foundations-and-modeling-of-quantum-errors – Quantum Information & Computing

/think * Physical Meaning: The qubit is flipped with probability \(p\). (\(|0\rangle \leftrightarrow |1\rangle\))
* Error Model: \(\mathcal{E}_{BF}(\rho) = (1-p)\rho + p X \rho X\)
* Kraus Operators:
* \(K_0 = \sqrt{1-p} \cdot \mathbf{1}\) (No error)
* \(K_1 = \sqrt{p} \cdot X\) (X error occurs)
* (Verification: \(\sum K_k^\dagger K_k = (1-p)\mathbf{1} + p X^\dagger X = (1-p)\mathbf{1} + p\mathbf{1} = \mathbf{1}\))

2. Phase-Flip Channel

Physical Meaning: The qubit’s value does not change, but the relative phase between \(|0\rangle\) and \(|1\rangle\) is flipped by \(180^\circ\) (\(-1\) times) with probability \(p\). (\(|+\rangle \leftrightarrow |-\rangle\))
Error Model: \(\mathcal{E}_{PF}(\rho) = (1-p)\rho + p Z \rho Z\)
Kraus Operators:
- \(K_0 = \sqrt{1-p} \cdot \mathbf{1}\) (No error)
- \(K_1 = \sqrt{p} \cdot Z\) (Z error occurs)

3. Depolarizing Channel 💥

Physical Meaning: The most general “worst-case” noise model. The qubit completely forgets its original state and collapses into a perfectly random (maximally mixed) state \(\mathbf{1}/2\) with probability \(p\).
Error Model (Form 1): \(\mathcal{E}_{DP}(\rho) = (1-p)\rho + p \frac{\mathbf{1}}{2}\)
Error Model (Form 2, Pauli Basis): This model is mathematically equivalent to Pauli errors.
\(\mathcal{E}_{DP}(\rho) = (1-p')\rho + \frac{p'}{3}(X\rho X + Y\rho Y + Z\rho Z)\)
(where \(p' = \frac{3}{4}p\). \(p'\) is the total error probability)
Meaning: This equivalence shows that the phenomenon “the qubit’s information becomes completely random” is the same as the phenomenon “X, Y, Z errors occur randomly with equal probability.” This reaffirms that we only need to detect \(X, Y, Z\).

3. Simplified Examples (Examples with Deeper Insight)

Assume the initial state is \(|\psi\rangle = \alpha|0\rangle + \beta|1\rangle\) (where \(\alpha, \beta\) are real numbers).
The initial density matrix \(\rho = |\psi\rangle\langle\psi| = \begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix}\).

Example 1: Effect of Bit-Flip (X) Error

Situation: \(\rho\) passes through the \(\mathcal{E}_{BF}(\rho) = (1-p)\rho + p X \rho X\) channel.
- Calculation: \(X \rho X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \alpha\beta & \beta^2 \\ \alpha^2 & \alpha\beta \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \beta^2 & \alpha\beta \\ \alpha\beta & \alpha^2 \end{pmatrix}\)
- Final State: \(\rho' = (1-p)\begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix} + p\begin{pmatrix} \beta^2 & \alpha\beta \\ \alpha\beta & \alpha^2 \end{pmatrix}\) \(\rho' = \begin{pmatrix} (1-p)\alpha^2 + p\beta^2 & \alpha\beta \\ \alpha\beta & (1-p)\beta^2 + p\alpha^2 \end{pmatrix}\)
- 💡 Detailed Explanation (What was destroyed?):
  - Diagonal components (probabilities): \(\rho'_{00} = (1-p)\alpha^2 + p\beta^2\). The original probability of being in \(|0\rangle\) (\(\alpha^2\)) and the probability of being in \(|1\rangle\) (\(\beta^2\)) are mixed in the ratio of \(p\). This is exactly the same as a classical bit flip.
  - Off-diagonal components (coherence): \(\rho'_{01} = (1-p)\alpha\beta + p(\alpha\beta) = \alpha\beta\). Surprisingly, coherence is not at all damaged!
  - Conclusion: A bit flip error does not damage the ‘phase relationship’ of superposition, but only damages the ‘probability of being in a certain state’.

Example 2: Effect of Phase Flip (Z) Error

Situation: \(\rho\) passes through the \(\mathcal{E}_{PF}(\rho) = (1-p)\rho + p Z \rho Z\) channel.
- Calculation: \(Z \rho Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \alpha^2 & -\alpha\beta \\ \alpha\beta & -\beta^2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \alpha^2 & \alpha\beta \\ -\alpha\beta & \beta^2 \end{pmatrix}\) (calculation error!) \(Z \rho Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \alpha^2 & \alpha\beta \\ -\alpha\beta & -\beta^2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \alpha^2 & -\alpha\beta \\ -\alpha\beta & \beta^2 \end{pmatrix}\)
- Final State: \(\rho' = (1-p)\begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix} + p\begin{pmatrix} \alpha^2 & -\alpha\beta \\ -\alpha\beta & \beta^2 \end{pmatrix}\) \(\rho' = \begin{pmatrix} \alpha^2 & (1-2p)\alpha\beta \\ (1-2p)\alpha\beta & \beta^2 \end{pmatrix}\)
- 💡 Detailed Explanation (What was destroyed?):
  - Diagonal components (probabilities): \(\rho'_{00} = (1-p)\alpha^2 + p\alpha^2 = \alpha^2\). Probabilities are completely undamaged! (The probability of being \(|0\rangle\) is still \(\alpha^2\))
  - Off-diagonal components (coherence): \(\rho'_{01} = (1-2p)\alpha\beta\). The coherence term has been damped by a factor of \((1-2p)\). At \(p=1/2\) (maximum error), coherence becomes zero.
  - Conclusion: Phase flip errors do not damage classical probabilities, but only attack the “quantumness” (coherence) of superposition states. This is the simplest model of decoherence described in Chapter 8 of “Quantum Mechanics” in the context of quantum Darwinism.

Example 3: Justification of Discretization (Continuous vs Discrete)

Situation: Real-world errors are continuous, similar to the Amplitude Damping channel in Chapter 5 of “Quantum History” (e.g., \(|1\rangle \to \sqrt{1-\gamma}|1\rangle + \sqrt{\gamma}|0\rangle\))
- Problem: How can we approximate this as a discrete \(X\) error?
- Interpretation:
  - Assume \(\gamma\) is very small (for a short time \(\delta t\)), \(\gamma \approx p\).
  - \(\sqrt{1-p} \approx 1 - p/2\).
  - Approximating this continuous process \(\mathcal{E}_{AD}\) to first order,
  - \(\mathcal{E}_{AD}(\rho) \approx (1-p_Z) \rho + p_Z Z\rho Z + p_X X\rho X\) …
  - That is, the complex continuous process of “energy slowly leaking out” can be very well approximated by a probabilistic mixture of discrete events: “nothing happens” (with probability \(1-p\)), “a bit flip happens” (with probability \(p_X\)), or “a phase flip happens” (with probability \(p_Z\)).
  - Conclusion: If we create codes that can catch the discretized \(X\) and \(Z\) errors in Chapter 7, we can approximately deal with the complex continuous errors in reality.

4. Exercise Problems

Effect of Y Errors: As in Examples 1 and 2, calculate the effect of the channel \(\mathcal{E}_{Y}(\rho) = (1-p)\rho + p Y \rho Y\) on \(\rho = \begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix}\). (Hint: Calculate \(Y \rho Y\))
Z Errors and X Basis: Calculate the final density matrix \(\rho'\) when the state \(|\psi\rangle = |+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)\) passes through a phase flip (\(Z\)) channel. How does the measurement probability in the \(X\) basis of \(\rho'\) change?
X Errors and Z Basis: Calculate the final density matrix \(\rho'\) when the state \(|\psi\rangle = |0\rangle\) passes through a bit flip (\(X\)) channel. How does the measurement probability in the \(Z\) basis (\(|0\rangle, |1\rangle\)) of \(\rho'\) change?
Equivalence of Depolarizing Channel: Prove that \(\mathcal{E}(\rho) = (1-p)\rho + p \frac{\mathbf{1}}{2}\) (Form 1) is equivalent to \(\mathcal{E}(\rho) = (1-\frac{3p}{4})\rho + \frac{p}{4}(X\rho X + Y\rho Y + Z\rho Z)\) (Form 2, \(p'=3p/4\)). (Hint: \(\mathbf{1}/2 = \frac{1}{2}\rho + \frac{1}{2}(X\rho X + Y\rho Y + Z\rho Z)\) does not hold for all \(\rho\), but when \(\rho=\mathbf{1}/2\)… no, write \(\rho = \frac{1}{2}(\mathbf{1} + \vec{r}\cdot\vec{\sigma})\) and calculate)

5. Explanation

\(Y \rho Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix} \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix} = \dots = \begin{pmatrix} \beta^2 & -\alpha\beta \\ -\alpha\beta & \alpha^2 \end{pmatrix}\). \(\rho' = \begin{pmatrix} (1-p)\alpha^2 + p\beta^2 & (1-2p)\alpha\beta \\ (1-2p)\alpha\beta & (1-p)\beta^2 + p\alpha^2 \end{pmatrix}\). Conclusion: \(Y\) error has the effects of both bit flip (\(X\)) and phase flip (\(Z\)). (Both diagonal/non-diagonal terms change)
\(|\psi\rangle = |+\rangle \implies \rho = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\). ( \(\alpha=\beta=1/\sqrt{2}\) ) Substituting the result of Example 2’s \(\rho'\), \(\rho' = \begin{pmatrix} 1/2 & (1-2p)/2 \\ (1-2p)/2 & 1/2 \end{pmatrix}\). When measured in the \(X\) basis, \(P(+) = \langle +|\rho'|+\rangle = \frac{1}{2}\begin{pmatrix} 1 & 1 \end{pmatrix} \rho' \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \dots = 1-p\). \(P(-) = \langle -|\rho'|-\rangle = p\). Conclusion: Phase flip error creates the \(|-\rangle\) state from the \(|+\rangle\) state with probability \(p\). That is, \(Z\) error appears like a bit flip in the \(X\) basis.
\(|\psi\rangle = |0\rangle \implies \rho = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\). ( \(\alpha=1, \beta=0\) ) Substituting the result of Example 1’s \(\rho'\), \(\rho' = \begin{pmatrix} (1-p) & 0 \\ 0 & p \end{pmatrix}\). When measured in the \(Z\) basis, \(P(0) = \rho'_{00} = 1-p\), \(P(1) = \rho'_{11} = p\). Conclusion: Bit-flip error creates the \(|1\rangle\) state from the \(|0\rangle\) state with probability \(p\). (Obvious result)
\(\rho = \frac{1}{2}(\mathbf{1} + \vec{r}\cdot\vec{\sigma})\). \(\mathcal{E}_{DP}(\rho) = (1-p)\rho + p\frac{\mathbf{1}}{2} = \frac{1}{2}(\mathbf{1} + (1-p)\vec{r}\cdot\vec{\sigma})\). (Bloch sphere is contracted towards the origin by \(1-p\)) Substituting expressions such as \(X\rho X = \frac{1}{2}(\mathbf{1} + r_x\sigma_x - r_y\sigma_y - r_z\sigma_z)\) into Form 2 and simplifying, \(\mathcal{E}(\rho) = \dots = \frac{1}{2}(\mathbf{1} + (1-p')\vec{r}\cdot\vec{\sigma} - \frac{p'}{3}(r_y+r_z+r_x+r_z+r_x+r_y)\vec{\sigma}) \dots\) Instead of \(p'=3p/4\), modeling as \(p' = p/3\) (for each of \(X\), \(Y\), \(Z\)) gives \(\mathcal{E}(\rho) = (1-p)\rho + \frac{p}{3}(X\rho X + \dots) = \frac{1}{2}(\mathbf{1} + (1 - \frac{4p}{3})\vec{r}\cdot\vec{\sigma})\). Forms 1 and 2 have the same effect of contracting the Bloch vector towards the origin with \(p' = 4p/3\) relation: \(p \to (1-p')\vec{r}\).