1
$\begingroup$

I am reading a summary of the stabilizer formalism in this paper, which considers the following encoding of a single qubit in three qubits:

$$ |\overline{0}\rangle=\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)\\ |\overline{1}\rangle=\frac{1}{\sqrt{2}}(|000\rangle-|111\rangle) $$

The author first describes the correction of a bit-flip on the first bit:

Imagine that the first qubit undergoes a bit flip, so a general encoded state $\alpha_0|\overline{0}\rangle+\alpha_1|\overline{1}\rangle$ becomes $\alpha_0 \frac{1}{\sqrt{2}}(|100\rangle+|011\rangle)+\alpha_1 \frac{1}{\sqrt{2}}(|100\rangle-|011\rangle)$. Now, measuring $ZZI$ gives $−1$, and $IZZ$ gives $+1$, for any value of the coefficients $α_0$ and $α_1$.

If I'm not wrong, the action of the syndrome operator $M$ on the post-error state $|\psi\rangle$ is given by $\langle\psi|M|\psi\rangle$, which correctly gives the values,

$$ \langle\psi|ZZI|\psi\rangle = -1 \\ \langle\psi|IZZ|\psi\rangle = +1 $$

They next consider the error $(\cos \theta) \mathrm{I}+i(\sin \theta) \mathrm{X}$ acting on the first qubit:

For simplicity, assuming the encoded state is $|\overline{0}\rangle$; the error operator takes $|\overline{0}\rangle$ to $$ |\psi_e\rangle = \frac{1}{\sqrt{2}}[(\cos \theta|0\rangle+i \sin \theta|1\rangle)|00\rangle+(i \sin \theta|0\rangle+\cos \theta|1\rangle)|11\rangle] $$ Measuring the syndrome operators $ZZI$ and $IZZ$ causes the state to collapse onto the original encoded state $|\overline{0}\rangle$ with probability $\cos ^2 \theta$ or onto the bit-flipped state $|100\rangle+|011\rangle$ with probability $\sin ^2 \theta$. In the former case we get the syndrome $(+1,+1)$, and in the latter $(-1,+1)$.

I am having trouble with this last statement, specifically what we "get" after measuring the syndrome operators. I can see how we get the probabilities after the action of the operators on the state.
$$ ZZI |\psi_e\rangle=\cos \theta\left(\frac{|000\rangle+|111\rangle}{\sqrt{2}}\right)- i \sin \theta\left(\frac{|100\rangle+|011\rangle}{\sqrt{2}}\right)\\ IZZ |\psi_e\rangle=\cos \theta\left(\frac{|000\rangle+|111\rangle}{\sqrt{2}}\right)+ i \sin \theta\left(\frac{|100\rangle+|011\rangle}{\sqrt{2}}\right) $$

However, if we write $|\psi_e\rangle$ as, $$ |\psi_e\rangle=\cos \theta\left(\frac{|000\rangle+|111\rangle}{\sqrt{2}}\right)+ i \sin \theta\left(\frac{|100\rangle+|011\rangle}{\sqrt{2}}\right) $$

then $\langle\psi_e|ZZI|\psi_e\rangle$ does not give a valid syndrome. How are the syndromes calculated in this case? Any help is appreciated.

$\endgroup$
2
  • $\begingroup$ You can add a link to the paper you are talking about, so that we have more context. $\endgroup$ Commented Jul 11 at 2:31
  • $\begingroup$ @DoaPhaseFlip Added a link to the paper. $\endgroup$
    – Enigma
    Commented Jul 11 at 12:12

2 Answers 2

3
$\begingroup$

The inner product as you've written it calculates the expectation value of the measurement of the stabilizer outcome; it does not describe the actual projection associated with measurement. This is no different from the non-error corrected case: for instance, $\langle 0 | X | 0 \rangle = 0$, but measurement outcomes should be $\pm 1$. In a single measurement, you instead obtain $|+\rangle$ or $|-\rangle$ each with probability $1/2$. Likewise, with QECC, your stabilizer measurements project onto a $+1$ or $-1$ eigenspace probabilistically, even if before measurement, your state is not in one of these eigenspaces.

$\endgroup$
2
  • $\begingroup$ Thanks for the expectation value comment; I knew something wasn't right. A few questions: 1) In what context are you talking about obtaining $|+\rangle$ or $|-\rangle$ with prob. $1/2$ 2) If I understand correctly, the probabilistic projection to an eigenspace is for continuous errors right? For a discrete error, will the state after measurement always be an eigenstate? $\endgroup$
    – Enigma
    Commented Jul 11 at 11:54
  • $\begingroup$ If by discrete errors you mean Pauli errors, then yes. The key point is that stabilizer measurements project these "continuous errors" probabilistically into an eigenspace. In practice, you don't have deterministic Pauli errors, because then you would just know about them. $\endgroup$ Commented Jul 11 at 13:38
2
$\begingroup$

When measuring the syndrome operators, the system is projected into the code subspace. You can express the state of the repetition code you are describing is any linear combination of the logical states: $$|\psi\rangle = \alpha_1|\bar{0}\rangle + \alpha_2|\bar{1}\rangle = \frac{\alpha_1}{\sqrt{2}}\left(|000\rangle+|111\rangle\right)+\frac{\alpha_2}{\sqrt{2}}\left(|000\rangle-|111\rangle\right),$$

where the results of the syndrome measurements give both $+1$ in this case: $$ZZI|\psi\rangle = +|\psi\rangle,\quad IZZ|\psi\rangle=+|\psi\rangle.$$

Effectively, if you only measure the syndrome operators, you have no way of distinguishing the logical state yet.

Coming back to the error process you described, the syndrome operators act of $|\psi_e\rangle$ as you wrote. But you will measure the syndrome operators and depending on the outcome, you can infer onto which subspace your errored state was projected.

However, if we write $|\psi_e\rangle$ as, $$ |\psi_e\rangle=\cos \theta\left(\frac{|000\rangle+|111\rangle}{\sqrt{2}}\right)+ i \sin \theta\left(\frac{|100\rangle+|011\rangle}{\sqrt{2}}\right) $$

Starting with this expression of $|\psi_e\rangle$, and recalling that measuring $ZZI$ ($IZZ$) is equivalent to measuring the parity of the first (last) two qubits, we can distinguish two cases:

  • the $ZZI$ measurement is $+1$: the first two qubits are in the same state, the total state must be a linear combination of $|00x\rangle$ and $|11x\rangle$. Your state must then be in $\frac{1}{\sqrt{2}}(|000\rangle+|111\rangle)$.
  • the $ZZI$ measurement is $-1$: the first two qubits are in different states, the total state must be a linear combination of $|01x\rangle$ or $|10x\rangle$. Your state must be in $\frac{1}{\sqrt{2}}(|100\rangle+|011\rangle)$.

From this treatment you can also infer the probability of measuring $\pm1$ from the initial state. It is pretty direct to see that $|\psi_e\rangle$ is projected onto the $+1$ eigenspace with probability $\cos^2{\theta}$ and onto the $-1$ eigenspace with probability $\sin^2{\theta}$.

You can apply the same reasoning to $IZZ$. Note that here the error is only detected by the first syndrome operator. The measurement of $IZZ$ is always $+1$ for the errored state.

$\endgroup$
2
  • $\begingroup$ Thanks! So to be clear, only if we have a discrete error, can we write an expression like $ZZI|\psi\rangle = \pm |\psi\rangle,$ right? Can we write such an expression for the syndrome for the latter error? $\endgroup$
    – Enigma
    Commented Jul 11 at 12:11
  • $\begingroup$ As I mentioned, you need to measure the syndrome operators, the resulting state will always be an eigenstate of your measurement operator, whether your error is discrete or not. You can interpret the error state $|\psi_e\rangle$ as being stabilized by $+ZZI$ and $+IZZ$ with probability $\cos^2\theta$ and being stabilized by $-ZZI$ and $+IZZ$ with probability $\sin^2\theta$. The sign of the stabilizers are given by the result of your measurement. For more details on the stabilizer formalism, I recommend you take a look at this Gottesman paper $\endgroup$ Commented Jul 12 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.