# Error syndromes and recovery procedure in bit flip code

This question relates to exercise 10.4 in Nielsen and Chuang.

For syndrome diagnosis, the textbook provides an example where one has four projectors, by which, you can identify where a one qubit error has occurred. In this scheme, the syndrome measurement does not cause any change to the state.

The exercise asks you to write down the projectors for the eight computational basis states, which is easy enough. $$|0/1 \,\, 0/1 \,\, 0/1 \rangle \langle0/1 \,\, 0/1 \,\, 0/1 |$$.

I think the point of the question is for one to say that when you project these onto your one qubit corrupted state, you can tell where the one qubit error has occurred. However, I thought projective measurement needs to be repeated multiple times to be able to reconstruct your full state (for example, if my state is $$\alpha |000\rangle + \beta|111\rangle$$, in order for me to actually find out what $$\alpha$$ and $$\beta$$ are, I need to repeat projection of the state on to $$|000\rangle$$ and $$|111\rangle$$ multiple times to form a statistic.

So, the bottom line is, I don't see how projections on to the eight computational basis states can be used to diagnose the error syndrome. Since projecting your state onto any of the eight computational basis states will mess up your state.

(Even before getting to the question above, how does one perform the syndrome measurement without causing any change to the state? On page 428, it's claimed that if the corrupted state is $$\alpha |100\rangle + \beta|011\rangle$$, it remains $$\alpha |100\rangle + \beta|011\rangle$$ even after the syndrome measurement. I'm having a hard time wrapping my head around the idea of constructing such syndrome projectors in terms of designing an actual experiment.)

You quoted the first part of the exercise; your question is exactly the second part of the exercise :-)

Exercise 10.4 (2) asks you to show that the described recovery procedure (the one which projects the state on one of the 8 basis states) works only for computational basis states.

Indeed, if you have a corrupted state $$\alpha |100\rangle + \beta|011\rangle$$, projecting it onto one of the basis states will yield $$|100\rangle$$ or $$|011\rangle$$, telling you that there was indeed an error on the first qubit, but at the same time destroying the superposition and ruining the calculation. This is why you have to do a smarter measurement than that to preserve superposition.

For the second part of the question, for me it is easier to think about these measurements in terms of measuring $$Z_1Z_2$$ and $$Z_2Z_3$$ observables (see page 430 and exercise 10.3). To measure an observable $$Z_1Z_2$$, you can do the following:

• allocate an extra qubit $$a$$ in state $$|0\rangle$$,
• do a CNOT with $$q1$$ as control and $$a$$ as target,
• do a CNOT with $$q2$$ as control and $$a$$ as target,
• measure $$a$$ in computational basis.

If you do the math, you'll see that the measurement will give 1 if qubits 1 and 2 have different parity and 0 if they have the same parity (which is exactly what you're looking for) without destroying the superposition (since in each term of the superposition the parity of qubits 1 and 2 is the same, even if qubit states differ).

• Thank you so much! On page 428 (this is about the last bit of my post in parentheses), it says the syndrome measurement (acting any of P0, P1, P2, and P3 I assume) does not cause any change to the state; you get the same corrupted state back). Do you see how you can design a physical experiment that performs such a measurement? – Blackwidow Jun 8 '19 at 16:22
• Off the top of my head it seems to be the easiest to show that those measurements are equivalent to parity measurements and proceed with parity measurements. I could try to come up with a measurement for the given projectors, but that will take me a bit of time... – Mariia Mykhailova Jun 8 '19 at 17:33