What are the projectors corresponding to the error syndrome measurement performed on a three qubit bit flip code?

Question

This question relates to exercise 10.4 in Nielson and Chuang's textbook "Quantum Computation and Quantum Information".

The question asks us to consider the three qubit bit flip code. It asks us to suppose that we have performed the error syndrome measurement by measuring the 8 orthogonal projectors corresponding to projections onto the eight computational basis states.

I am confused as to what these 8 orthogonal projectors are.

Are they:

$$|000\rangle\langle000| + |111\rangle \langle 111|, \quad |100\rangle\langle 100| + |011\rangle\langle 011|,$$ $$|010\rangle\langle 010| + |101\rangle\langle 101|, \quad |001\rangle\langle 001| + |110\rangle\langle 110|,$$ $$|110\rangle\langle 110| + |001\rangle\langle 001|, \quad |101\rangle\langle 101| + |010\rangle\langle 010|,$$ $$|011\rangle\langle 011| + |100\rangle\langle 100|, \quad |111\rangle\langle 111| + |000\rangle\langle 000|.$$

Or is there a difference between "projectors" and "projection operators"?

Thanks.

Answer 1

1. I don't think there's a difference between "projectors" and "projection operators". Recall "projectors" are defined on page 70, equation 2.35.

2. The projectors you listed aren't what N&C are going for; this can be seen since you have repeats, e.g. $$|000\rangle\langle000| + |111\rangle\langle111| = |111\rangle\langle111| + |000\rangle\langle000|$$ The operators intended "correspond to the eight computational basis states"; that is, use each computational basis state to form an operator. This gives you $$|000\rangle\langle000|, \quad |001\rangle\langle001|, \quad |010\rangle\langle010|, \quad |011\rangle\langle011|,$$ $$|100\rangle\langle100|, \quad |101\rangle\langle101|, \quad |110\rangle\langle110|, \quad |111\rangle\langle111|.$$ You can see these are orthogonal by multiplying them to each other; as an example, $$|010\rangle\langle010||001\rangle\langle001| = |010\rangle \cdot (0) \cdot \langle001| = 0 \cdot |010\rangle \langle001| = 0_{3 \times 3}.$$ This results directly from the computational basis being orthogonal.