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--- cp-1 [May 23, 2026 at 19:33] – Ivan Janevski
+++ cp-1 [June 13, 2026 at 03:13] (current) – external edit 127.0.0.1
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 # $\mathbb{CP}^1$
-**Complex projective line** (or $\mathbb{CP}^1$) is the Hilbert space where pure qubits live in. It is defined as the quotient space of $\mathbb{C}^2$ and $\mathbb{C}$, both of which are punctured in their respective zero elements.
+**Complex projective line** (or $\mathbb{CP}^1$) is the space in which pure qubit states live. It is defined as the quotient of $\mathbb{C}^2 \setminus \{0\}$ by $\mathbb{C}^\times$.
 $$\mathbb{CP}^1 \stackrel{\text{def}}{=} (\mathbb{C}^2 \setminus\{0\})/\mathbb{C}^\times\qquad \mathbb{C}^\times\stackrel{\text{def}}{=}\mathbb{C} \setminus \{0\}$$
 ## Motivation
-To do quantum computing, we usually use the Hilbert space $\mathbb{C}^2$. That's the Hilbert space in which all qubit state vectors (or qubit "kets") $\lvert\psi\rangle$ live in. However, many elements  in this space correspond to something unphysical. For example, it contains state vectors whose probabilities that don't add up to 100%. There are also elements which we would consider to be duplicates of each other -- up to a global phase.
+To do quantum computing, we usually use the Hilbert space $\mathbb{C}^2$. That's the Hilbert space in which all qubit state vectors (or qubit "kets") $\lvert\psi\rangle$ live in. However, many elements in this space correspond to something unphysical. For example, it contains state vectors whose probabilities don't add up to 100%. There are also elements which we would consider to be duplicates of each other -- up to a global phase.
 Here are the 3 major issues with the space $\mathbb{C}^2$
-) The zero vector is in $\mathbb{C}^2$ but it's unphysical. Applying the "Born rule" yields 0% + 0% = 0%. Meaning, the qubit doesn't even exist. This violates the second axiom of probability which states that probabilities must add up to 100%.
+) The zero vector is in $\mathbb{C}^2$ but it's unphysical. Applying the [[born-rule|Born rule]] yields 0% + 0% = 0%. Meaning, the qubit doesn't even exist. This violates the second axiom of probability which states that probabilities must add up to 100%.
 $$\lvert\psi\rangle = \begin{pmatrix}0\\ 0 \end{pmatrix}\qquad \lvert\psi\rangle\in\mathbb{C}^2 \qquad |0|^2 + |0|^2 = 0$$
-) There are many other states which violate the "Born rule" and are thus considered unphysical. For example, the following is a state which is in state $\lvert 0\rangle$ with 300% probability and in state $\lvert 1\rangle$ with 200% yielding the total probability of 500%. But according to the second axiom of probabilitiy, the probabilities must always add up to exactly 100%.
+) There are many other states which violate the [[born-rule|Born rule]] and are thus considered unphysical. For example, the following is a state which is in state $\lvert 0\rangle$ with 300% probability and in state $\lvert 1\rangle$ with 200% yielding the total probability of 500%. But according to the second axiom of probability, the probabilities must always add up to exactly 100%.
 $$\lvert\psi\rangle = \begin{pmatrix}\sqrt{3} \\ \sqrt{2}\end{pmatrix}\qquad \lvert\psi\rangle\in\mathbb{C}^2\qquad |\sqrt{3}|^2 + |\sqrt{2}|^2 = 3 + 2 = 5$$
-) Two states can differ by some phase $e^{i\phi}$ called the global phase. This phase does not have any effect on  -- in fact it is completely unobservable. Therefore, we
+) Two states can differ by some phase $e^{i\phi}$ called the [[global-phase|global phase]]. This phase has no effect on measurement outcomes — it is completely unobservable. Therefore, we identify all vectors that differ by a global phase as representing the same physical state.
-$$\lvert\psi_1\rangle = \begin{pmatrix}1\\ 0\end{pmatrix}\qquad \lvert\psi_2\rangle = \begin{pmatrix}i \\ 0\end{pmatrix}\qquad \lvert\psi_1\rangle = e^{i\pi/2}\lvert\psi_2\rangle\qquad \lvert\psi\rangle \sim e^{i\phi}\lvert\psi\rangle$$.
+$$\lvert\psi_1\rangle = \begin{pmatrix}1\\ 0\end{pmatrix}\qquad \lvert\psi_2\rangle = \begin{pmatrix}i \\ 0\end{pmatrix}\qquad \lvert\psi_2\rangle = e^{i\pi/2}\lvert\psi_1\rangle\qquad \lvert\psi\rangle \sim e^{i\phi}\lvert\psi\rangle$$
@@ Line 22: / Line 22: @@
 We construct $\mathbb{CP}^1$ by resolving each of these issues individually.
-) We request that the zero state vector is not in $\mathbb{CP}^1$
+) We require that the zero vector is not in $\mathbb{CP}^1$
 $$\lvert\psi\rangle = \begin{pmatrix}0\\0\end{pmatrix}\qquad \lvert\psi\rangle\notin\mathbb{CP}^1$$
-) We can always normalize
+) If probabilities don't add up to 100% we can always normalize the state vector by scaling it with some real number $A$. We just require $A\neq 0$ since that would turn the state vector into a zero vector which we just removed in 1). Therefore, we identify all vectors that differ by some constant $A\in\mathbb{R}\setminus\{0\}$ i.e. we say they represent the same state. The following is an example for state $\lvert0\rangle$.
-$$\lvert 0\rangle = \left\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}2\\0\end{pmatrix}, \begin{pmatrix}-1\\ 0\end{pmatrix}\right\}\qquad \lvert \psi\rangle\sim A\lvert\psi\rangle, A\in\mathbb{R},A\neq 0$$
+$$\lvert 0\rangle \sim \left\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}2\\0\end{pmatrix}, \begin{pmatrix}-1\\ 0\end{pmatrix}, \cdots\right\}\qquad \lvert \psi\rangle\sim A\lvert\psi\rangle\qquad  A\in\mathbb{R}\setminus\{0\}$$
-)
+) We can always factor the global phase $e^{i\phi}$ out of any state vector. Therefore we identify all vectors that differ by some arbitrary global phase $e^{i\phi}$. The following is an example for state $\lvert 1\rangle$.
+$$\lvert 1\rangle \sim \left\{\begin{pmatrix}0\\1\end{pmatrix},\begin{pmatrix}0\\i\end{pmatrix}, \begin{pmatrix}0\\e^{i\pi/4}\end{pmatrix}\cdots\right\}\qquad \lvert\psi\rangle\sim e^{i\phi}\lvert\psi\rangle \qquad \phi\in(-\pi,\pi]$$
+From 1), we see the overarching space we're looking for is the following. Here, 0 refers to the zero vector in $\mathbb{C}^2$.
+$$\mathbb{C}^2 \setminus \{0\}$$
+From 2) and 3), we can multiply $A$ and $e^{\phi}$ and get a new complex number $Ae^{i\phi}$ which is an element of complex plane $\mathbb{C}$. However we required $A\neq 0$ in order to avoid the collapsing vectors into a zero vector. Therefore the number $Ae^{i\phi}$ is actually a member of $\mathbb{C}\setminus\{0\}$ otherwise written as $\mathbb{C}^\times$. Here, 0 refers to the zero complex number $0 + i0$. This is sometimes called a "complex plane with a puncture" and it's the space we want to quotient out from 1).
+$$\mathbb{C}^\times = \mathbb{C}\setminus\{0\}$$
+Putting it all together we constructed the following:
+$$\mathbb{CP}^1 = (\mathbb{C^2}\setminus\{0\})/\mathbb{C}^\times\qquad \mathbb{C}^\times = \mathbb{C}\setminus\{0\}$$