Add missing section in classical shadows example

docs/source/examples/shadows_tutorial.md

@@ -104,6 +104,63 @@ In the main function, the quantum measurement process is encapsulated within the
 shadow_quantum_processing(test_circuits, cirq_executor, 2)
 ```
 
+## 3. Obtain Snapshot and Classical Shadows.
+
+This random measurement contains valuable information about $\rho$ in expectation:
+\begin{equation}
+    \mathbb{E}[U^\dagger |\hat{b}\rangle\langle\hat{b}|U]=\mathcal{M}(\rho),
+\end{equation}
+the expectation in the first expression has the form $\mathbf{Pr}[\hat{{b}}={b}]=\langle {b}|U\rho U^\dagger|b\rangle$. For any unitary ensemble $\mathcal{U}$, the expected value of the outer product of the classical snapshot corresponds to the operation of the quantum channel $\mathcal{M}$ on the quantum state $\rho$. If the measurements we sample from are tomographically complete, then the protocol $\mathcal{M}$ defines an invertible linear transformation $\mathcal{M}^{-1}$, which may not be a quantum channel, since it is not CP, which means that it could not be performed in the lab. But it will only be performed on the classical data stored in
+a classical memory. If we apply $\mathcal{M}$ to all the snapshots, the expected value of these inverted snapshots equations with the density operator as defined by the protocol,
+
+\begin{equation}
+\hat{\rho}=\mathcal{M}^{-1}\left(U^\dagger|\hat{b}\rangle\langle\hat{b}|U\right)
+\end{equation}
+which has been named a single copy of **classical shadow**. Based on *Schur's Lemma* the quantum channel $\mathcal{M}$ is a depolarizing channel $\mathcal{D}_p$ with $p=\frac{1}{2^n+1}$. It is easy to solve for the inverted map 
+
+\begin{equation}
+\mathcal{M}^{-1}(\cdot)=[(2^n +1)-\mathbb{I}\cdot\mathrm{Tr}](\cdot),
+\end{equation}
+ which is indeed unitary, however, not CP, so it is not a physical map as expected.
+
+In the case of random Pauli measurement, the unitary could be represented by the tensor product of all qubits, so it is with the state $|\hat{b}\rangle\in\{0,1\}^{\otimes n}$, i.e. $U^\dagger|\hat{b}\rangle=\bigotimes_{i\leq n}U_i|\hat{b}_i\rangle$. Therefore, based on Schur's Lemma, a snapshot would takes the form:
+\begin{equation}
+\hat{\rho}=\bigotimes_{i=1}^{n}\left(3U_i^\dagger|\hat{b}_i\rangle\langle\hat{b}_i|U_i-\mathbb{I}\right),\qquad|\hat{b}_i\rangle\in\{0,1\}.
+\end{equation}
+which is a tensor product of $n$ qubits, each of which is a classical state. This step is realized by `classical_snapshot` function. Repeating this procedure $N$ times results in an array of $N$ independent classical snapshots of $\rho$:
+\begin{equation}
+    S(\rho,\; N)=\left\{\hat{\rho}_1=\mathcal{M}^{-1}\left(U_1^\dagger |\hat{b}_1\rangle\langle\hat{b}_1| U_1\right),\dots,\mathcal{M}^{-1}\left(U_N^\dagger |\hat{b}_N\rangle\langle\hat{b}_N| U_N\right)\right\} .
+\end{equation}
+
+## 4. State Reconstruction from Classical Shadows
+### 4.1 State Reconstruction
+The classical shadows state reconstruction are then obtained by taking the average of the snapshots, this process is designed to reproduce the underlying state $\rho$ exactly in expectation:
+\begin{equation}
+   \rho= \mathbb{E}[\hat{\rho}],
+\end{equation}
+this is realized in the function `state_reconstruction`. In the main function `classical_post_processing`, we take the output of `shadow_quantum_processing`, then apply the inverse channel to obtain the snapshots, and finally take the average of the snapshots to obtain the reconstructed state if *state_reconstruction =* **True**. **In the current notebook, we don't preform Pauli twirling calibration, and we set** *rshadow* = **False**.
+
+#### 4.1.1 Error Analysis
+We can take a visualization of the elementwise difference between the reconstructed state and the original state. 
+\begin{equation}
+\Delta\rho_{ij}=|\rho^{\mathrm{shadow}}_{ij}-\rho_{ij}|
+\end{equation}
+The difference is very small, which means that the classical shadow is a good approximation of the original state even in the sence of state tomography. 
+
+It is anticipated that the fidelity will not necessarily be lower than 1, as the state reconstructed through classical shadow estimation is not guaranteed to be a physical quantum state, given that $\mathcal{M}^{-1}$ is not a quantum channel. 
+
+Fidelity is defined by $F(\rho,\sigma)=\mathrm{Tr}\sqrt{\rho^{1/2}\sigma\rho^{1/2}}$, when $\rho=|v\rangle\langle v|$ is a pure state $F(\rho,\sigma)=\langle v|\sigma|v\rangle$.
+Based on the theorem, if the error rate of fedelity is $\epsilon$, i.e.
+\begin{equation}
+|F(\rho,\sigma)-1|\leq\epsilon,
+\end{equation}
+then the minimum number of measurements $N$ (number of snapshots) should be:
+\begin{equation}
+N = \frac{34}{\epsilon^2}\left\|\rho-\mathrm{Tr}(\rho)/{2^n}\mathbb{I}\right\|_{\mathrm{shadow}}^2
+\end{equation}
+with the shadow norm upper bound of the random Pauli measurement $\left\|\cdot\right\|_{\mathrm{shadow}}\leq 2^k\|\cdot\|_\infty$ when the operator acting on $k$ qubits, we have $N\leq 34\epsilon^{-2}2^{2n}+\mathcal{O}(e^{-n})$. Based on Fuchs–van de Graaf inequalities and properties of $L_p$ norm, $\|\rho-\sigma\|_2\leq \|\rho-\sigma\|_1 \leq (1-F(\rho,\sigma))^{1/2}$, the $L_2$ norm distance between the state reconstructed through classical shadow estimation and the state prepared by the circuit is upperbounded by the fidelity error rate $\epsilon$. The dependency of the bound number of measurements $N$ to achieve the error rate $\epsilon$ is depicked in function `n_measurements_tomography_bound`.
+
+
 ```{code-cell} ipython3
 # error rate of state reconstruction epsilon < 1.
 epsilon = 1
@@ -480,7 +537,6 @@ plt.xscale("log")
 plt.show()
 ```
 
-
 **Acknowledgements**
 
 This project contains code adapted from PennyLane's tutorial on Classical Shadows. We would like to acknowledge the original authors of the tutorial, PennyLane developers Brian Doolittle and Roeland Wiersema. The tutorial can be found at [this link](https://pennylane.ai/qml/demos/tutorial_classical_shadows).