The GHZ Experiment

A GHZ gate

Consider a coin that can represent the behaviour of a quantum particle. This quantum coin [1] can be placed into an equal superposition of heads and tails, which is represented in Dirac (Ket) notation as follows:

\[(1) \hspace{5 mm}|\psi\rangle = \frac{1}{\sqrt2}(|H\rangle + |T\rangle)\]

\(|\psi\rangle\) (read: psi) represents the state of the quantum coin. \(|H\rangle\) represents the head component and \(|T\rangle\) represents the tails component. 1/√2 is the probability amplitude for each component of the superposition. Squaring this number gives the probability that a particular component will be the result when measured. That is, there is a (1/√2)² = 0.5 (or 50%) probability that heads will be measured. Similarly, in this case, for tails.

Quantum coins can also be entangled. The following state, which represents two entangled quantum coins, is called a Bell state (see also the quantum circuit):

\[(2) \hspace{5 mm}|\psi\rangle = \frac{1}{\sqrt2}(|HT\rangle - |TH\rangle)\]

When the coins are entangled, they are not in independent superpositions, but in a collective superposition. Which is to say, if one of the coins is measured as heads then the other coin, when measured, will be tails.

Quantum coins can be measured from any angle (and are still measured as only heads or tails - there are no 'side of coin' measurements, so-to-speak). To rotate the state in equation (2) from the default z-axis onto the x-axis, a Hadamard gate is applied to each coin (note: a Hadamard gate is conventionally symbolized as \(\boldsymbol H\), not to be confused with the ket \(|H\rangle\) representing heads in this post) . The Hadamard gate rotates the \(|H\rangle\) component to \(\frac{1}{\sqrt2}(|H\rangle + |T\rangle)\) and the \(|T\rangle\) component to \(\frac{1}{\sqrt2}(|H\rangle - |T\rangle)\).[2] This results in the following state (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H)}(|HT\rangle - |TH\rangle) \\ &= (|H\rangle + |T\rangle)(|H\rangle - |T\rangle) - ((|H\rangle - |T\rangle)(|H\rangle + |T\rangle) \\ &= |HH\rangle - |HT\rangle + |TH\rangle - |TT\rangle - (|HH\rangle + |HT\rangle - |TH\rangle - |TT\rangle) \\ &= |HH\rangle - |HT\rangle + |TH\rangle - |TT\rangle - |HH\rangle - |HT\rangle + |TH\rangle + |TT\rangle \\ &= -(|HT\rangle - |TH\rangle)\end{aligned}\]

This is the same Bell state as in equation (2) except that a global phase (the minus sign, which is equivalent to π radians, or a 180° rotation around the axis origin) has been added. This global phase makes no difference to the calculated probabilities. A similar result occurs if the state is rotated onto the y-axis, or onto any arbitrary axis. In other words, that Bell state (called the singlet state) is rotationally invariant, ignoring the global phase which has no physical meaning. The rotation onto the y-axis uses the Phase gate (conventionally symbolized as \(\boldsymbol S\)) and Hadamard gate (conventionally symbolized as \(\boldsymbol H\), as noted above) to rotate the \(|H\rangle\) component to \(\frac{1}{\sqrt2}(|H\rangle + |T\rangle)\) and the \(|T\rangle\) component to \(\frac{i}{\sqrt2}(|H\rangle - |T\rangle)\).[3] This results in the following state (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H)(S \otimes S)}(|HT\rangle-|TH\rangle) \\ &= (|H\rangle + |T\rangle)(i|H\rangle - i|T\rangle) - ((i|H\rangle - i|T\rangle)(|H\rangle + |T\rangle) \\ &= i|HH\rangle - i|HT\rangle + i|TH\rangle + -i|TT\rangle - (i|HH\rangle + i|HT\rangle + -i|TH\rangle + -i|TT\rangle) \\ &= i|HH\rangle - i|HT\rangle + i|TH\rangle + -i|TT\rangle - i|HH\rangle - i|HT\rangle + i|TH\rangle + i|TT\rangle \\ &= -i(|HT\rangle - |TH\rangle)\end{aligned}\]

Again, this is the same Bell state as in equation (2) except that a global phase (the imaginary −i, which is equivalent to 3*π/2 radians, or a 270° rotation around the axis origin) has been added.

Now how does the entanglement work? For example, if Alice and Bob take one of two quantum coins entangled as per equation (2) to distant locations, and measure their coin on a predetermined axis (say, the y-axis), how do they end up being measured in opposite orientations? This is the question that the EPR paradox raises. Einstein's answer was that there must be hidden "elements of reality" corresponding to the coin's orientation on any given axis. I discuss Bell's Theorem which demonstrates the constraints on that position - namely, that hidden variables can't be used to explain the measurements without also violating locality (i.e., cause/effect not greater than the speed of light). In this post, I want to outline an intuitive demonstration of those constraints via the GHZ experiment.

To begin, a three qubit entangled state is created, termed the GHZ state, as follows:

Figure 1: The GHZ state

In terms of our quantum coins, this can be represented as:

\[(3) \hspace{5 mm}|\psi\rangle = \frac{1}{\sqrt2}(|HHH\rangle + |TTT\rangle)\]

If Alice, Bob and Charlie each take one of the three quantum coins, measure it, and then compare results, they will find that they have all measured heads, or have all have measured tails.

Figure 2: xxx basis

Now suppose that Alice, Bob and Charlie instead rotate their respective coins onto the x-axis.[4] The resulting state is:

\(\begin{aligned}(4) \hspace{5 mm}|\psi\rangle = \frac{1}{2}(&|HHH\rangle + |HTT\rangle + \\ & |THT\rangle + |TTH\rangle)\end{aligned}\)

If we assign heads the value of 1 and tails the value of -1 then, for each component, the product of the three coin measurements will be 1. For example, \(|HTT\rangle\) gives 1 * -1 * -1 = 1. Similarly for the other components. This is a mathematical representation of the fact that for each of the four measurement possibilities, there will be an odd number of heads measured (1 or 3) and an even number of tails measured (0 or 2).

Figure 3: yyx basis

Now suppose that only Alice rotates her coin onto the x-axis, and instead Bob and Charlie rotate their coins onto the y-axis.[5] The resulting state is:

\(\begin{aligned}(5) \hspace{5 mm}|\psi\rangle = \frac{1}{2}(&|HHT\rangle + |HTH\rangle + \\ & |THH\rangle + |TTT\rangle)\end{aligned}\)

In this case, if we assign heads the value of 1 and tails the value of -1 then, for each component, the product of the three coin measurements will be -1. For example, \(|HHT\rangle\) gives 1 * 1 * -1 = -1. Similarly for the other components. This is a mathematical representation of the fact that for each of the four measurement possibilities, there will be an even number of heads measured (0 or 2) and an odd number of tails measured (1 or 3). Note that this is the opposite of the possible outcomes for state (4).

Symmetrically, if it is only Bob or Charlie, rather than Alice that rotates their coin onto the x-axis, while the others instead rotate their coins onto the y-axis, the resulting state is the same as equation (5). We can now list the product equations for each of the quantum states:

\((6) \hspace{5 mm}Alice_x * Bob_x * Charlie_x = 1 \hspace{5 mm}[from (4)]\)

\((7) \hspace{5 mm}Alice_x * Bob_y * Charlie_y = -1 \hspace{5 mm}[from (5)]\)

\((8) \hspace{5 mm}Alice_y * Bob_x * Charlie_y = -1 \hspace{5 mm}[from (5)]\)

\((9) \hspace{5 mm}Alice_y * Bob_y * Charlie_x = -1 \hspace{5 mm}[from (5)]\)

Now the product of the left-sides of the above equations is:

\(\begin{aligned}LHS = & Alice_x * Bob_x * Charlie_x * \\ & Alice_x * Bob_y * Charlie_y * \\ & Alice_y * Bob_x * Charlie_y * \\ & Alice_y * Bob_y * Charlie_x\end{aligned}\)

As each element appears twice, the equation can be rearranged as:

\(LHS = {Alice_x}^2 * {Bob_x}^2 * {Charlie_x}^2 * {Alice_y}^2 * {Bob_y}^2 * {Charlie_y}^2\)

Since LHS is the product of squares, LHS = 1. But the product of the RHS is:

\(RHS = 1 * -1 * -1 * -1 = -1\)

Thus the equations can't be jointly satisfied. This proves that there can't be hidden elements of reality (i.e., unknown predefined values) for each possible measurement combination since that would lead to contradiction. That is, quantum mechanics experimentally rules out local hidden variable theories.[6]

--

[1] Our quantum coin is a qubit, where \(|0\rangle\) and \(|1\rangle\) have been replaced by \(|H\rangle\) and \(|T\rangle\).

[2] See the Bloch Sphere in Figure 5 of this earlier blog post to visualize the equivalent qubit rotation.

[3] Derivation for the matrix product of \(\boldsymbol H\) and \(\boldsymbol S\):

\[\hspace{10 mm}\boldsymbol{HS} = \begin{bmatrix} 1 & 1 \\ 1 & −1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 1*1 + 0*1 & 0*1 + i*1 \\ 1*1 + 0*−1 & 0*1 + i*−1 \end{bmatrix} = \begin{bmatrix} 1 & i \\ 1 & −i \end{bmatrix}\]

[4] Derivation for Alice, Bob, and Charles rotating their coins from the z-axis onto the x-axis (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H \otimes H)}(|HHH\rangle + |TTT\rangle) \\ & = (|H\rangle + |T\rangle)(|H\rangle + |T\rangle))(|H\rangle + |T\rangle) + ((|H\rangle - |T\rangle)(|H\rangle - |T\rangle)(|H\rangle - |T\rangle)) \\ & = |HHH\rangle + |HHT\rangle + |HTH\rangle + |HTT\rangle + |THH\rangle + |THT\rangle + |TTH\rangle + |TTT\rangle + \\ & \hspace{5 mm} |HHH\rangle - |HHT\rangle - |HTH\rangle + |HTT\rangle - |THH\rangle + |THT\rangle + |TTH\rangle - |TTT\rangle \\ & = |HHH\rangle + |HTT\rangle + |THT\rangle + |TTH\rangle\end{aligned}\]

[5] Derivation for Alice, Bob, and Charles rotating their coins from the z-axis onto the y-axis, y-axis, and x-axis respectively (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H \otimes H)(S \otimes S\otimes I)}(|HHH\rangle + |TTT\rangle) \\ & = (|H\rangle + |T\rangle)(|H\rangle + |T\rangle))(|H\rangle + |T\rangle) + \\ & \hspace{5 mm} ((i|H\rangle + -i|T\rangle)(i|H\rangle + -i|T\rangle)(|H\rangle - |T\rangle)) \\ & = |HHH\rangle + |HHT\rangle + |HTH\rangle + |HTT\rangle + \\ & \hspace{5 mm}|THH\rangle + |THT\rangle + |TTH\rangle + |TTT\rangle + \\ & \hspace{5 mm} ii|HHH\rangle + -ii|HHT\rangle + -ii|HTH\rangle + --ii|HTT\rangle + \\ & \hspace{5 mm} -ii|THH\rangle + --ii|THT\rangle + --ii|TTH\rangle + ---ii|TTT\rangle \\ & = |HHH\rangle + |HHT\rangle + |HTH\rangle + |HTT\rangle + \\ & \hspace{5 mm} |THH\rangle + |THT\rangle + |TTH\rangle + |TTT\rangle + \\ & \hspace{5 mm} -|HHH\rangle + |HHT\rangle + |HTH\rangle + -|HTT\rangle + \\ & \hspace{5 mm} |THH\rangle + -|THT\rangle + -|TTH\rangle + |TTT\rangle \\ & = |HHT\rangle + |HTH\rangle + |THH\rangle + |TTT\rangle\end{aligned}\]

[6] Note that quantum mechanics rules out predefined measurement values for non-local hidden variable theories as well. However, if a theory is non-local then, once a measurement has been made for one coin, non-local variables can potentially be updated instantaneously thus potentially affecting subsequent measurements of the other coins.