Math ~ Quantum and Butterflies

The math is simple and speaks for itself. It's the inelegant that is vulnerable to reality. Words and explanations are a dime a dozen.

We are sitting on a number of solved problems that have yet to be tied cohesively; it's in the tying that a new dimension is unlocked. There are aspects that go beyond the particle and reach into every domain of life, and I will cover some of them below.

The rotation of nature

I would call this is an accumulation-free turnover of a system that remains integrated at every moment in time.

A fallen tree shelters insects, feeds fungi, and becomes soil for the forest that follows. A predator’s unfinished kill sustains ravens, beetles, and microbes; even fire makes way for something new. It is cyclical, and without waste.

The math shows this beautifully.

First, $e^x$ is special because its derivative is itself; it is growth through self-propagating change. But Euler’s formula adds $i$ to the mix, and if we take the derivative of this formula, we can see that change is turned, phase by phase, until it returns, and keeps cycling like that.

\begin{gathered} e^{ix} = \cos x + i \sin x \\[8pt] z = e^{ix} \longrightarrow z’ = iz \end{gathered}

Now, the time-dependent Schrödinger equation already tells us that quantum state evolves according to its Hamiltonian. We can use this to arrive at something that transforms an entire closed quantum system through time.

\begin{gathered} |\Psi(t)\rangle = U(t)|\Psi(0)\rangle \\[12pt] i\hbar\frac{d}{dt}|\Psi(t)\rangle = H|\Psi(t)\rangle \quad \longrightarrow \quad U(t) = e^{-\frac{i}{\hbar}Ht} \\[12pt] |\Psi(t)\rangle = e^{-\frac{i}{\hbar}Ht}|\Psi(0)\rangle \\[12pt] \end{gathered}

This something takes on the same fundamental form of the exponential to the imaginary, $e^{i\theta}$ , and this phase turning is in part dictated by its time-independent Hamiltonian.

We can then see that this, when applied to a closed quantum system, preserves the total probability while continuously evolving the relative phases of the components in accordance with time.

\begin{gathered} H|E_n\rangle = E_n|E_n\rangle \\[8pt] U(t)|E_n\rangle = e^{-\frac{i}{\hbar}Ht}|E_n\rangle = e^{-\frac{i}{\hbar}E_nt}|E_n\rangle \\[12pt] |\Psi(t)\rangle = U(t)|\Psi(0)\rangle = e^{-\frac{i}{\hbar}Ht} \sum_n c_n|E_n\rangle = \sum_n c_n e^{-\frac{i}{\hbar}E_nt}|E_n\rangle \\[8pt] \frac{ c_m e^{-\frac{i}{\hbar}E_mt} }{ c_n e^{-\frac{i}{\hbar}E_nt} } = \frac{c_m}{c_n} e^{-\frac{i}{\hbar}(E_m-E_n)t} \end{gathered}

\begin{gathered} H = \frac{\hbar\omega}{2}\sigma_z = \frac{\hbar\omega}{2} \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \\[8pt] U(t) = e^{-\frac{i}{\hbar}Ht} = \begin{pmatrix} e^{-i\omega t/2} & 0\\ 0 & e^{+i\omega t/2} \end{pmatrix} \\[8pt] |\Psi(0)\rangle = \alpha|0\rangle+\beta|1\rangle = \begin{pmatrix} \alpha\\ \beta \end{pmatrix} \\[8pt] |\Psi(t)\rangle = U(t)|\Psi(0)\rangle \quad\longrightarrow\quad \alpha e^{-i\omega t/2}|0\rangle + \beta e^{+i\omega t/2}|1\rangle \\[8pt] \frac{\beta(t)}{\alpha(t)} = \frac{\beta}{\alpha}e^{i\omega t} \end{gathered}

The whole is encoded in the coherent relations among its parts, and the next move is incorporated into the present.

Tropical Drip — interactive panoramic tour — Enter

In the world above, it is integration $\mathcal{I}$ , not time, that turns the loop.

\begin{gathered}   U=\min(S,C),   \qquad   L=|S-C|,   \qquad   \mathcal{I}=\frac{U}{U+L} \\[12pt] \end{gathered}

$S$ is what is available to be received; C is the capacity to receive and harness it.

$U$ is what turns through, and $L$ is what remains unmatched.

$S$ and $C$ are measured through the qualities of warmth, transformation, fluidity, continuity, movement, exchange, substance, and support. We can group these into Fire, Water, Air, and Earth. These are qualities are fundamental, and in flux.

What is received in one form can return in another: Water receives the Sun’s warmth, Fire, and rises as vapour, carrying that warmth into the Air. Condensation releases the warmth; rain returns Water to Earth. What one receives becomes what another can receive, and the loop turns.

If:

\begin{gathered}   S=C>0 \quad\Longrightarrow\quad \mathcal{I}=1 \end{gathered}

... then what is received can be transformed and offered again in the next cycle. Complete, present-tense integration.

For simplicity, we can maintain the same relationship regardless of scale:

\mathcal{I} = \frac{\sum_j U_j} {\sum_j U_j + \sum_j L_j}

Pure and relative, inside and out

There is something only present in the Whole and cannot be grasped merely by part, and I see parallels of this across life. Below is yet another parallel, but perhaps the most fundamental of them all.

Relational structure among alternatives within the system (coherence) can, depending on the basis, be harnessed to produce outcomes of varying certainty (interference). And there is a geometric reason that certainty too is basis dependent. More on this later.

Zoom out out of the single particle and you'll find relational structure among subsystems with a larger whole ~ entanglement.

Consider the pure entangled state with environmental overlap $\gamma$ :

\begin{gathered} |\Psi\rangle=\alpha |A\rangle |E_A\rangle+\beta |B\rangle |E_B\rangle,\qquad|\alpha|^2+|\beta|^2=1 \\[8pt] \gamma=\langle E_B|E_A\rangle \end{gathered}

Define the path predictability (certainty) $\mathcal P$ , the local coherence $\mathcal C$ in the path basis, and the system-environment entanglement $\mathcal E$ :

\begin{gathered} \mathcal P = \left| |\alpha|^2-|\beta|^2 \right| \\[8pt] \mathcal C = 2|\alpha\beta|\,|\gamma| \\[8pt] E = 2|\alpha\beta| \sqrt{1-|\gamma|^2} \end{gathered}

This satisfies:

\begin{gathered} \mathcal P^2+\mathcal C^2+\mathcal E^2=1 \end{gathered}

\begin{gathered} \mathcal P^2 = \left(|\alpha|^2-|\beta|^2\right)^2 \quad\longrightarrow\quad 4|\alpha\beta|^2 = 1-\mathcal P^2 \\[8pt] \mathcal C^2+\mathcal E^2 = 4|\alpha\beta|^2 \left( |\gamma|^2+1-|\gamma|^2 \right) = 4|\alpha\beta|^2 = 1-\mathcal P^2 \end{gathered}

For an equal superposition:

\begin{gathered} |\alpha|=|\beta|=\frac{1}{\sqrt 2}, \quad \mathcal P=0 \\[16pt] \therefore\quad\mathcal C=|\gamma|, \qquad \mathcal E=\sqrt{1-|\gamma|^2} \\[12pt] \therefore\quad{ \mathcal C^2+\mathcal E^2=1 } \end{gathered}

If the environment cannot distinguish the paths, the particle retains its local coherence:

\begin{gathered} |\gamma|=1 \\[8pt] \therefore\quad\mathcal C=1, \qquad \mathcal E=0 \end{gathered}

If the environment perfectly distinguishes the paths, coherence is delocalized:

\begin{gathered} |\gamma|=0 \\[8pt] \therefore\quad\mathcal C=0, \qquad \mathcal E=1 \end{gathered}

This is an entangled state. Each particle alone contains less information than the pair contains together. Specifically, no particle holds the coherence terms that contain information about the phase that belongs to and compares two branches of the entangled state.

\begin{gathered} |\Phi^{\pm}\rangle = \frac{|00\rangle \pm |11\rangle}{\sqrt{2}} \\[12pt] |\Phi^{+}\rangle \neq |\Phi^{-}\rangle \\[12pt] \rho_A^{\pm} = \rho_B^{\pm} = \frac{I}{2} = \frac12 \left( |0\rangle\langle0| + |1\rangle\langle1| \right) \\[16pt] \left\{ |00\rangle\langle11|, \; |11\rangle\langle00| \right\} \subset \operatorname{terms}\!\left(\rho_{AB}^{\pm}\right) \\[12pt] \langle00|\rho_{AB}^{\pm}|11\rangle = \pm\frac12 \neq 0 ,\quad \langle11|\rho_{AB}^{\pm}|00\rangle = \pm\frac12 \neq 0 \end{gathered}

The entangled state is pure while each part is mixed, and it also cannot be factorized.

\begin{gathered} S(\rho_{AB})=0 \\[8pt] S(\rho_A)>0, \qquad S(\rho_B)>0 \\[8pt] |\Psi\rangle_{AB} \neq |\psi\rangle_A \otimes |\epsilon\rangle_B \end{gathered}

This is how we connect purity, coherence, certainty, and entanglement.

And I mentioned earlier that there is a geometric reason that certainty changes with the basis. A measurement basis is a particular physical question being asked of the state. Let's close the loop on this.

For a pure state represented by Bloch vector $\mathbf r$ , measured along axis $\mathbf n$ :

\begin{gathered} \rho = \frac{1}{2} \left( I+\mathbf r\cdot\boldsymbol{\sigma} \right) \\[8pt] \Pi_{+\mathbf n} = \frac{1}{2} \left( I+\mathbf n\cdot\boldsymbol{\sigma} \right) , \quad \Pi_{-\mathbf n} = \frac{1}{2} \left( I-\mathbf n\cdot\boldsymbol{\sigma} \right) \\[8pt] P(\pm_{\mathbf n}) = \operatorname{Tr} \left( \rho\Pi_{\pm\mathbf n} \right) = \frac{1}{4} \operatorname{Tr} \left[ \left( I+\mathbf r\cdot\boldsymbol{\sigma} \right) \left( I\pm\mathbf n\cdot\boldsymbol{\sigma} \right) \right] \\[8pt] \operatorname{Tr}(I)=2, \qquad \operatorname{Tr}(\sigma_i)=0, \qquad \operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij} \\[12pt] P(+_{\mathbf n}) = \frac{1+\mathbf r\cdot\mathbf n}{2}, \qquad P(-_{\mathbf n}) = \frac{1-\mathbf r\cdot\mathbf n}{2} \end{gathered}

\begin{gathered} \text{Three perpendicular axes of the Bloch sphere} \\[8pt] Z\text{-axis} \longleftrightarrow |0\rangle,\ |1\rangle \\[8pt] X\text{-axis} \longleftrightarrow |+_x\rangle,\ |-_x\rangle \\[8pt] Y\text{-axis} \longleftrightarrow |+_y\rangle,\ |-_y\rangle \end{gathered}

Therefore, if the state is pure, then we can know ... that the degree of alignment between $\mathbf r$ and $\mathbf n$ directly informs the certainty of measurement outcomes.

\begin{gathered} \mathbf r=\mathbf n \quad\Longrightarrow\quad P(+_{\mathbf n})=1 \\[8pt] \mathbf r=-\mathbf n \quad\Longrightarrow\quad P(-_{\mathbf n})=1 \\[8pt] \mathbf r\cdot\mathbf n=0 \quad\Longrightarrow\quad P(+_{\mathbf n}) = P(-_{\mathbf n}) = \frac12 \end{gathered}

Purity belongs to the state, while certainty belongs to the relation between the state and the way it is met.