Wavefunction Interpretation: What Does It Mean for the Reinforcement Field?¶

Motivation¶

"The reinforcement field is a wavefunction over augmented state-action space."

This raises important questions:

What exactly is a wavefunction in quantum mechanics?
What does it represent and predict?
How should we interpret this claim for the reinforcement field?

This document provides the precise conceptual grounding.

1. What Is the Wavefunction in Quantum Mechanics?¶

In standard (non-relativistic) quantum mechanics, the wavefunction $\psi(x,t)$ is:

A complete mathematical representation of the physical state of a system, expressed in a particular basis (the position basis).

Simple Analogy First: 3D Vectors and Coordinates¶

Before the formal definition, let's build intuition with a familiar example.

Consider a vector in 3D space, like a velocity: $\mathbf{v}$.

The vector itself is a geometric object—an arrow with direction and magnitude. This exists independent of any coordinate system.

But to work with it numerically, we express it in coordinates:

In Cartesian coordinates $(x, y, z)$:

\[\mathbf{v} = \begin{bmatrix} 3 \\ 4 \\ 0 \end{bmatrix}\]

This means: "3 units in the $x$ direction, 4 in $y$, 0 in $z$."

In polar coordinates $(r, \theta, z)$:

\[\mathbf{v} = \begin{bmatrix} 5 \\ 53.1° \\ 0 \end{bmatrix}\]

Key insight: The vector $\mathbf{v}$ is the same geometric object in both cases. Only its coordinate representation changed.

The Quantum Version: State Vector vs. Wavefunction¶

The same idea applies in quantum mechanics:

Formal definition:

The state of a quantum system is a vector $|\psi\rangle$ in a complex Hilbert space $\mathcal{H}$ (like the geometric vector $\mathbf{v}$)
The wavefunction $\psi(x)$ is the coordinate representation of that vector in the position basis $\{|x\rangle\}$

The relationship is given by an inner product (projection):

\[\psi(x) = \langle x | \psi \rangle\]

What this means in plain English:

The wavefunction $\psi(x)$ tells you "how much" of the state $|\psi\rangle$ "points in the direction" of position $x$.

It's exactly like asking: "What is the $x$-component of velocity $\mathbf{v}$?" Answer: 3.

Concrete Example: Two-Level System (Qubit)¶

Let's work through this with actual numbers.

Setup: A qubit has a 2-dimensional Hilbert space with computational basis:

\[|0\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad |1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\]

These are basis vectors, analogous to $\mathbf{e}_x = [1, 0]$ and $\mathbf{e}_y = [0, 1]$ in 2D Cartesian coordinates.

State vector (the quantum system itself):

\[|\psi\rangle = \frac{1}{\sqrt{2}} |0\rangle + \frac{1}{\sqrt{2}} |1\rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Analogy: Just like $\mathbf{v} = 3\mathbf{e}_x + 4\mathbf{e}_y = [3, 4]$, here $|\psi\rangle$ is a linear combination of basis vectors $|0\rangle$ and $|1\rangle$.

Question: What are the "wavefunction values" (coordinates) in the $\{|0\rangle, |1\rangle\}$ basis?

Answer: Compute the inner products (projections)!

\[\psi_0 = \langle 0 | \psi \rangle = \begin{bmatrix} 1 & 0 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}}\]

\[\psi_1 = \langle 1 | \psi \rangle = \begin{bmatrix} 0 & 1 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}}\]

What we computed: Project the state $|\psi\rangle$ onto each basis vector.

Result: The wavefunction in this basis is $\left[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$.

Interpretation:

"How much of $|\psi\rangle$ points in the $|0\rangle$ direction?" → $\frac{1}{\sqrt{2}}$
"How much of $|\psi\rangle$ points in the $|1\rangle$ direction?" → $\frac{1}{\sqrt{2}}$

Analogy: If $\mathbf{v} = [3, 4]$, the "$x$-coordinate" is 3 (how much of $\mathbf{v}$ is in the $\mathbf{e}_x$ direction).

Now let's use a DIFFERENT coordinate system:

So far, we've been working in the computational basis $\{|0\rangle, |1\rangle\}$—think of this as our "Cartesian coordinates."

Now let's define a second coordinate system, the Hadamard basis:

\[|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

\[|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}\]

Important: These coordinates $[1/\sqrt{2}, 1/\sqrt{2}]$ are expressed in the computational basis. We're defining new basis vectors by specifying their coordinates in the old basis.

Analogy: Like defining polar coordinates by saying: $\hat{r} = \cos\theta \, \mathbf{e}_x + \sin\theta \, \mathbf{e}_y$ (new basis vectors in terms of old).

Key observation: Our state $|\psi\rangle$ and the basis vector $|+\rangle$ have the same coordinates in the computational basis:

\[|\psi\rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = |+\rangle\]

Wait, so they're equal? Yes! As vectors, $|\psi\rangle = |+\rangle$. But they play different roles in our current discussion:

Symbol	Role in This Example
$\\|\psi\rangle$	The state we're analyzing (the "subject")
$\\|+\rangle$	A basis vector in the Hadamard coordinate system (the "axis")

Analogy: The vector $[1, 0]$ could be:

"The velocity we're analyzing" ← role: subject
"The x-axis of our coordinate system" ← role: reference axis

Same vector, different roles!

Now the question: What is the wavefunction of $|\psi\rangle$ in the Hadamard basis $\{|+\rangle, |-\rangle\}$?

What we're computing: Express our state $|\psi\rangle$ using the Hadamard basis vectors $|+\rangle$ and $|-\rangle$ as coordinates.

Component in the $|+\rangle$ direction:

\[\psi_+ = \langle + | \psi \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{2}(1 + 1) = 1\]

Why is this 1? Because $|\psi\rangle = |+\rangle$! The state is perfectly aligned with the $|+\rangle$ basis vector. It's like asking: "How much of $\mathbf{e}_x$ is in the $\mathbf{e}_x$ direction?" Answer: 1 (all of it).

Component in the $|-\rangle$ direction:

\[\psi_- = \langle - | \psi \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & -1 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{2}(1 - 1) = 0\]

Why is this 0? Because $|\psi\rangle$ is orthogonal to $|-\rangle$. It has zero component in that direction.

Result: In the Hadamard basis, the wavefunction is $[1, 0]$.

What this means:

Basis Used	Coordinates of $\\|\psi\rangle$	Interpretation
Computational $\{\\|0\rangle, \\|1\rangle\}$	$[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$	Equal mix of both directions
Hadamard $\{\\|+\rangle, \\|-\rangle\}$	$[1, 0]$	Fully in $\\|+\rangle$ direction, nothing in $\\|-\rangle$

Same state, different coordinates! Like how $\mathbf{v} = [3, 4]$ in Cartesian is $[5, 53.1°]$ in polar.

Common Confusion: Which Direction Is the Projection?¶

Question: "Isn't $\langle + | \psi \rangle$ just the projection of the basis vector $|+\rangle$ onto the state $|\psi\rangle$?"

Answer: No! The notation $\langle + | \psi \rangle$ means:

"Project the state $|\psi\rangle$ onto the basis vector $|+\rangle$"

Analogy: In 3D, if you have:

Vector: $\mathbf{v} = [3, 4, 0]$
Basis vector: $\mathbf{e}_x = [1, 0, 0]$

The "$x$-coordinate" is: $$x = \mathbf{e}_x \cdot \mathbf{v} = [1, 0, 0] \cdot [3, 4, 0] = 3$$

You're asking: "How much of $\mathbf{v}$ is in the $\mathbf{e}_x$ direction?"

Same in QM: $\langle + | \psi \rangle$ asks: "How much of $|\psi\rangle$ is in the $|+\rangle$ direction?"

Why the bra-ket notation?

The notation $\langle + |$ is the "bra" (row vector), $|\psi\rangle$ is the "ket" (column vector):

\[\langle + | \psi \rangle = \text{row vector} \times \text{column vector} = \text{scalar}\]

This is the inner product that gives you the component!

Summary: Two Bases, One State¶

Let's be crystal clear about what just happened:

1. We have ONE state (the thing we're analyzing):

\[|\psi\rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}_{\text{computational}}\]

2. We expressed it in TWO different coordinate systems:

Coordinate System 1: Computational Basis $\{|0\rangle, |1\rangle\}$ - Basis vectors (the "axes"): $|0\rangle = [1, 0]$, $|1\rangle = [0, 1]$ - State coordinates: $|\psi\rangle$ has wavefunction $[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$

Coordinate System 2: Hadamard Basis $\{|+\rangle, |-\rangle\}$ - Basis vectors (the "axes"): $|+\rangle = [\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]_{\text{computational}}$, $|-\rangle = [\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}]_{\text{computational}}$ - State coordinates: $|\psi\rangle$ has wavefunction $[1, 0]$

3. Key observation:

In computational coordinates, $|\psi\rangle$ and $|+\rangle$ are the same vector: $[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$.

But they play different roles:

$|\psi\rangle$ = the state (subject of analysis)
$|+\rangle$ = one of the axes in the Hadamard coordinate system

Analogy: If your velocity is $\mathbf{v} = [1, 0]$ m/s (heading east), you could also use that same vector $[1, 0]$ as the x-axis of a new coordinate system. Same vector, different roles!

The key insight:

The state $|\psi\rangle$ is the same geometric object in both cases—it's the same quantum system!

But its wavefunction (coordinate representation) is different:

Coordinate System (Basis)	State $\\|\psi\rangle$ Coordinates
Computational $\{\lvert 0\rangle, \lvert 1\rangle\}$	$\left[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$
Hadamard $\{\lvert +\rangle, \lvert -\rangle\}$	$[1, 0]$

Analogy: Same vector $\mathbf{v}$, different coordinate systems:

Coordinate System	Coordinates
Cartesian $(x, y)$	$[3, 4]$
Polar $(r, \theta)$	$[5, 53.1°]$

Same object, different numbers!

Infinite-Dimensional Case: Position Basis¶

In standard quantum mechanics, position can be any real number, so the Hilbert space is infinite-dimensional.

State: $|\psi\rangle$ (abstract vector in infinite-dimensional Hilbert space)

Position basis: $\{|x\rangle : x \in \mathbb{R}\}$

What is $|x\rangle$?

The symbol $|x\rangle$ is a basis vector representing "the state where the particle is definitely at position $x$."

Analogy to finite case:

Finite (Qubit)	Infinite (Position)
2 basis vectors: $\\|0\rangle, \\|1\rangle$	Infinite basis vectors: $\\|x\rangle$ for every $x \in \mathbb{R}$
$\\|0\rangle$ = "definitely in state 0"	$\\|x\rangle$ = "definitely at position $x$"
Discrete index: 0, 1	Continuous index: $x \in \mathbb{R}$

Critical distinction: $|x\rangle$ is NOT a position vector like $\mathbf{r} = [x, y, z]$ in classical mechanics. It's a basis vector in Hilbert space that represents a particular position eigenstate.

Important: $|\psi\rangle$ is NOT a basis vector—it's the state (the thing we're analyzing), expressed as a combination of basis vectors!

Wavefunction: For each position $x$, compute the projection:

\[\psi(x) = \langle x | \psi \rangle\]

What this means:

"Project the state $|\psi\rangle$ onto the basis vector $|x\rangle$ (the state 'definitely at position $x$')"

Result: This gives a function $\psi: \mathbb{R} \to \mathbb{C}$ that tells you the "component" of $|\psi\rangle$ in each position direction.

Analogy: Just like we computed:

$\psi_0 = \langle 0 | \psi \rangle$ for the $|0\rangle$ direction
$\psi_1 = \langle 1 | \psi \rangle$ for the $|1\rangle$ direction

Now we compute:

$\psi(x=0) = \langle 0 | \psi \rangle$ for position $x=0$
$\psi(x=1) = \langle 1 | \psi \rangle$ for position $x=1$
$\psi(x=2.5) = \langle 2.5 | \psi \rangle$ for position $x=2.5$
... and so on for every real number $x$

Since $x$ is continuous, we get a continuous function $\psi(x)$!

Concrete Example: Gaussian Wavepacket

Suppose our state $|\psi\rangle$ has the following wavefunction in the position basis:

\[\psi(x) = \langle x | \psi \rangle = \frac{1}{(\pi \sigma^2)^{1/4}} \exp\left(-\frac{x^2}{2\sigma^2}\right)\]

This is a bell curve! Let's evaluate it at specific positions (say $\sigma = 1$):

Position $x$	Basis Vector	Component $\psi(x) = \langle x \\| \psi \rangle$
$x = 0$	$\\|0\rangle$	$\psi(0) = \frac{1}{(\pi)^{1/4}} \approx 0.75$ (maximum)
$x = 1$	$\\|1\rangle$	$\psi(1) = \frac{1}{(\pi)^{1/4}} e^{-1/2} \approx 0.46$
$x = 3$	$\\|3\rangle$	$\psi(3) = \frac{1}{(\pi)^{1/4}} e^{-9/2} \approx 0.008$ (nearly zero)

Interpretation:

The state $|\psi\rangle$ has large component in the $|x=0\rangle$ direction (particle likely near origin)
Medium component in the $|x=1\rangle$ direction
Tiny component in the $|x=3\rangle$ direction (particle unlikely far from origin)

Visual:

ψ(x)
  ^
  |     *
  |    ***
  |   *****
  |  *******
  | *********
  |***********
  +---------------> x
 -3  -1  0  1  3

The wavefunction $\psi(x)$ tells you how much of $|\psi\rangle$ "points in the direction" of each position basis vector $|x\rangle$.

Clarification: $|\psi\rangle$ vs. $|x\rangle$ - Which Is the Basis?¶

Common confusion: "Are both $|\psi\rangle$ and $|x\rangle$ basis vectors?"

Answer: NO! This is a critical distinction:

Symbol	What It Is	Role
$\\|\psi\rangle$	The state (thing being analyzed)	Like the vector $\mathbf{v} = [3, 4]$ you're analyzing
$\\|x\rangle$	A basis vector (coordinate axis)	Like $\mathbf{e}_x = [1, 0]$ or $\mathbf{e}_y = [0, 1]$

$|\psi\rangle$ is expressed AS A COMBINATION of basis vectors $|x\rangle$:

\[|\psi\rangle = \int_{-\infty}^{\infty} \psi(x) \, |x\rangle \, dx\]

This is like writing: $\mathbf{v} = 3\mathbf{e}_x + 4\mathbf{e}_y$

Your Classical Intuition Is Right—But Quantum Is Different!¶

You said: "When I hear 'basis vector', it gives me the impression of [0, 0, 1], [0, 1, 0], [1, 0, 0] - linearly independent vectors."

You're absolutely right! That's exactly what basis vectors are in classical mechanics.

The quantum twist:

In quantum mechanics, $|x\rangle$ plays a role analogous to $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$, but there's a crucial difference:

Classical Position	Quantum State
Position vector: $\mathbf{r} = [x, y, z]$	State vector: $\\|\psi\rangle$
Expressed using basis: $\mathbf{r} = x\mathbf{e}_x + y\mathbf{e}_y + z\mathbf{e}_z$	Expressed using basis: $\\|\psi\rangle = \int \psi(x) \\|x\rangle dx$
Basis vectors: $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$ (3 of them)	Basis vectors: $\\|x\rangle$ (one for each $x \in \mathbb{R}$, infinitely many)
Coordinates: $x, y, z$ (numbers)	Coordinates: $\psi(x)$ (the wavefunction!)

The Big Difference: What We're Representing¶

Classical mechanics:

Thing: Position of a particle
Representation: $\mathbf{r} = x\mathbf{e}_x + y\mathbf{e}_y + z\mathbf{e}_z$
Basis vectors: $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$ (spatial directions)

Quantum mechanics:

Thing: State of a particle
Representation: $|\psi\rangle = \int \psi(x) |x\rangle dx$
Basis vectors: $|x\rangle$ (not spatial directions, but quantum states!)

Key insight: In QM, $|x\rangle$ doesn't represent "the direction x" in space. It represents the quantum state "particle is at position x".

Concrete Example: Finite Case First¶

Let's make this super concrete with a qubit:

Basis vectors (the "axes"):

$|0\rangle = [1, 0]$
$|1\rangle = [0, 1]$

These are linearly independent, just like $\mathbf{e}_x$ and $\mathbf{e}_y$!

State (the thing we're analyzing): $$|\psi\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

Is $|\psi\rangle$ a basis vector? NO! It's a combination of basis vectors, like $\mathbf{v} = 3\mathbf{e}_x + 4\mathbf{e}_y$.

Now the Infinite Case¶

Basis vectors (the "axes"):

$|x\rangle$ for every $x \in \mathbb{R}$

State (the thing we're analyzing): $$|\psi\rangle = \int_{-\infty}^{\infty} \psi(x) \, |x\rangle \, dx$$

where $\psi(x) = \langle x | \psi \rangle$ are the coordinates (wavefunction).

Is $|\psi\rangle$ a basis vector? NO! It's a continuous combination of basis vectors $|x\rangle$.

Classical vs. Quantum Interpretation¶

Common expectation: "An arbitrary position x should be expressible in terms of chosen basis vectors."

Classical mechanics (YES): Position $\mathbf{r}$ is expressed as $\mathbf{r} = x\mathbf{e}_x + y\mathbf{e}_y + z\mathbf{e}_z$.

Quantum mechanics (DIFFERENT!): We're not expressing positions—we're expressing states!

The state $|\psi\rangle$ is expressed in the position basis as: $$|\psi\rangle = \int \psi(x) |x\rangle dx$$

The basis vectors $|x\rangle$ are NOT expressing positions—they ARE quantum states (states of "definitely at position x").

Why This Is Confusing¶

The notation $|x\rangle$ looks like it should mean "position x", but it actually means:

"The quantum state where the particle is definitely located at position x"

So $|x\rangle$ is not a point in space—it's a vector in Hilbert space representing a specific quantum state.

Better notation (if we could redesign QM):

$|x\rangle$ → $|\text{definitely at } x\rangle$ (clearer!)

But physicists use $|x\rangle$ as shorthand.

Summary: Classical vs. Quantum Representation¶

Aspect	Classical Mechanics	Quantum Mechanics
What we represent	Position of particle	State of particle
The thing	$\mathbf{r}$ (position vector)	$\\|\psi\rangle$ (state vector)
Basis vectors	$\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$	$\\|x\rangle$ for each $x \in \mathbb{R}$
What basis vectors mean	Spatial directions	Quantum states ("at position x")
Number of basis vectors	3 (in 3D space)	Infinite (one per position)
Representation	$\mathbf{r} = x\mathbf{e}_x + y\mathbf{e}_y + z\mathbf{e}_z$	$\\|\psi\rangle = \int \psi(x) \\|x\rangle dx$
Coordinates	$x, y, z$ (position components)	$\psi(x)$ (wavefunction)
"Is position a basis vector?"	No, position uses basis vectors	No, $\\|x\rangle$ IS a basis vector

Key takeaway:

Classical: Position $\mathbf{r}$ is expressed using basis vectors $\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$
Quantum: Basis vector $|x\rangle$ represents the state "particle at position x"

This is why quantum mechanics feels weird! We're not representing positions anymore—we're representing states, and the basis vectors themselves encode "where the particle is."

Key Distinction (Now Clear!)¶

State $|\psi\rangle$ = The quantum system itself (basis-independent)
Wavefunction $\psi(x)$ = Coordinate representation in position basis

Summary Table¶

Concept	What It Is	Analogy
State vector $\lvert\psi\rangle$	The quantum system (abstract)	The geometric vector $\mathbf{v}$
Wavefunction $\psi(x)$	Coordinate representation	Cartesian coordinates $[3, 4, 0]$
Inner product $\langle x \lvert \psi \rangle$	Component in direction $\lvert x\rangle$	How much of $\mathbf{v}$ is in $x$-direction?
Different basis	Different coordinate system	Cartesian vs. polar
Same state, different wavefunction	Same $\lvert\psi\rangle$, different basis	Same $\mathbf{v}$, different coordinates

Why this matters for GRL:

In GRL, the reinforcement field $Q^+(z)$ is like a wavefunction—it's the coordinate representation of a state vector in RKHS, expressed in the kernel-induced basis $\{k(z_i, \cdot)\}$.

This maps cleanly onto RKHS language, as we'll see in Section 5.

2. What Does the Wavefunction Represent?¶

The wavefunction does not represent:

❌ A probability
❌ A physical wave in space
❌ Ignorance in the Bayesian sense

Instead, it represents probability amplitudes.

The Born Rule¶

The Born rule tells us how to extract observable predictions:

\[p(x) = |\psi(x)|^2\]

Key properties:

$\psi(x)$ can be positive, negative, or complex
Interference arises because amplitudes add before squaring
Probabilities are derived, not primitive

The Fundamental Move¶

This is the single most important structural move quantum mechanics makes:

QM is not a probabilistic theory.
It is an amplitude theory from which probabilities are derived.

This alone justifies:

Spectral methods
Interference-like effects
Superposition-based reasoning

—without invoking physics mysticism.

3. Why "One State, Many Wavefunctions"?¶

From the examples above, you've seen that one state $|\psi\rangle$ can have different coordinate representations depending on the basis.

This is why physicists sometimes say "the wavefunction" (singular) and sometimes "wavefunctions" (plural):

Singular: "The Wavefunction"¶

When we say "the wavefunction," we usually mean:

The position-basis representation $\psi(x) = \langle x | \psi \rangle$

This is the most common choice because position is directly measurable.

Plural: "Different Wavefunctions"¶

When we say "different wavefunctions," we mean different basis representations of the same state:

Position basis: $$\psi(x) = \langle x | \psi \rangle \quad \text{(position wavefunction)}$$

Momentum basis: $$\tilde{\psi}(p) = \langle p | \psi \rangle \quad \text{(momentum wavefunction)}$$

Energy basis: $$c_n = \langle E_n | \psi \rangle \quad \text{(energy amplitudes)}$$

These are not different physical states—they are different coordinate charts on the same object, like Cartesian vs. polar coordinates for the same vector.

Connection to GRL¶

In GRL, when we talk about "wavefunction-like amplitude fields," we mean:

The reinforcement field $Q^+(z)$ is one representation of the state in RKHS, specifically the representation in the kernel-induced basis $\{k(z_i, \cdot)\}$.

We could also express the same state in different bases (e.g., Fourier basis, wavelet basis), just like quantum states have position and momentum representations.

4. Operators, Observables, and Prediction¶

In quantum mechanics, nothing observable comes directly from the wavefunction.

All predictions are mediated by operators.

Observables as Hermitian Operators¶

Observables are Hermitian operators $\hat{O}$
Expected value:

\[\langle O \rangle = \langle \psi | \hat{O} | \psi \rangle\]

Measurement Probabilities¶

Measurement probabilities arise from projection operators:

\[p(o) = |\hat{P}_o |\psi\rangle|^2\]

The Workflow¶

Quantum mechanics follows this structure:

state → operator → expectation / distribution

Not:

state → probability

This is a deep conceptual alignment with GRL's formulation.

5. Mapping Back to GRL: State vs. Representation¶

Let's translate each component with discipline.

(a) What Corresponds to the Quantum State?¶

In GRL, the reinforcement field is:

\[Q^+(\cdot) = \sum_i w_i \, k(z_i, \cdot)\]

This is best interpreted as:

The GRL state is the entire reinforcement field as an element of RKHS

That is:

\[Q^+ \in \mathcal{H}_k\]

This is the analogue of $|\psi\rangle$, not of $\psi(x)$.

(b) What Corresponds to the Wavefunction?¶

The wavefunction analogue appears only after choosing a query point.

Given a "query configuration" $z = (s, \theta)$, the scalar:

\[Q^+(z) = \langle Q^+, k(z,\cdot) \rangle_{\mathcal{H}_k}\]

is exactly analogous to:

\[\psi(x) = \langle x | \psi \rangle\]

Summary of the Mapping¶

Quantum Mechanics	GRL
State vector $\\|\psi\rangle \in \mathcal{H}$	Reinforcement field $Q^+ \in \mathcal{H}_k$
Wavefunction $\psi(x) = \langle x \\| \psi \rangle$	Value at query $Q^+(z) = \langle Q^+, k(z,\cdot) \rangle$
Position basis $\\|x\rangle$	Kernel basis $k(z, \cdot)$
Probability $p(x) = \\|\psi(x)\\|^2$	Policy $\pi(a\\|s) \propto \exp(\beta Q^+(s,a))$

Key insight:

$Q^+$ is the state
$Q^+(z)$ is the coordinate representation of that state in the kernel-induced basis

6. One Reinforcement Field or Many?¶

Now we can answer this precisely.

Strict Answer¶

There is one reinforcement field state $Q^+$.

But there are many induced wavefunction-like representations, depending on:

Which subspace you project onto
Which action slice you fix
Which kernel basis you query
Which abstraction level you operate at

Examples of Different Representations¶

Fixing state $s$: $Q^+(s, \cdot)$ → action-amplitude field

Fixing action parameters $\theta$: $Q^+(\cdot, \theta)$ → state-amplitude field

Projecting onto a concept subspace: → concept-level amplitude field

Marginalizing over actions: $V(s) = \mathbb{E}_\theta[Q^+(s, \theta)]$ → state value function

All of these are representations, not distinct states.

This mirrors quantum mechanics exactly.

7. Implications for Concept Discovery (Section V)¶

This interpretation does important conceptual work:

What Concepts Are¶

Functional clusters are not mixtures of policies
They are coherent subspaces of a single state
Spectral clustering identifies approximate eigenstates
Hierarchies correspond to coarse-graining of observables

Concept Formation as Spectral Decomposition¶

Concept formation becomes:

Identifying stable subspaces under the action of GRL's implicit operators

This is far stronger than "kernel clustering" in the usual ML sense.

Connection to Part II¶

Part II (Emergent Structure & Spectral Abstraction) leverages this:

Spectral methods reveal the natural decomposition of $Q^+$
Eigenmodes of the kernel matrix are "concept basis states"
Hierarchical structure emerges from nested spectral decompositions

8. Refined Terminology¶

Based on this analysis, we should refine our language.

Instead of:¶

"The reinforcement field is a wavefunction over augmented state-action space."

Use:¶

"The reinforcement field is a state vector in RKHS, whose projections onto kernel-induced bases yield wavefunction-like amplitude fields over augmented state-action space."

Why this is better:

Distinguishes state (abstract) from representation (coordinate)
Prevents over-interpretation
Preserves the structural claim
Aligns precisely with quantum mechanics terminology

9. What This Opens Up¶

Once this is conceptually clean, several things become almost unavoidable:

For Theory (Part II)¶

Section V-C: Frame concepts as approximately invariant subspaces
Hierarchies: Nested spectral decompositions
World models: Operators acting on the GRL state
Complex RKHS: Introduces phase (not just probability)
Interference: Meaningful without metaphysics

For Implementation¶

Query the reinforcement field at different points → different "wavefunctions"
Spectral decomposition reveals concept structure
Projections onto subspaces enable hierarchical reasoning
Phase relationships (in complex RKHS) encode temporal/contextual structure

For Understanding¶

Nothing here requires claiming GRL is quantum mechanics.

Only that it lives in the same mathematical universe.

That's not mysticism—it's functional analysis doing what it always does.

10. Next Steps: What Operators Does GRL Define?¶

The natural next move is to formalize which operators GRL implicitly defines, because that's where the analogy becomes productive rather than decorative.

Candidate Operators in GRL¶

1. Value Functional $$\hat{V}: Q^+ \mapsto \mathbb{E}_\theta[Q^+(\cdot, \theta)]$$

2. MemoryUpdate as State Transition $$\hat{M}: Q^+_t \mapsto Q^+_{t+1}$$

3. Concept Projection $$\hat{P}_c: Q^+ \mapsto \text{proj}_{\text{concept}_c}(Q^+)$$

4. Action Selection $$\hat{A}: (Q^+, s) \mapsto \theta^* = \arg\max_\theta Q^+(s, \theta)$$

Each of these operators acts on the reinforcement field state, producing either:

Another state (state transition)
An expectation value (observable)
A projection (reduced representation)

This is exactly how observables work in quantum mechanics.

11. Summary¶

Question	Answer
What is the wavefunction?	Coordinate representation of a state vector in a chosen basis
What does it represent?	Probability amplitudes (not probabilities directly)
Why "one" wavefunction?	One state, many representations (different bases)
What is the GRL state?	The reinforcement field $Q^+ \in \mathcal{H}_k$
What is the GRL "wavefunction"?	The value $Q^+(z)$ at a query point (coordinate representation)
One field or many?	One state, many projections (action-fields, state-fields, concept-fields)
What does this enable?	Spectral methods, interference, hierarchical concepts, operator formalism

The Core Insight¶

Quantum mechanics and GRL share the same mathematical structure:

State = vector in Hilbert space
Observations = inner products
Probabilities = derived from amplitudes
Dynamics = operators on the state
Structure = revealed by spectral decomposition

This is not analogy—it is mathematical identity.

Basis Used	Coordinates of \(\\|\psi\rangle\)	Interpretation
Computational \(\{\\|0\rangle, \\|1\rangle\}\)	\([\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]\)	Equal mix of both directions
Hadamard \(\{\\|+\rangle, \\|-\rangle\}\)	\([1, 0]\)	Fully in \(\\|+\rangle\) direction, nothing in \(\\|-\rangle\)

Coordinate System (Basis)	State \(\\|\psi\rangle\) Coordinates
Computational \(\{\lvert 0\rangle, \lvert 1\rangle\}\)	\(\left[\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]\)
Hadamard \(\{\lvert +\rangle, \lvert -\rangle\}\)	\([1, 0]\)

Coordinate System	Coordinates
Cartesian \((x, y)\)	\([3, 4]\)
Polar \((r, \theta)\)	\([5, 53.1°]\)

Finite (Qubit)	Infinite (Position)
2 basis vectors: \(\\|0\rangle, \\|1\rangle\)	Infinite basis vectors: \(\\|x\rangle\) for every \(x \in \mathbb{R}\)
\(\\|0\rangle\) = "definitely in state 0"	\(\\|x\rangle\) = "definitely at position \(x\)"
Discrete index: 0, 1	Continuous index: \(x \in \mathbb{R}\)

Position \(x\)	Basis Vector	Component \(\psi(x) = \langle x \\| \psi \rangle\)
\(x = 0\)	\(\\|0\rangle\)	\(\psi(0) = \frac{1}{(\pi)^{1/4}} \approx 0.75\) (maximum)
\(x = 1\)	\(\\|1\rangle\)	\(\psi(1) = \frac{1}{(\pi)^{1/4}} e^{-1/2} \approx 0.46\)
\(x = 3\)	\(\\|3\rangle\)	\(\psi(3) = \frac{1}{(\pi)^{1/4}} e^{-9/2} \approx 0.008\) (nearly zero)

Symbol	Role in This Example
\(\\|\psi\rangle\)	The state we're analyzing (the "subject")
\(\\|+\rangle\)	A basis vector in the Hadamard coordinate system (the "axis")

Symbol	What It Is	Role
\(\\|\psi\rangle\)	The state (thing being analyzed)	Like the vector \(\mathbf{v} = [3, 4]\) you're analyzing
\(\\|x\rangle\)	A basis vector (coordinate axis)	Like \(\mathbf{e}_x = [1, 0]\) or \(\mathbf{e}_y = [0, 1]\)

Classical Position	Quantum State
Position vector: \(\mathbf{r} = [x, y, z]\)	State vector: \(\\|\psi\rangle\)
Expressed using basis: \(\mathbf{r} = x\mathbf{e}_x + y\mathbf{e}_y + z\mathbf{e}_z\)	Expressed using basis: \(\\|\psi\rangle = \int \psi(x) \\|x\rangle dx\)
Basis vectors: \(\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z\) (3 of them)	Basis vectors: \(\\|x\rangle\) (one for each \(x \in \mathbb{R}\), infinitely many)
Coordinates: \(x, y, z\) (numbers)	Coordinates: \(\psi(x)\) (the wavefunction!)

Concept	What It Is	Analogy
State vector \(\lvert\psi\rangle\)	The quantum system (abstract)	The geometric vector \(\mathbf{v}\)
Wavefunction \(\psi(x)\)	Coordinate representation	Cartesian coordinates \([3, 4, 0]\)
Inner product \(\langle x \lvert \psi \rangle\)	Component in direction \(\lvert x\rangle\)	How much of \(\mathbf{v}\) is in \(x\)-direction?
Different basis	Different coordinate system	Cartesian vs. polar
Same state, different wavefunction	Same \(\lvert\psi\rangle\), different basis	Same \(\mathbf{v}\), different coordinates

Quantum Mechanics	GRL
State vector \(\\|\psi\rangle \in \mathcal{H}\)	Reinforcement field \(Q^+ \in \mathcal{H}_k\)
Wavefunction \(\psi(x) = \langle x \\| \psi \rangle\)	Value at query \(Q^+(z) = \langle Q^+, k(z,\cdot) \rangle\)
Position basis \(\\|x\rangle\)	Kernel basis \(k(z, \cdot)\)
Probability \(p(x) = \\|\psi(x)\\|^2\)	Policy \(\pi(a\\|s) \propto \exp(\beta Q^+(s,a))\)

Wavefunction Interpretation: What Does It Mean for the Reinforcement Field?¶

Motivation¶

1. What Is the Wavefunction in Quantum Mechanics?¶

Simple Analogy First: 3D Vectors and Coordinates¶

The Quantum Version: State Vector vs. Wavefunction¶

Concrete Example: Two-Level System (Qubit)¶

Common Confusion: Which Direction Is the Projection?¶

Summary: Two Bases, One State¶

Infinite-Dimensional Case: Position Basis¶

Clarification: \(|\psi\rangle\) vs. \(|x\rangle\) - Which Is the Basis?¶

Your Classical Intuition Is Right—But Quantum Is Different!¶

The Big Difference: What We're Representing¶

Concrete Example: Finite Case First¶

Now the Infinite Case¶

Classical vs. Quantum Interpretation¶

Why This Is Confusing¶

Summary: Classical vs. Quantum Representation¶

Key Distinction (Now Clear!)¶

Summary Table¶

2. What Does the Wavefunction Represent?¶

The Born Rule¶

The Fundamental Move¶

3. Why "One State, Many Wavefunctions"?¶

Singular: "The Wavefunction"¶

Plural: "Different Wavefunctions"¶

Connection to GRL¶

4. Operators, Observables, and Prediction¶

Observables as Hermitian Operators¶

Measurement Probabilities¶

The Workflow¶

5. Mapping Back to GRL: State vs. Representation¶

(a) What Corresponds to the Quantum State?¶

(b) What Corresponds to the Wavefunction?¶

Summary of the Mapping¶

6. One Reinforcement Field or Many?¶

Strict Answer¶

Examples of Different Representations¶

7. Implications for Concept Discovery (Section V)¶

What Concepts Are¶

Concept Formation as Spectral Decomposition¶

Connection to Part II¶

8. Refined Terminology¶

Instead of:¶

Use:¶

9. What This Opens Up¶

For Theory (Part II)¶

For Implementation¶

For Understanding¶

10. Next Steps: What Operators Does GRL Define?¶

Candidate Operators in GRL¶

11. Summary¶

The Core Insight¶

Further Reading¶

Within This Tutorial¶

Quantum Mechanics Foundations¶

RKHS and Functional Analysis¶

GRL Original Paper¶