## 7.8 *Remarks and exercises*

### 7.8.1 Measuring parity

Suppose you have a two-qubit stabiliser *not* measure the bit values

Mathematically speaking, the parity measurement

In terms of circuits, if we want to measure the parity of two qubits prepared in some state

which is different from the circuit which effects the measurement of the individual bit values of the two qubits, namely

When dealing with stabilisers, we prefer to think of the controlled-

But this is a bit confusing, since we are measuring the

This is a quantum version of the two-bit parity measurement.
When the auxiliary qubit (now in the top register) is found in state

This scheme can be generalised and used to implement any sequence of Pauli measurements.
For example, the circuit below shows to consecutive measurements:

### 7.8.2 The Pauli group of three qubits

Consider the three-qubit Pauli group

The elements of *not* in the stabiliser.
All together, there are

Having pictured the cosets of

### 7.8.3 Half commuting

Any Pauli matrix that is not the identity commutes with exactly half of all the Pauli matrices: namely, with the identity and with itself.
For example,

Extend this observation to any non-identity element in

### 7.8.4 One out of four stabilisers

Explain why, if

### 7.8.5 Stabilisers and projectors

Let

- Show that
\frac{1}{2}(\mathbf{1}\pm G_j) is the projector onto the\pm1 -eigenspace ofG_j for anyj=1,\ldots,r . - Show that the projector
P = \frac{1}{2}(\mathbf{1}+G_1)\frac{1}{2}(\mathbf{1}+G_2)\ldots\frac{1}{2}(\mathbf{1}+G_r) onto the simultaneous+1 -eigenspace of the generatorsG_1,\ldots,G_r can be written asP = \frac{1}{2^r}(S_1+S_2+\ldots+S_{2^r}) where the sum contains all elementsS_i of\mathcal{S} . - The fact that independent generators consecutively bisect the total Hilbert space relies on the fact that, if
G_1 andG_2 are independent generators, thenG_2 restricted to the+1 -eigenspace ofG_1 bisects it into two subspaces of equal dimension. Explain how this fact follows from\operatorname{tr}\left[ \frac{1}{2}(\mathbf{1}+G_1)G_2\frac{1}{2}(\mathbf{1}+G_1) \right] = 0 and prove that this trace is indeed zero. Why do the two generatorsG_1 andG_2 have to be independent?

### 7.8.6 Abelian Pauli quotients

Given any group **commutator**

Now let

**Theorem.** The quotient

Using this theorem (or otherwise),

- Prove that
\mathcal{P}_n/N(\mathcal{S}) is abelian for any Pauli stabiliser\mathcal{S}\leqslant\mathcal{P}_n . - Prove that
\mathcal{P}_n/C_4 is abelian, whereC_4\cong\mathbb{Z}/4\mathbb{Z} is given by the global phase.

### 7.8.7 Equivalent projective measurements

In Section 7.4 we described a generic method for constructing projective measurements of Pauli observables. However, sometimes it may be easier to use equivalent, simpler constructions. For example, the following two circuit identities are often used:

These both follows from the fact that *both* qubits are in state