I am reading a paper on Quantum Secret Sharing Quantum Secret Sharing using GHZ states

I have doubts regarding the initial phase of the paper, which are: Let me state what things I read and understood

  1. Alice, Bob, Charlie share a GHZ state $|\psi\rangle=\dfrac{|000\rangle+|111\rangle}{\sqrt{2}}$ and are given one particle each from this state, well this is quite clear to me, proceeding next

  2. They each choose at random whether to measure their particle in the $x$ or $y$ direction They then announce publicly in which direction they have made a measurement, but not the results of their measurements. Half the time, Bob and Charlie, by combining the results of their measurements, can determine what the result of Alice’s measurement was. This allows Alice to establish a joint key with Bob and Charlie, which she can then use to send her message.

  3. The $x$ and $y$ eigenstates are defined as $|+x\rangle=\dfrac{|0\rangle+|1\rangle}{\sqrt{2}}, |-x\rangle=\dfrac{|0\rangle-|1\rangle}{\sqrt{2}}$ , $|+y\rangle=\dfrac{|0\rangle+i|1\rangle}{\sqrt{2}}, |-y\rangle=\dfrac{|0\rangle-i|1\rangle}{\sqrt{2}}$.
    Now my questions start from this point.

Q1. What does it mean to measure in $x$ and $y$ direction?

Q2. The $x$ eigenstates and $y$ eigenstates are just the two different basis of a Hilbert space of dimension $2$ which are actually the eigenvectors of the unitary matrices $\begin{bmatrix} 0 &1\\ 1&0\end{bmatrix}$ and $\begin{bmatrix} 0 &-i\\ i&0\end{bmatrix}$ expressed in ket notation, so any Qubit can be expressed in terms of these two basis is what I understand. But what is the results when say measured in the $x$ basis, and what is the system after that measurement? I have a few more doubts but first i want to understand this from a mathematical point of view. Can somebody help?

Edit: After @DaftWullie's answer

So let's say I want to measure what is the probability that upon measurement of the third qubit I get a $|+\rangle$. So, I calculate $$p_{+}=\left(\dfrac{\langle 000|+\langle 111|}{\sqrt{2}}\right)(I\otimes I\otimes|+x\rangle\langle +x|)\left(\dfrac{ |000\rangle+|111\rangle }{\sqrt{2}}\right).$$ Now I express the third qubit in terms of the Pauli basis to get $|0\rangle=\dfrac{|+x\rangle + |-x\rangle}{2}$ and $|1\rangle=\dfrac{|+x\rangle - |-x\rangle}{2} $ and substitute these values in place of third qubit above to obtain $$ p_{+}=\left(\dfrac{\langle 00|(\langle +x|+\langle -x|)+\langle 11|(\langle +x|-\langle -x|)}{\sqrt{2}\sqrt{2}}\right)(I\otimes I\otimes|+x\rangle\langle +x|)\left(\dfrac{ |00\rangle( |+x\rangle + |-x\rangle) +|11\rangle (|+x\rangle - |-x\rangle) }{\sqrt{2}\sqrt{2}}\right)$$ The left part of my expression evaluates to $$ \dfrac{\langle 00|\left( \langle+x|+x\rangle + \langle-x|+x\rangle\right) + \langle 11|\left( \langle+x|+x\rangle - \langle-x|+x\rangle\right)}{2}$$ $$ = \dfrac{\langle 00|(1+0) +\langle 11|(1-0)}{2}= \dfrac{\langle 00| +\langle 11|}{2}$$ Similarly the right part evaluates to $$\dfrac{|00\rangle+|11\rangle}{2}$$ so i get $$p_+=\left(\dfrac{\langle 00| +\langle 11|}{2}\right)\left( \dfrac{|00\rangle+|11\rangle}{2}\right)=\dfrac{1}{4}+ \dfrac{1}{4}=\dfrac{1}{2}$$

  • $\begingroup$ The first issue is your expression for $|0\rangle$: it's $(|+x\rangle+|-x\rangle)/\sqrt{2}$. This will immediately improve the factors. $\endgroup$
    – DaftWullie
    Commented Jun 14, 2019 at 7:33
  • $\begingroup$ The second issue must be something about how you've simplified the expression for $p_+$, but since you don't give any intermediate steps, it's hard to diagnose. Basically I think you're missing the fact that the inner product gives two terms, one for the 00 on the left matching with the 00 on the right, and another corresponding to 11 pairs. $\endgroup$
    – DaftWullie
    Commented Jun 14, 2019 at 7:37
  • $\begingroup$ Sir, thanx for the pointing out the mistake, i got the answer, but the maths seems messy even for $3$ qubits. $\endgroup$
    – Upstart
    Commented Jun 14, 2019 at 7:54
  • $\begingroup$ You've still not quite got the maths right, even though you're not getting the right answer. The maths is a little messy - it gets exponentially harder with the number of qubits you're using. $\endgroup$
    – DaftWullie
    Commented Jun 14, 2019 at 7:57
  • $\begingroup$ Or, more accurately, you've got the maths good enough, but there's probably an explanatory step missing that holds it all together.... $\endgroup$
    – DaftWullie
    Commented Jun 14, 2019 at 7:58

1 Answer 1


This question is actually entirely about the basics of measurement on a quantum system, and nothing to do with secret sharing.

Let's state the measurement postulate of quantum mechanics as it applies to projective measurements:

A measurement is described by a set of projectors $P_i$ satisfying $\sum_iP_i=1$. If a state $|\psi\rangle$ is being measured, the outcome $i$ is obtained with probability $p_i=\langle\psi|P_i|\psi\rangle$, and if outcome $i$ is obtained, the state after measurement is $P_i|\psi\rangle/\sqrt{p_i}$.

So, all you need to know are the projectors. Giving an operator such as Pauli $X$ is one way of specifying the measurement as an operator decomposes into projectors onto the subspaces of different eigenvalues. e.g. $X=|+x\rangle\langle+x|-|-x\rangle\langle-x|$, so it says use projectors $P_+=|+x\rangle\langle+x|$, $P_-=|-x\rangle\langle-x|$ on a single qubit. If you're measuring multiple qubits, you pad all the other qubits with identity matrices. So a measurement in the $X$ basis of qubit 2 means $1\otimes X\otimes 1$, which has projectors $1\otimes P_+\otimes 1$ and $1\otimes P_-\otimes 1$.

  • $\begingroup$ Sir to measure a quantum state or some specific qubits with respect to certain basis like this one i.e in Pauli basis, do we first have to express those qubits in that basis and then perform the measurement? $\endgroup$
    – Upstart
    Commented Jun 13, 2019 at 11:13
  • $\begingroup$ @Upstart That is one way to do it which often makes the maths quite easy, but it isn't necessary to do it like that. Whatever basis you've got things expressed with respect to, you can just mechanically apply the statement that I made about the measurement postulate. $\endgroup$
    – DaftWullie
    Commented Jun 13, 2019 at 12:13

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