Notes › PHYS 3571: Quantum Computing Lecture 8

Error Correction

calendar_today February 19, 2026 article 803 words access_time 6 min replay Modified September 4, 2026

Table of Contents

Errors

Kinds of errors
- Bit flip
- Phase flip
- Both error
  - Need to use Shor’s Error-correcting code

Bit Flip

Partial bit flip
- Occurs when one of the points on the Bloch sphere moves towards the other point instead of completely swapping states
- Assuming $\psi$ was unchanged and $\theta$ small:

$$\ket{\psi} \approx\frac{\ket{0} + \frac{\theta}{2} \ket{1} }{\sqrt{ 1 + \left( \frac{\theta}{2} \right)^2 } }$$

To get an exact mapping of how the partial bit flip affects $\ket{\psi}$, we can parameterize the universal gate operator
- Partial bit flip = rotation about the $x$-axis by angle $\theta$, so $\hat{n} = (1, 0, 0)$
- Choose global phase $\alpha = \pi/2$, so the general rotation operator reduces to:

$$\hat{U}_{\text{partial flip}} = i\cos\left(\frac{\theta}{2}\right)\hat{I} + \sin\left(\frac{\theta}{2}\right)\hat{X}$$

Definition 08.1 (Partial Bit Flip Operator).

Let $\varepsilon = \sin(\theta/2)$, so $\cos(\theta/2) = \sqrt{1 - \varepsilon^2}$. The partial bit flip operator is:

$$i\sqrt{1-\varepsilon^2}\,\hat{I} + \varepsilon\hat{X} = \begin{pmatrix} i\sqrt{1-\varepsilon^2} & \varepsilon \\ \varepsilon & i\sqrt{1-\varepsilon^2} \end{pmatrix}$$

Proposition 08.2 (Action of Partial Bit Flip on Basis States).

$$\ket{0} \to i\sqrt{1-\varepsilon^2}\,\ket{0} + \varepsilon\ket{1}$$$$\ket{1} \to \varepsilon\ket{0} + i\sqrt{1-\varepsilon^2}\,\ket{1}$$

Definition 08.3 (Error Parameter $\varepsilon$).

$$\varepsilon = \sin\left(\frac{\theta}{2}\right)$$

where $\theta$ is the Bloch sphere rotation angle of the partial flip.

Small physical disturbance $\implies$ small $\theta$ $\implies$ $\varepsilon \ll 1$
$\varepsilon^2$: probability ancilla measurement collapses into error branch
$1 - \varepsilon^2$: probability ancilla measurement collapses into no-error branch

Corollary 08.4 (Action on General State).

For $\ket{\psi} = \alpha\ket{0} + \beta\ket{1}$ with $\alpha, \beta \in \mathbb{C}$:

$$\ket{\psi} \to \left(i\sqrt{1-\varepsilon^2}\,\alpha + \varepsilon\beta\right)\ket{0} + \left(\varepsilon\alpha + i\sqrt{1-\varepsilon^2}\,\beta\right)\ket{1}$$

When $\theta = \pi$, $\varepsilon = 1$, recovering the full $\hat{X}$ gate: $\alpha\ket{0}+\beta\ket{1} \to \beta\ket{0}+\alpha\ket{1}$

Phase Flip

A rotation through angle $\theta$ around the $z$-axis

Solutions

Problems with Naive Approaches

Can’t use a parity bit because that would require that you measure the physical state of the qubit which would collapse its wave function
You also can’t copy the state from one qubit to the other because of the 4.4.4 No-Cloning Theorem [Wong, Introduction to Classical and Quantum Computing, p. 177]

Error-Correcting Codes

Assign 3 logical qubits that we use to encode a real qubit

$$\begin{align} \ket{0_{L}} &= \ket{0 \ 0 \ 0} \\ \ket{1_{L}} &= \ket{1 \ 1 \ 1} \\ \ket{\psi_{L}} &=\alpha \ket{0_{L}} + \beta \ket{1_{L}} \end{align}$$

Example 08.5 (Left bit-flip).

Consider a left-qubit bit-flip:

$$\ket{\psi} = \alpha \ket{1 \ 0 \ 0} + \beta \ket{0 \ 1 \ 1}$$

How can we encode the given state?

We know there’s an error if the parity between consecutive qubits differs! This is done by chaining the line with $\ket{\psi}$ to all the other qubits with a CNOT gate (which is like XOR in classical computing).

$$\text{CNOT}(\ket{a}\ket{b}) = \ket{a}\ket{a \oplus b} $$

Ancilla Qubits

To carry out error detection in general, we need ancilla qubits
- If both qubits flipped, then the error correction scheme
Based on the results of the measurement, you apply the rotation (depending on which qubit flipped), then apply a 180 degree rotation to $\theta$, and flip it

Detecting Partial Bit Flip Errors With Ancilla Qubits

Ancilla qubits initialized to $\ket{00}$, so the full initial state is:

$$\ket{00} \otimes \ket{\psi_L} = \ket{00}\left(\alpha\ket{000} + \beta\ket{111}\right)$$

Apply a partial bit flip to the first (leftmost) logical qubit using Action on General State:

$$\ket{000} = \ket{0}_{\text{flip}}\ket{00} \to \left(i\sqrt{1-\varepsilon^2}\ket{0} + \varepsilon\ket{1}\right)\ket{00} = i\sqrt{1-\varepsilon^2}\ket{000} + \varepsilon\ket{100}$$$$\ket{111} = \ket{1}_{\text{flip}}\ket{11} \to \left(\varepsilon\ket{0} + i\sqrt{1-\varepsilon^2}\ket{1}\right)\ket{11} = \varepsilon\ket{011} + i\sqrt{1-\varepsilon^2}\ket{111}$$

Applying linearity to the full logical state and tensoring in the ancilla:

$$\boxed{\ket{\Psi} = i\alpha\sqrt{1-\varepsilon^2}\ket{00}\ket{000} + \alpha\varepsilon\ket{00}\ket{100} + \beta\varepsilon\ket{00}\ket{011} + i\beta\sqrt{1-\varepsilon^2}\ket{00}\ket{111}}$$

Proposition 08.6 (Interpretation of Outcomes).

$i\alpha\sqrt{1-\varepsilon^2}\ket{00}\ket{000}$: no error on $\alpha$ branch
$\alpha\varepsilon\ket{00}\ket{100}$: first qubit flipped on $\alpha$ branch
$\beta\varepsilon\ket{00}\ket{011}$: first qubit flipped on $\beta$ branch
$i\beta\sqrt{1-\varepsilon^2}\ket{00}\ket{111}$: no error on $\beta$ branch
When $\varepsilon = 1$ (full $\hat{X}$), the no-error terms vanish and full flip is recovered

We can entangle each ancilla qubit with each consecutive pair in the logical qubit to get the following readings
- Ancilla reads $\ket{0 \ 0}$: no error; state collapses to $\alpha \ket{0 \ 0 \ 0} + \beta \ket{1 \ 1 \ 1}$
- Ancilla reads $\ket{1 \ 0}$: first qubit flipped
- Ancilla reads $\ket{1 \ 1}$: second qubit flipped
- Ancilla reads $\ket{0 \ 1}$: third qubit flipped

Proposition 08.7 (Ancilla Measurement Outcomes (First Qubit Partial Flip)).

$a_{01}$	$a_{12}$	Probability	Post-measurement $\ket{\psi}$
0	0	$1-\varepsilon^2$	$\alpha\ket{000} + \beta\ket{111}$
0	1	$0$	—
1	0	$\varepsilon^2$	$\alpha\ket{100} + \beta\ket{011}$
1	1	$0$	—
Apply $\hat{X}$ to the first qubit if ancilla reads $\ket{10}$ to recover $\alpha\ket{000}+\beta\ket{111}$.

Proposition 08.8 (Role of Measurement in Error Correction).

The ancilla measurement forces the partial rotation to commit to one of two outcomes:

No-error branch ($\ket{00}$): state collapses to $\alpha\ket{000}+\beta\ket{111}$, up to global phase $i\sqrt{1-\varepsilon^2}$, which has no physical consequence
Error branch ($\ket{10}$): state collapses to $\alpha\ket{100}+\beta\ket{011}$, corrected by applying $\hat{X}$ to the first qubit

This is why error correction works: measuring the ancilla collapses the superposition into a definite, correctable form without ever revealing $\alpha$ or $\beta$.

Sources

Wong, Introduction to Classical and Quantum Computing

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