, a _classical bit_ may take the values `0` and `1`.
We can model this with a two-sided coin, one face of which is labeled `0`, and the other, `1`.
#v(2mm)
Of course, if we toss such a "bit-coin," we'll get either `0` or `1`.
We'll denote the probability of getting `0` as $p_0$, and the probability of getting `1` as $p_1$.
As with all probabilities, $p_0 + p_1$ must be equal to 1.
#v(1fr)
#definition()
Say we toss a "bit-coin" and don't observe the result. We now have a _probabilistic bit_, with a probability $p_0$ of being `0`, and a probability $p_1$ of being `1`.
#v(2mm)
We'll represent this probabilistic bit's _state_ as a vector: $mat(p_0; p_1)$
We do *not* assume this coin is fair, and thus $p_0$ might not equal $p_1$.
#note[This may seem a bit redundant: since $p_0 + p_1 = 1$, we can always calculate one probability given the other. We'll still include both probabilities in the state vector, since this provides a clearer analogy to quantum bits.]
#v(1fr)
#definition()
The simplest probabilistic bit states are of course $[0]$ and $[1]$, defined as follows:
- $[0] = mat(1; 0)$
- $[1] = mat(0; 1)$
That is, $[0]$ represents a bit that we known to be `0`, and $[1]$ represents a bit we know to be `1`.
#v(1fr)
#definition()
$[0]$ and $[1]$ form a _basis_ for all possible probabilistic bit states:
Every other probabilistic bit can be written as a _linear combination_ of $[0]$ and $[1]$:
$ mat(p_0; p_1) = p_0 mat(1; 0) + p_1 mat(0; 1) = p_0 [0] + p_1 [1] $
#v(1fr)
#pagebreak()
#problem()
Every possible state of a probabilistic bit is a two-dimensional vector.
Draw all possible states on the axis below.
#table(
columns: (1fr,),
align: center,
stroke: none,
align(center, cetz.canvas({
import cetz.draw: *
set-style(content: (frame: "rect", stroke: none, fill: none, padding: .25))
scale(200%)
line(
(0, 1.5),
(0, 0),
(1.5, 0),
stroke: black + 0.25mm,
)
mark((0, 1.5), (0, 2), symbol: ")>", fill: black)
mark((1.5, 0), (2, 0), symbol: ")>", fill: black)
content((0, 1.5), $p_1$, anchor: "south")
content((1.5, 0), $p_0$, anchor: "west")
circle((0, 0), radius: 0.6mm, fill: black, name: "00")
content("00.south", $mat(0; 0)$, anchor: "north")
circle((0, 1), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.west", $[1]$, anchor: "east")
circle((1, 0), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.south", $[0]$, anchor: "north")
})),
)
#solution[
#table(
columns: (1fr,),
align: center,
stroke: none,
align(center, cetz.canvas({
import cetz.draw: *
set-style(content: (
frame: "rect",
stroke: none,
fill: none,
padding: .25,
))
scale(200%)
line(
(0, 1.5),
(0, 0),
(1.5, 0),
stroke: black + 0.25mm,
)
mark((0, 1.5), (0, 2), symbol: ")>", fill: black)
mark((1.5, 0), (2, 0), symbol: ")>", fill: black)
content((0, 1.5), $p_1$, anchor: "south")
content((1.5, 0), $p_0$, anchor: "west")
line(
(1, 0),
(0, 1),
stroke: ored + 1mm,
)
circle((0, 0), radius: 0.6mm, fill: black, name: "00")
content("00.south", $mat(0; 0)$, anchor: "north")
circle((0, 1), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.west", $[1]$, anchor: "east")
circle((1, 0), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.south", $[0]$, anchor: "north")
})),
)
]
#v(1fr)
#pagebreak()
= Measuring Probabilistic Bits
#definition()
As we noted before, a probabilistic bit represents a coin we've tossed but haven't looked at.
We do not know whether the bit is `0` or `1`, but we do know the probability of both of these outcomes.
#v(2mm)
If we _measure_ (or _observe_) a probabilistic bit, we see either `0` or `1`—and thus our knowledge of its state is updated to either $[0]$ or $[1]$, since we now certainly know what face the coin landed on.
#v(2mm)
Since measurement changes what we know about a probabilistic bit, it changes the probabilistic bit's state. When we measure a bit, its state _collapses_ to either $[0]$ or $[1]$, and the original state of the bit vanishes. We _cannot_ recover the state $[x_0, x_1]$ from a measured probabilistic bit.
#definition("Multiple bits")
Say we have two probabilistic bits, $x$ and $y$, with states $[x] = [x_0, x_1]$ and $[y] = [y_0, y_1]$
#v(2mm)
The _compound state_ of $[x]$ and $[y]$ is exactly what it sounds like: It is the probabilistic two-bit state $|x y angle.r$, where the probabilities of the first bit are determined by $[x]$, and the probabilities of the second are determined by $[y]$.
#problem(label: "firstcompoundstate")
Say $[x] = [2/3, 1/3]$ and $[y] = [3/4, 1/4]$.
- If we measure $x$ and $y$ simultaneously, what is the probability of getting each of `00`, `01`, `10`, and `11`?
- If we measure $y$ first and observe `1`, what is the probability of getting each of `00`, `01`, `10`, and `11`?
#note[*Note:* $[x]$ and $[y]$ are column vectors, but I've written them horizontally to save space.]
#v(1fr)
#problem()
Say $[x] = [2/3, 1/3]$ and $[y] = [3/4, 1/4]$.
What is the probability that $x$ and $y$ produce different outcomes?
#v(1fr)
#pagebreak()
= Tensor Products
#definition("Tensor Products")
The _tensor product_ of two vectors is defined as follows:
$
mat(x_1; x_2) times.circle mat(y_1; y_2) = mat(x_1 mat(y_1; y_2); x_2 mat(y_1; y_2)) = mat(x_1 y_1; x_1 y_2; x_2 y_1; x_2 y_2)
$
That is, we take our first vector, multiply the second vector by each of its components, and stack the result. You could think of this as a generalization of scalar multiplication, where scalar multiplication is a tensor product with a vector in $RR^1$:
$
a mat(x_1; x_2) = mat(a_1) times.circle mat(y_1; y_2) = mat(a_1 mat(y_1; y_2)) = mat(a_1 y_1; a_1 y_2)
$
#problem()
Say $x in RR^n$ and $y in RR^m$.
What is the dimension of $x times.circle y$?
#v(1fr)
#problem(label: "basistp")
What is the following pairwise tensor product?
#v(4mm)
$
{mat(1; 0; 0), mat(0; 1; 0), mat(0; 0; 1)}
times.circle
{mat(1; 0), mat(0; 1)}
$
#v(4mm)
#hint[Distribute the tensor product between every pair of vectors.]
#v(1fr)
#problem()
What is the _span_ of the vectors we found in @basistp?
In other words, what is the set of vectors that can be written as linear combinations of the vectors above?
#v(1fr)
#pagebreak()
#problem()
Say $[x] = [2/3, 1/3]$ and $[y] = [3/4, 1/4]$.
What is $[x] times.circle [y]$? How does this relate to @firstcompoundstate?
#v(1fr)
#problem()
The compound state of two vector-form bits is their tensor product.
Compute the following. Is the result what we'd expect?
- $[0] times.circle [0]$
- $[0] times.circle [1]$
- $[1] times.circle [0]$
- $[1] times.circle [1]$
#hint[Remember that $[0] = mat(1; 0)$ and $[1] = mat(0; 1)$.]
#v(1fr)
#problem(label: "fivequant")
Writing $[0] times.circle [1]$ is a bit tedious. We'll shorten this notation to $[01]$.
In fact, we could go further: if we wanted to write the set of bits $[1] times.circle [1] times.circle [0] times.circle [1]$, \
we could write $[1101]$—but a shorter alternative is $[13]$, since $13$ is `1101` in binary.
#v(2mm)
Write $[5]$ as a three-bit probabilistic state.
#solution[
$[5] = [101] = [1] times.circle [0] times.circle [1] = [0,0,0,0,0,1,0,0]^T$ \
Notice how we're counting from the top, with $[000] = [1,0,...,0]$ and $[111] = [0, ..., 0, 1]$.
]
#v(1fr)
#problem()
Write the three-bit states $[0]$ through $[7]$ as column vectors.
#hint[You do not need to compute every tensor product. Do a few and find the pattern.]
#v(1fr)
#pagebreak()
= Operations on Probabilistic Bits
Now that we can write probabilistic bits as vectors, we can represent operations on these bits with linear transformations—in other words, as matrices.
#definition()
Consider the NOT gate, which operates as follows:
- $"NOT"[0] = [1]$
- $"NOT"[1] = [0]$
What should NOT do to a probabilistic bit $[x_0, x_1]$?
If we return to our coin analogy, we can think of the NOT operation as flipping a coin we have already tossed, without looking at its state. Thus,
$ "NOT" mat(x_0; x_1) = mat(x_1; x_0) $
#review_box("Review: Multiplying vectors by matrices")[
#v(2mm)
$
A v = mat(1, 2; 3, 4) mat(v_0; v_1) = mat(1 v_0 + 2 v_1; 3 v_0 + 4 v_1)
$
#v(2mm)
Note that each element of $A v$ is the dot product of a row in $A$ and a column in $v$.
]
#problem()
Compute the following product:
$ mat(1, 0.5; 0, 1) mat(3; 2) $
#v(1fr)
#remark()
Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. We often use the terms _matrix_, _transformation_, and _linear map_ interchangeably.
#pagebreak()
#problem()
Find the matrix that represents the NOT operation on one probabilistic bit.
#solution[
$
mat(0, 1; 1, 0)
$
]
#v(1fr)
#problem("Extension by linearity")
Say we have an arbitrary operation $M$.
If we know how $M$ acts on $[1]$ and $[0]$, can we compute $M[x]$ for an arbitrary state $[x]$?
Say $[x] = [x_0, x_1]$.
- What is the probability we observe $0$ when we measure $x$?
- What is the probability that we observe $M[0]$ when we measure $M x$?
#v(1fr)
#problem(label: "linearextension")
Write $M[x_0, x_1]$ in terms of $M[0]$, $M[1]$, $x_0$, and $x_1$.
#solution[
$
M mat(x_0; x_1) = x_0 M mat(1; 0) + x_1 M mat(0; 1) = x_0 M[0] + x_1 M[1]
$
]
#v(1fr)
#remark() Every matrix represents a _linear_ map, so the following is always true:
$ A times (p x + q y) = p A x + q A y $
@linearextension is just a special case of this fact.

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#import "@local/handout:0.1.0": *
#import "@preview/cetz:0.4.2"
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
= One Qubit
Quantum bits (or _qubits_) are very similar to probabilistic bits, but have one major difference: probabilities are replaced with _amplitudes_.
#v(2mm)
Of course, a qubit can take the values `0` and `1`, which are denoted $#ket("0")$ and $#ket("1")$.
Like probabilistic bits, a quantum bit is written as a linear combination of $#ket("0")$ and $#ket("1")$:
$ #ket([$psi$]) = psi_0 #ket("0") + psi_1 #ket("1") $
Such linear combinations are called _superpositions_.
#v(2mm)
The $#ket("")$ you see in the expressions above is called a "ket," and denotes a column vector.
$#ket("0")$ is pronounced "ket zero," and $#ket("1")$ is pronounced "ket one." This is called bra-ket notation.
#note[*Note:* $#bra("0")$ is called a "bra," but we won't worry about that for now.]
#v(2mm)
This is very similar to the "box" $[#h(1.5mm)]$ notation we used for probabilistic bits.
As before, we will write $#ket("0") = mat(1; 0)$ and $#ket("1") = mat(0; 1)$.
#v(8mm)
Recall that probabilistic bits are subject to the restriction that $p_0 + p_1 = 1$.
Quantum bits have a similar condition: $psi_0^2 + psi_1^2 = 1$.
Note that this implies that $psi_0$ and $psi_1$ are both in $[-1, 1]$.
Quantum amplitudes may be negative, but probabilistic bit probabilities cannot.
#v(2mm)
If we plot the set of valid quantum states on our plane, we get a unit circle centered at the origin:
#table(
columns: (1fr,),
align: center,
stroke: none,
align(center, cetz.canvas({
import cetz.draw: *
set-style(content: (frame: "rect", stroke: none, fill: none, padding: .25))
scale(150%)
line(
(0, 1.5),
(0, 0),
(1.5, 0),
stroke: black + 0.25mm,
)
mark((0, 1.5), (0, 2), symbol: ")>", fill: black)
mark((1.5, 0), (2, 0), symbol: ")>", fill: black)
circle((0, 0), radius: 1, stroke: (
paint: black,
thickness: 0.25mm,
dash: "dashed",
))
content((0, 1.5), $p_1$, anchor: "south")
content((1.5, 0), $p_0$, anchor: "west")
circle((0, 1), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.west", $#ket("1")$, anchor: "east")
circle((1, 0), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.south", $#ket("0")$, anchor: "north")
circle((0.87, 0.5), radius: 0.6mm, fill: ored, stroke: ored, name: "00")
content("00.east", $#ket(math.psi)$, anchor: "west")
})),
)
Recall that the set of probabilistic bits forms a line instead:
#table(
columns: (1fr,),
align: center,
stroke: none,
align(center, cetz.canvas({
import cetz.draw: *
set-style(content: (frame: "rect", stroke: none, fill: none, padding: .25))
scale(150%)
line(
(0, 1.5),
(0, 0),
(1.5, 0),
stroke: black + 0.25mm,
)
mark((0, 1.5), (0, 2), symbol: ")>", fill: black)
mark((1.5, 0), (2, 0), symbol: ")>", fill: black)
line(
(1, 0),
(0, 1),
stroke: ored + 1mm,
)
content((0, 1.5), $p_1$, anchor: "south")
content((1.5, 0), $p_0$, anchor: "west")
circle((0, 0), radius: 0.6mm, fill: black, name: "00")
content("00.south", $mat(0; 0)$, anchor: "north")
circle((0, 1), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.west", $[1]$, anchor: "east")
circle((1, 0), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.south", $[0]$, anchor: "north")
})),
)
#problem()
In the above unit circle, the counterclockwise angle from $#ket("0")$ to $#ket([$psi$])$ is $30°$.
Write $#ket([$psi$])$ as a linear combination of $#ket("0")$ and $#ket("1")$.
#v(1fr)
#pagebreak()
#definition("Measurement I")
Just like a probabilistic bit, we must observed $#ket("0")$ or $#ket("1")$ when we measure a qubit.
If we were to measure $#ket([$psi$]) = psi_0 #ket("0") + psi_1 #ket("1")$, we'd observe either $#ket("0")$ or $#ket("1")$, with the following probabilities:
- $cal(P)(#ket("1")) = psi_1^2$
- $cal(P)(#ket("0")) = psi_0^2$
#note[Note that $cal(P)(#ket("0")) + cal(P)(#ket("1")) = 1$.]
#v(2mm)
As before, $#ket([$psi$])$ _collapses_ when it is measured: its state becomes that which we observed in our measurement, leaving no trace of the previous superposition.
#problem()
- What is the probability we observe $#ket("0")$ when we measure $#ket([$psi$])$?
- What can we observe if we measure $#ket([$psi$])$ a second time?
- What are these probabilities for $#ket([$phi$])$?
#table(
columns: (1fr,),
align: center,
stroke: none,
align(center, cetz.canvas({
import cetz.draw: *
set-style(content: (frame: "rect", stroke: none, fill: none, padding: .25))
scale(200%)
line(
(0, 1.5),
(0, 0),
(1.5, 0),
stroke: black + 0.25mm,
)
mark((0, 1.5), (0, 2), symbol: ")>", fill: black)
mark((1.5, 0), (2, 0), symbol: ")>", fill: black)
circle((0, 0), radius: 1, stroke: (
paint: black,
thickness: 0.25mm,
dash: "dashed",
))
content((0, 1.5), $p_1$, anchor: "south")
content((1.5, 0), $p_0$, anchor: "west")
circle((0, 1), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.west", $#ket("1")$, anchor: "east")
circle((1, 0), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.south", $#ket("0")$, anchor: "north")
circle((0.87, 0.5), radius: 0.6mm, fill: ored, stroke: ored, name: "00")
content("00.east", $#ket(math.psi)$, anchor: "west")
arc(
(0, 0),
start: 0deg,
stop: -135deg,
anchor: "origin",
radius: 0.3,
name: "a135",
stroke: gray,
)
mark(
"a135.end",
135deg,
symbol: ")>",
fill: gray,
stroke: gray,
)
content("a135.center", text(fill: gray)[$135degree$], anchor: "north")
line((0, 0), (-0.607, -0.607), stroke: (
paint: gray,
thickness: 0.4mm,
dash: "dotted",
))
mark(
(-0.627, -0.627),
(-0.708, -0.708),
symbol: ")>",
fill: gray,
stroke: gray,
)
arc(
(0, 0),
start: 0deg,
stop: 30deg,
anchor: "origin",
radius: 0.6,
name: "a30",
stroke: gray,
)
mark(
"a30.end",
120deg,
symbol: ")>",
fill: gray,
stroke: gray,
)
content("a30.end", text(fill: gray)[$30degree$], anchor: "south")
line((0, 0), (0.87, 0.5), stroke: (
paint: gray,
thickness: 0.4mm,
dash: "dotted",
))
mark(
(0.80, 0.46),
(0.87, 0.5),
symbol: ")>",
fill: gray,
stroke: gray,
)
circle(
(-0.707, -0.707),
radius: 0.6mm,
fill: ored,
stroke: ored,
name: "00",
)
content("00.west", $#ket(math.phi)$, anchor: "east")
})),
)
#v(1fr)
As you may have noticed, we don't need two coordinates to fully define a qubit's state. We can get by with one coordinate just as well.
Instead of referring to each state using its cartesian coordinates $psi_0$ and $psi_1$, we can address it using its _polar angle_ $theta$, measured from $#ket("0")$ counterclockwise:
#table(
columns: (1fr,),
align: center,
stroke: none,
align(center, cetz.canvas({
import cetz.draw: *
set-style(content: (frame: "rect", stroke: none, fill: none, padding: .25))
scale(180%)
line(
(0, 1.5),
(0, 0),
(1.5, 0),
stroke: black + 0.25mm,
)
mark((0, 1.5), (0, 2), symbol: ")>", fill: black)
mark((1.5, 0), (2, 0), symbol: ")>", fill: black)
circle((0, 0), radius: 1, stroke: (
paint: black,
thickness: 0.25mm,
dash: "dashed",
))
content((0, 1.5), $p_1$, anchor: "south")
content((1.5, 0), $p_0$, anchor: "west")
circle((0, 1), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.west", $#ket("1")$, anchor: "east")
circle((1, 0), radius: 0.6mm, fill: oblue, stroke: oblue, name: "00")
content("00.south", $#ket("0")$, anchor: "north")
circle((0.87, 0.5), radius: 0.6mm, fill: ored, stroke: ored, name: "00")
content("00.east", $#ket(math.psi)$, anchor: "west")
arc(
(0, 0),
start: 0deg,
stop: 30deg,
anchor: "origin",
radius: 0.6,
name: "a30",
stroke: gray,
)
mark(
"a30.end",
120deg,
symbol: ")>",
fill: gray,
stroke: gray,
)
content("a30.mid", text(fill: gray)[$theta$], anchor: "west")
line((0, 0), (0.87, 0.5), stroke: (
paint: gray,
thickness: 0.4mm,
dash: "dotted",
))
mark(
(0.80, 0.46),
(0.87, 0.5),
symbol: ")>",
fill: gray,
stroke: gray,
)
})),
)
#problem()
Find $psi_0$ and $psi_1$ in terms of $theta$ for an arbitrary qubit $psi$.
#v(1fr)
#pagebreak()
#problem()
Consider the following qubit states:
#grid(
columns: (1fr, 1fr),
$ #ket("+") = (#ket("0") + #ket("1"))/sqrt(2) $,
$ #ket("-") = (#ket("0") - #ket("1"))/sqrt(2) $,
)
- Where are these on the unit circle?
- What are their polar angles?
- What are the probabilities of observing $#ket("0")$ and $#ket("1")$ when measuring $#ket("+")$ and $#ket("-")$?
#table(
columns: (1fr,),
align: center,
stroke: none,
align(center, cetz.canvas({
import cetz.draw: *
set-style(content: (frame: "rect", stroke: none, fill: none, padding: .25))
scale(300%)
line(
(0, 1.3),
(0, 0),
(1.3, 0),
stroke: black + 0.25mm,
)
mark((0, 1.3), (0, 2), symbol: ")>", fill: black)
mark((1.3, 0), (2, 0), symbol: ")>", fill: black)
circle((0, 0), radius: 1, stroke: (
paint: black,
thickness: 0.25mm,
dash: "dashed",
))
content((0, 1.3), $p_1$, anchor: "south")
content((1.3, 0), $p_0$, anchor: "west")
circle((0, 1), radius: 0.4mm, fill: oblue, stroke: oblue, name: "00")
content("00.west", $#ket("1")$, anchor: "east")
circle((1, 0), radius: 0.4mm, fill: oblue, stroke: oblue, name: "00")
content("00.south", $#ket("0")$, anchor: "north")
})),
)
#v(1fr)
#v(1fr)
#v(1fr)
#pagebreak()
= Operations on One Qubit
We may apply transformations to qubits just as we apply transformations to probabilistic bits. Again, we'll represent transformations as $2 times 2$ matrices, since we want to map one qubit state to another.
#note[In other words, we want to map elements of $RR^2$ to elements of $RR^2$.]
We will call such maps _quantum gates,_ since they are the quantum equivalent of classical logic gates.
#v(2mm)
There are two conditions a valid quantum gate $G$ must satisfy:
- For any valid state $#ket([$psi$])$, $G #ket([$psi$])$ is a valid state. Namely, $G$ must preserve the length of any vector it is applied to. Recall that the set of valid quantum states is the set of unit vectors in $RR^2$
- Any quantum gate must be _invertible_. We'll skip this condition for now, and return to it later.
In short, a quantum gate is a linear map that maps the unit circle to itself. There are only two kinds of linear maps that do this: reflections and rotations.
#problem()
The $X$ gate is the quantum analog of the `not` gate, defined by the following table:
- $X #ket("0") = #ket("1")$
- $X #ket("1") = #ket("0")$
Find the matrix $X$.
#solution[
$
mat(0, 1; 1, 0)
$
]
#v(1fr)
#problem()
What is $X #ket("+")$ and $X #ket("-")$?
#hint[Remember that all matrices are linear maps. What does this mean?]
#solution[
$X #ket("+") = #ket("+")$ and $X #ket("-") = - #ket("-")$ (that is, a negative ket-minus). \
Most notably, rememver that $G(a#ket("0") + b #ket("1")) = a G #ket("0") + b G #ket("1")$.
]
#v(1fr)
#problem()
In terms of geometric transformations, what does $X$ do to the unit circle?
#solution[
It is a reflection about the $45degree$ axis.
]
#v(1fr)
#pagebreak()
#problem()
Let $Z$ be a quantum gate defined by the following table:
- $Z #ket("0") = #ket("0")$,
- $Z #ket("1") = -#ket("1")$.
What is the matrix $Z$? What are $Z #ket("+")$ and $Z #ket("-")$?
What is $Z$ as a geometric transformation?
#v(1fr)
#problem()
Is the map $B$ defined by the table below a valid quantum gate?
- $B #ket("0") = #ket("0")$
- $B #ket("1") = #ket("+")$
#hint[Find a $#ket([$psi$])$ so that $B #ket([$psi$])$ is not a valid qubit state]
#solution[
$ B #ket("+") = (1 + sqrt(2))/(2) #ket("0") + 1/2 #ket("1") $
This has a non-unit length of
$
(sqrt(2) + 1)/(2)
$
]
#v(1fr)
#problem("Rotation")
As we noted earlier, any rotation about the center is a valid quantum gate. Let's derive all transformations of this form.
- Let $U_theta$ be the matrix that represents a counterclockwise rotation of $theta$ degrees. What is $U #ket("0")$ and $U #ket("1")$?
- Find the matrix $U_theta$ for an arbitrary $theta$.
#v(1fr)
#problem()
Say we have a qubit that is either $#ket("+")$ or $#ket("-")$. We do not know which of the two states it is in.
Using one operation and one measurement, how can we find out, for certain, which qubit we received?
#v(1fr)

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#import "@local/handout:0.1.0": *
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
= Two Qubits
#definition()
Just as before, we'll represent multi-qubit states as linear combinations of multi-qubit basis states.
For example, a two-qubit state $#ket("ab")$ is the four-dimensional unit vector
$ mat(a; b; c; d) = a #ket("00") + b #ket("01") + c #ket("10") + d #ket("11") $
As always, multi-qubit states are unit vectors. Thus, $a^2 + b^2 + c^2 + d^2 = 1$ in the two-bit case above.
#problem()
Say we have two qubits $#ket([$psi$])$ and $#ket([$phi$])$.
Show that $#ket([$psi$]) times.circle #ket([$phi$])$ is always a unit vector (and is thus a valid quantum state).
#v(1fr)
#definition("Measurement II")<measureii>
Measurement of a two-qubit state works just like measurement of a one-qubit state:
If we measure $a #ket("00") + b #ket("01") + c #ket("10") + d #ket("11")$, we get one of the four basis states with the following probabilities:
- $cal(P)(#ket("00")) = a^2$
- $cal(P)(#ket("01")) = b^2$
- $cal(P)(#ket("10")) = c^2$
- $cal(P)(#ket("11")) = d^2$
As before, the sum of all the above probabilities is $1$.
#problem()
Consider the two-qubit state
$#ket([$psi$]) = 1/sqrt(2) #ket("00") + 1/2 #ket("01") + sqrt(3)/4 #ket("10") + 1/4 #ket("11")$
- If we measure both bits of $#ket([$psi$])$ simultaneously, what is the probability of getting each of $#ket("00")$, $#ket("01")$, $#ket("10")$, and $#ket("11")$?
- If we measure the ONLY the first qubit, what is the probability we get $#ket("0")$? How about $#ket("1")$?
#hint[There are two basis states in which the first qubit is $#ket("0")$.]
- Say we measured the second bit and read $#ket("1")$. If we now measure the first bit, what is the probability of getting $#ket("0")$?
#v(1fr)
#pagebreak()
#problem()
Again, consider the two-qubit state
$#ket([$psi$]) = 1/sqrt(2) #ket("00") + 1/2 #ket("01") + sqrt(3)/4 #ket("10") + 1/4 #ket("11")$
If we measure the first qubit of $#ket([$psi$])$ and get $#ket("0")$, what is the resulting state of $#ket([$psi$])$?
What would the state be if we'd measured $#ket("1")$ instead?
#v(1fr)
#problem()
Consider the three-qubit state $#ket([$psi$]) = c_0 #ket("000") + c_1 #ket("001") + ... + c_7 #ket("111")$.
Say we measure the first two qubits and get $#ket("00")$. What is the resulting state of $#ket([$psi$])$?
#solution[
We measure $#ket("00")$ with probability $c_0^2 + c_1^2$, and $#ket(math.psi)$ collapses to
#v(3mm)
$
(c_0 #ket("000") + c_1 #ket("001"))/(sqrt(c_0^2 + c_1^2))
$
]
#v(1fr)
#pagebreak()
#definition("Entanglement")
Some product states can be factored into a tensor product of individual qubit states. For example,
$
1/2 (#ket("00") + #ket("01") + #ket("10") + #ket("11")) = 1/sqrt(2) (#ket("0") + #ket("1")) times.circle 1/sqrt(2) (#ket("0") + #ket("1"))
$
Such states are called _product states._ States that aren't product states are called _entangled_ states.
#problem()
Factor the following product state:
$
1/(2sqrt(2)) (sqrt(3) #ket("00") - sqrt(3) #ket("01") + #ket("10") - #ket("11"))
$
#solution[
$
(1)/(2 sqrt(2)) (sqrt(3) #ket("00") - sqrt(3) #ket("01") + #ket("10") - #ket("11"))
= (sqrt(3)/2 #ket("0") + 1/2 #ket("1") )
times.circle
( 1/sqrt(2) #ket(0) - 1/sqrt(2) #ket("1"))
$
]
#v(1fr)
#problem()
Show that the following is an entangled state.
$ 1/sqrt(2) #ket("00") + 1/sqrt(2) #ket("11") $
#solution[
$
mat(a_0; a_1)
times.circle
mat(b_0; b_1)
=
a_0b_0 #ket(00) + a_0b_1 #ket(01) + a_1b_0 #ket(10) + a_1b_1 #ket(11)
$
#v(2mm)
So, we have that $a_1b_1 = a_0b_0 = sqrt(2)^(-1)$ \
But $a_0b_1 = a_1b_0 = 0$, so one of $a_0$ and $b_1$ must be zero. \
We thus have a contradiction.
]
#v(1fr)

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#import "@local/handout:0.1.0": *
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
= Logic Gates
#definition("Matrices")
Throughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix, and that every matrix represents a linear map. For example, if $f: RR^2 -> RR^2$ is a linear map, we can write it as follows:
$
f(#ket("x")) = mat(m_1, m_2; m_3, m_4) mat(x_1; x_2) = mat(m_1 x_1 + m_2 x_2; m_3 x_1 + m_4 x_2)
$
#definition()
Before we discussing multi-qubit quantum gates, we need to review to classical logic.
Of course, a classical logic gate is a linear map from ${0,1}^m$ to ${0,1}^n$
#problem()<notgatex>
The `not` gate is a map defined by the following table:
- $X #ket("0") = #ket("1")$
- $X #ket("1") = #ket("0")$
Write the `not` gate as a matrix that operates on single-bit vector states.
That is, find a matrix $X$ so that $X mat(1; 0) = mat(0; 1)$ and $X mat(0; 1) = mat(1; 0)$
#solution[
$
X = mat(0, 1; 1, 0)
$
]
#v(1fr)
#problem()
The `and` gate is a map $BB^2 -> BB$ defined by the following table:
#align(center, table(
columns: 3,
stroke: none,
table.hline(),
[`a`], [`b`], [`a` and `b`],
table.hline(),
[0], [0], [0],
[0], [1], [0],
[1], [0], [0],
[1], [1], [1],
table.hline(),
))
Find a matrix $A$ so that $A #ket("ab")$ works as expected.
#hint[Remember, we write bits as vectors.]
#solution[
$
A = mat(1, 1, 1, 0; 0, 0, 0, 1)
$
#instructornote[
Because of the way we represent bits here, we also have the following property: \
The columns of $A$ correspond to the output for each input---i.e, $A$ is just a table of outputs. \
#v(2mm)
For example, if we look at the first column of $A$ (which is $[1, 0]$), we see: \
$A#ket(00) = A[1,0,0,0] = [1,0] = #ket(0)$
#v(2mm)
Also with the last column (which is $[0,1]$): \
$A#ket(00) = A[0,0,0,1] = [0,1] = #ket(1)$
]
]
#v(1fr)
#pagebreak()
#remark()
The way a quantum circuit handles information is a bit different than the way a classical circuit does. We usually think of logic gates as _functions_: they consume one set of bits, and return another/
// TODO: and gate (input a, input b, output)
#v(2mm)
This model, however, won't work for quantum logic. If we want to understand quantum gates, we need to see them not as _functions_, but as _transformations_. This distinction is subtle, but significant:
- functions _consume_ a set of inputs and _produce_ a set of outputs
- transformations _change_ a set of objects, without adding or removing any elements
#v(2mm)
Our usual logic circuit notation models logic gates as functions—we thus can't use it. We'll need a different diagram to draw quantum circuits.
#v(1fr)
First, we'll need a set of bits. For this example, we'll use two, drawn in a vertical array. We'll also add a horizontal time axis, moving from left to right:
#align(center)[
// Quantum circuit diagram showing two qubits over time
#box(width: 10cm, height: 4cm)[
_[Quantum circuit diagram with time axis would go here]_
]
]
In the diagram above, we didn't change our bits—so the labels at the start match those at the end.
#v(1fr)
Thus, our circuit forms a grid, with bits ordered vertically and time horizontally. If we want to change our state, we draw transformations as vertical boxes. Every column represents a single transformation on the entire state:
#align(center)[
// Quantum circuit with transformations
#box(width: 10cm, height: 4cm)[
_[Quantum circuit with transformations $T_1$, $T_2$, $T_3$ would go here]_
]
]
Note that the transformations above span the whole state. This is important: we cannot apply transformations to individual bits—we always transform the _entire_ state.
#v(1fr)
#pagebreak()
*Setup:* Say we want to invert the first bit of a two-bit state. That is, we want a transformation $T$ so that
#align(center)[
// Circuit showing bit flip
#box(width: 8cm, height: 3cm)[
_[Circuit diagram showing first bit flip would go here]_
]
]
In other words, we want a matrix $T$ satisfying the following equalities:
- $T #ket("00") = #ket("10")$
- $T #ket("01") = #ket("11")$
- $T #ket("10") = #ket("00")$
- $T #ket("11") = #ket("01")$
#problem()
Find the matrix that corresponds to the above transformation.
#hint[Remember that $#ket("0") = mat(1; 0)$ and $#ket("1") = mat(0; 1)$. Also, we found earlier that $X = mat(0, 1; 1, 0)$, and of course $I = mat(1, 0; 0, 1)$.]
#v(1fr)
*Remark:* We could draw the above transformation as a combination $X$ and $I$ (identity) gate:
#align(center)[
// Circuit with X and I gates
#box(width: 6cm, height: 3cm)[
_[Circuit diagram with X gate on first qubit, I gate on second would go here]_
]
]
We can even omit the $I$ gate, since we now know that transformations affect the whole state:
#align(center)[
// Simplified circuit with just X gate
#box(width: 6cm, height: 3cm)[
_[Simplified circuit diagram with just X gate on first qubit would go here]_
]
]
We're now done: this is how we draw quantum circuits. Don't forget that transformations _always_ affect the whole state—even if our diagram doesn't explicitly state this.
#pagebreak()

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#import "@local/handout:0.1.0": *
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
= Quantum Gates
In the previous section, we stated that a quantum gate is a linear map. Let's complete that definition.
#definition()
A quantum gate is a _orthonormal matrix_, which means any gate $G$ satisfies $G G^T = I$.
This implies the following:
- $G$ is square. In other words, it has as many rows as it has columns.
#note[If we think of $G$ as a map, this means that $G$ has as many inputs as it has outputs. This is to be expected: we stated earlier that quantum gates do not destroy or create qubits.]
- $G$ preserves lengths; i.e $|x| = |G x|$.
#note[This ensures that $G #ket([$psi$])$ is always a valid state.]
(You will prove all these properties in any introductory linear algebra course. This isn't a lesson on linear algebra, so you may take them as given today.)
*Remark:* Let $G$ be a quantum gate. Since quantum gates are, by definition, _linear_ maps, the following holds:
$ G(a_0 #ket("0") + a_1 #ket("1")) = a_0 G #ket("0") + a_1 G #ket("1") $
#problem()<cnot>
Consider the _controlled not_ (or _cnot_) gate, defined by the following table:
- $X_c #ket("00") = #ket("00")$
- $X_c #ket("01") = #ket("01")$
- $X_c #ket("10") = #ket("11")$
- $X_c #ket("11") = #ket("10")$
In other words, the cnot gate inverts its second bit if its first bit is $#ket("1")$.
Find the matrix that applies the cnot gate.
#v(1fr)
#pagebreak()

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#import "@local/handout:0.1.0": *
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
= HXH
Let's return to the quantum circuit diagrams we discussed a few pages ago. Keep in mind that we're working with quantum gates and proper qubits—not classical bits, as we were before.
#definition("Controlled Inputs")
A _control input_ or _inverted control input_ may be attached to any gate. These are drawn as filled and empty circles in our circuit diagrams:
#align(center)[
#grid(columns: (1fr, 1fr),
[
// Non-inverted control circuit diagram would go here
#box(width: 6cm, height: 4cm)[
_[Non-inverted control input circuit would go here]_
]
],
[
// Inverted control circuit diagram would go here
#box(width: 6cm, height: 4cm)[
_[Inverted control input circuit would go here]_
]
]
)]
#v(2mm)
An $X$ gate with a (non-inverted) control input behaves like an $X$ gate if _all_ its control inputs are $#ket("1")$, and like $I$ otherwise. An $X$ gate with an inverted control inputs does the opposite, behaving like $I$ if its input is $#ket("1")$ and like $X$ otherwise. The two circuits above illustrate this fact—take a look at their inputs and outputs.
#v(2mm)
Of course, we can give a gate multiple controls. An $X$ gate with multiple controls behaves like an $X$ gate if...
- all non-inverted controls are $#ket("1")$, and
- all inverted controls are $#ket("0")$
...and like $I$ otherwise.
#problem()
What are the final states of the qubits in the diagram below?
#align(center)[
// Multi-control circuit diagram would go here
#box(width: 8cm, height: 6cm)[
_[Multi-control circuit diagram would go here]_
]
]
#v(1fr)
#pagebreak()

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#import "@local/handout:0.1.0": *
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
= Superdense Coding
Consider the following entangled two-qubit states, called the _bell states_:
- $#ket([$Phi^+$]) = 1/sqrt(2) #ket("00") + 1/sqrt(2) #ket("11")$
- $#ket([$Phi^-$]) = 1/sqrt(2) #ket("00") - 1/sqrt(2) #ket("11")$
- $#ket([$Psi^+$]) = 1/sqrt(2) #ket("01") + 1/sqrt(2) #ket("10")$
- $#ket([$Psi^-$]) = 1/sqrt(2) #ket("01") - 1/sqrt(2) #ket("10")$
#problem()
The probabilistic bits we get when measuring any of the above may be called _anticorrelated bits_.
If we measure the first bit of any of these states and observe $1$, what is the resulting compound state?
What if we observe $0$ instead?
Do you see why we can call these bits anticorrelated?
#v(1fr)
#problem()
Show that the bell states are orthogonal
#hint[Dot product]
#v(1fr)
#problem()<bellmeasure>
Say we have a pair of qubits in one of the four bell states.
How can we find out which of the four states we have, with certainty?
#hint[$H #ket("+") = #ket("0")$, and $H #ket("-") = #ket("1")$]
#v(1fr)
#pagebreak()
#definition()
The $Z$ gate is defined as follows:
$ Z mat(psi_0; psi_1) = mat(psi_0; -psi_1) $
#v(1fr)
#pagebreak()

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src/Advanced/Introduction to Quantum/src/parts/08 teleport.typ Normal file
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#import "@local/handout:0.1.0": *
// Define quantum notation macros
#let ket(content) = $|#content angle.r$
#let bra(content) = $angle.l #content|$
= Quantum Teleportation
Superdense coding lets us convert quantum bandwidth into classical bandwidth. Quantum teleportation does the opposite, using two classical bits and an entangled pair to transmit a quantum state.
*Setup:* Again, suppose Alice and Bob each have half of a $#ket([$Phi^+$])$ state. We'll call the state Alice wants to teleport $#ket(math.psi) = psi_0 #ket("0") + psi_1 #ket("1")$.
#problem()
What is the three-qubit state $#ket(math.psi) #ket([$Phi^+$])$ in terms of $psi_0$ and $psi_1$?
#v(1fr)
#problem()
To teleport $#ket(math.psi)$, Alice applies the following circuit to her two qubits, where $#ket([$Phi^+_"A"$])$ is her half of $#ket([$Phi^+$])$. She then measures both qubits and sends the result to Bob.
#align(center)[
// Teleportation circuit diagram would go here
#box(width: 8cm, height: 4cm)[
_[Quantum teleportation circuit diagram would go here]_
]
]
What should Bob do so that $#ket([$Phi^+_"B"$])$ takes the state $#ket(math.psi)$ had initially?
#v(1fr)
#pagebreak()