# Conventions

## States and Bases

### Bases

A basis refers to a set of two eigenstates. The transition between
these two states is said to be addressed by a channel that targets that basis. Namely:

```{eval-rst}
.. list-table::
   :align: center
   :widths: 50 35 35
   :header-rows: 1

   * - Basis
     - Eigenstates
     - ``Channel`` type
   * - ``ground-rydberg``
     - :math:`|g\rangle,~|r\rangle`
     - ``Rydberg``
   * - ``digital``
     - :math:`|g\rangle,~|h\rangle`
     - ``Raman``
   * - ``XY``
     - :math:`|0\rangle,~|1\rangle`
     - ``Microwave``


```

### Qutrit state

The qutrit state combines the basis states of the `ground-rydberg` and `digital` bases,
which share the same ground state, $|g\rangle$. This qutrit state comes into play
in the digital approach, where the qubit state is encoded in $|g\rangle$ and
$|h\rangle$ but then the Rydberg state $|r\rangle$ is accessed in multi-qubit
gates.

The qutrit state's basis vectors are defined as:

$$
|r\rangle = (1, 0, 0)^T,~~|g\rangle = (0, 1, 0)^T, ~~|h\rangle = (0, 0, 1)^T.
$$

### Qubit states

:::{caution}
There is no implicit relationship between a state's vector representation and its
associated measurement value. To see the measurement value of a state for each
measurement basis, see {ref}`spam-table` .
:::

When using only the `ground-rydberg` or `digital` basis, the qutrit state is not
needed and is thus reduced to a qubit state. This reduction is made simply by tracing-out
the extra basis state, so we obtain

- `ground-rydberg`: $|r\rangle = (1, 0)^T,~~|g\rangle = (0, 1)^T$
- `digital`: $|g\rangle = (1, 0)^T,~~|h\rangle = (0, 1)^T$

On the other hand, the `XY` basis uses an independent set of qubit states that are
labelled $|0\rangle$ and $|1\rangle$ and follow the standard convention:

- `XY`: $|0\rangle = (1, 0)^T,~~|1\rangle = (0, 1)^T$

### Multi-partite states

The combined quantum state of multiple atoms respects their order in the `Register`.
For a register with ordered atoms `(q0, q1, q2, ..., qn)`, the full quantum state will be

$$
|q_0, q_1, q_2, ...\rangle = |q_0\rangle \otimes |q_1\rangle \otimes |q_2\rangle \otimes ... \otimes |q_n\rangle
$$

:::{note}
The atoms may be labelled arbitrarily without any inherent order, it's only the
order with which they are stored in the `Register` (as returned by
`Register.qubit_ids`) that matters .
:::


## State Preparation and Measurement

```{eval-rst}
.. list-table:: Initial State and Measurement Conventions
   :name: spam-table
   :align: center
   :widths: 60 40 75
   :header-rows: 1

   * - Basis
     - Initial state
     - Measurement
   * - ``ground-rydberg``
     - :math:`|g\rangle`
     - |
       | :math:`|r\rangle \rightarrow 1`
       | :math:`|g\rangle,|h\rangle \rightarrow 0`
   * - ``digital``
     - :math:`|g\rangle`
     - |
       | :math:`|h\rangle \rightarrow 1`
       | :math:`|g\rangle,|r\rangle \rightarrow 0`
   * - ``XY``
     - :math:`|0\rangle`
     - |
       | :math:`|1\rangle \rightarrow 1`
       | :math:`|0\rangle \rightarrow 0`
```

### Measurement samples order

Measurement samples are returned as a sequence of 0s and 1s, in
the same order as the atoms in the `Register` and in the multi-partite state.

For example, a four-qutrit state $|q_0, q_1, q_2, q_3\rangle$ that's
projected onto $|g, r, h, r\rangle$ when measured will record a count to
sample

- `0101`, if measured in the `ground-rydberg` basis
- `0010`, if measured in the `digital` basis

## Hamiltonians

:::{tip}
This section uses formulas that rely on the [Indexed Operator](#indexed-operator)
notation.
:::

Independently of the mode of operation, the Hamiltonian describing the system
can be written as

$$
H(t) = \sum_i \left (H^D_i(t) + \sum_{j<i}H^\text{int}_{ij} \right),
$$

where $H^D_i$ is the driving Hamiltonian for atom $i$ and
$H^\text{int}_{ij}$ is the interaction Hamiltonian between atoms $i$
and $j$. Note that, if multiple basis are addressed, there will be a
corresponding driving Hamiltonian for each transition.

### Driving Hamiltonian

The driving Hamiltonian describes the coherent excitation of an individual atom
between two energies levels, $|a\rangle$ and $|b\rangle$, with
Rabi frequency $\Omega(t)$, detuning $\delta(t)$ and phase $\phi(t)$.

:::{figure} files/two_level_ab.png
:align: center
:alt: The energy levels for the driving Hamiltonian.
:width: 200

The coherent excitation is driven between a lower energy level, $|a\rangle$, and a higher energy level,
$|b\rangle$, with Rabi frequency $\Omega(t)$ and detuning $\delta(t)$.
:::

:::{warning}
In this form, the Hamiltonian is **independent of the state vector representation of each basis state**,
but it still assumes that $|b\rangle$ **has a higher energy than** $|a\rangle$.
:::

$$
H^D(t) / \hbar = \frac{\Omega(t)}{2} e^{-i\phi(t)} |a\rangle\langle b| + \frac{\Omega(t)}{2} e^{i\phi(t)} |b\rangle\langle a| - \delta(t) |b\rangle\langle b|
$$

#### Pauli matrix form

A more conventional representation of the driving Hamiltonian uses Pauli operators
instead of projectors. However, this form now **depends on the state vector definition**
of $|a\rangle$ and $|b\rangle$.

##### Case 1. Descending energy order

:::{important}
**In `pulser-simulation`, this is the case for the `ground-rydberg` basis**, since

 $$
 E_r > E_g \rightarrow \ket{r} \equiv \ket{b}, \ket{g} \equiv \ket{a}
 $$

:::

When the basis vectors are defined in descending energy order, we have

$$
|b\rangle = (1, 0)^T,~~|a\rangle = (0, 1)^T
$$

Thus, the Pauli and excited state occupation operators are defined as

$$
\hat{\sigma}^x = |a\rangle\langle b| + |b\rangle\langle a|, \\
\hat{\sigma}^y = i|a\rangle\langle b| - i|b\rangle\langle a|, \\
\hat{\sigma}^z = |b\rangle\langle b| - |a\rangle\langle a|  \\
\hat{n} = |b\rangle\langle b| = (1 + \sigma_z) / 2
$$

and the driving Hamiltonian takes the form

$$
H^D(t) / \hbar = \frac{\Omega(t)}{2} \cos\phi(t) \hat{\sigma}^x
- \frac{\Omega(t)}{2} \sin\phi(t) \hat{\sigma}^y
- \delta(t) \hat{n}
$$

##### Case 2. Ascending energy order

:::{important}
**In `pulser-simulation`, this is the case for the `XY` basis**, since

 $$
 E_0 < E_1 \rightarrow \ket{0} \equiv \ket{a}, \ket{1} \equiv \ket{b}
 $$

:::

When the basis vectors are defined in ascending energy order, we have

$$
|a\rangle = (1, 0)^T,~~|b\rangle = (0, 1)^T
$$

This changes the operators and Hamiltonian definitions with respect to [Case 1](#case-1-descending-energy-order), as rewriten below with highlighted differences.

$$
\hat{\sigma}^x = |a\rangle\langle b| + |b\rangle\langle a|, \\
\hat{\sigma}^y = \textcolor{red}{-}i|a\rangle\langle b| \textcolor{red}{+}i|b\rangle\langle a|, \\
\hat{\sigma}^z = \textcolor{red}{-}|b\rangle\langle b| \textcolor{red}{+} |a\rangle\langle a|  \\
\hat{n} = |b\rangle\langle b| = (1 \textcolor{red}{-} \sigma_z) / 2
$$

$$
H^D(t) / \hbar = \frac{\Omega(t)}{2} \cos\phi(t) \hat{\sigma}^x
\textcolor{red}{+}\frac{\Omega(t)}{2} \sin\phi(t) \hat{\sigma}^y
- \delta(t) \hat{n}
$$

:::{note}
This is also the Hamiltonian one should use when trying to reconcile the basis states of the `ground-rydberg` basis
with the computational-basis state-vector convention, i.e.

$$
|0\rangle = |g\rangle = |a\rangle = (1, 0)^T,~~|1\rangle = |r\rangle = |b\rangle = (0, 1)^T
$$
:::


### Interaction Hamiltonian

The interaction Hamiltonian depends on the states involved in the sequence.
When working with the `ground-rydberg` and `digital` bases, atoms interact
when they are in the Rydberg state $|r\rangle$:

$$
H^\text{int}_{ij} = \frac{C_6}{R_{ij}^6} \hat{n}_i \hat{n}_j
$$

where $\hat{n}_i = |r\rangle\langle r|_i$ (the projector of
atom $i$ onto the Rydberg state), $R_{ij}^6$ is the distance
between atoms $i$ and $j$ and $C_6$ is a coefficient
depending on the specific Rydberg level of $|r\rangle$.

On the other hand, with the two Rydberg states of the `XY`
basis, the interaction Hamiltonian takes the form

$$
H^\text{int}_{ij} =  \frac{C_3}{R_{ij}^3} (|1\rangle\langle 0|_i |0\rangle\langle 1|_j + |0\rangle\langle 1|_i |1\rangle\langle 0|_j)
$$

where $C_3$ is a coefficient that depends on the chosen Ryberg states.

:::{note}
The definitions given for both interaction Hamiltonians are independent of the chosen state vector convention.
:::

## Notation

### Indexed Operator

Whenever an arbitrary operator is written with an index (typically $i$ or $j$), e.g. $\hat{O}_i$, it is implicit that $\hat{O}$ is applied *only* to qudit $i$ while the rest of the qudits are applied the identity operator, $\hat{I}$. Put another way,

$$ \hat{O}_i = \underset{(1)}{\hat{I}} \otimes \underset{(2)}{\hat{I}} \otimes ... \otimes\ \underset{(i)}{\hat{O}}\ \otimes ... \otimes \underset{(N)}{\hat{I}},$$

where $1 \leq i \leq N$.

This notation is extendable to multiple indices. Take for instance the case with two indices, $\hat{O}_{ij}$ – here, $\hat{O}$ is a two-qudit operator. A good example is the [interaction Hamiltonian](#interaction-hamiltonian) in the `ground-rydberg` basis, which we write as

$$H^\text{int}_{ij} = \frac{C_6}{R_{ij}^6} \hat{n}_i \hat{n}_j = \frac{C_6}{R_{ij}^6} \left( \underset{(1)}{\hat{I}} \otimes ... \otimes \ \underset{(j)}{\hat{n}}\ \otimes ... \otimes \ \underset{(i)}{\hat{n}} \ \otimes ... \otimes \underset{(N)}{\hat{I}}\right),$$

where $1 \leq j < i \leq N$.

Note that, generally, we cannot write $\hat{O}_{ij}$ in the form used above because $\hat{O}$ might not be separable in a tensor product of two single-qudit operators, but the operator is valid nonetheless.