May 1 · 8 min read

IBM Research created Entanglion, a board game about quantum computing. The game is open-source, and all of the images and rules are available online for free. The game was designed to familiarize players with basic concepts, such as entanglement and quantum gates. Technical knowledge is not needed to play the game. However, the game board is based on rigorously exact transformations of quantum states.

The creators of the game report that it “took three quantum scientists multiple attempts to perform the math to map out these states and reduce it to the form used in the game board.” The math that they performed has not, to my knowledge, previously been published. The goal of this article is to elucidate the mathematical substructure of the Entanglion game board. Without the underlying math, the game board appears as randomly constructed as Candyland or Chutes and Ladders.

To understand the hidden structure of the game board, we do not need to delve into the minutiae of the game mechanics. We need to know only that the game is for two players, who each control a spaceship placed on the board. The positions occupied by the two spaceships represent the quantum states of two particles. On a player’s turn, a card is played to move one or both spaceships. Each card represents a quantum operator, which transforms one quantum state into another. Thus, the interconnections between the spaces on the board represent the operations required to transform the particles from one state to another. There are only four operators: X (bit flip), CNOT (conditional NOT), H (Hadamard), and SWAP. In quantum computing, these operators are called gates.

One spaceship is blue, and the other is red. The spaceships move around the game board, which has 12 spaces. The eight yellow spaces, which form an octagon, represent entangled states. These spaces must be occupied jointly by both spaceships because the entangled states always involve both particles.

The remaining four spaces represent states of individual particles. The two spaceships move independently of each other across these four spaces. When the spaceships occupy any of these four spaces, the total state is factorable: it can be written as a product of the states of the two particles.

The two purple spaces on the far left are labeled ZERO and ONE, which I will write as |ZERO> and |ONE>. These represent “basis states,” written more compactly as |0> and |1>. If both spaceships are in |ZERO>, the total state is |0(blue)>|0(red)>. If the blue spaceship is in |ZERO> while the red spaceship is in |ONE>, the total state is |0(blue)>|1(red)>.

The two lowest spaces (colored green) represent superpositions of the basis states. Thus, the game board clearly distinguishes three types of states: basis states, superposition states, and entangled states.

# Operators

We now define the four operators that transform the quantum states and move the spaceships around the board. The simplest operator is the bit flip, X. This operator is a quantum NOT gate: It converts |0> to |1> and |1> to |0>. This operator affects only a single particle, so we need to specify which particle is targeted by the operator: X(blue) or X(red). If (i) represents either (blue) or (red), then

X(i)|0(i)> = |1(i)>, (Eq. 1a)

and

X(i)|1(i)> = |0(i)>. (Eq. 1b)

The (i) represents the player who plays the X card; that player’s own spaceship moves as a result. However, if the particles are entangled, X(i) causes both spaceships to move to a space representing a new entangled state.

The next operator is CNOT, conditional NOT. Its acts as a NOT on the target particle only if the other particle is in the |1> state. Just as X(i) indicates that player (i) plays the X card, CNOT(i) indicates that player (i) plays the CNOT card, targeting the (i) spaceship. If i represents the target particle, and j represents the other particle (the “control”),

CNOT(i)(|0,i>|0,j>) = |0,i>|0,j>, (Eq. 2a)

CNOT(i)(|0,i>|1,j>) = |1,i>|1,j>, (Eq. 2b)

CNOT(i)(|1,i>|0,j>) = |1,i>|0,j>, (Eq. 2c)

and

CNOT(i)(|1,i>|1,j>) = |0,i>|1,j>. (Eq. 2d)

X and CNOT are the two operators that connect |ZERO> to |ONE> on the game board. X always moves a spaceship between |ZERO> and |ONE>, as shown in Eq. (1). CNOT moves the spaceship only if the other spaceship occupies |ONE>, as implied in Eq. (2).

To form superpositions of basis states, we need the Hadamard operator, H. H, like X, operates on a single particle and leaves the other one alone. H(i), where (i) as always is either (blue) or (red), is defined as

H(i)|0,i> = (|0,i> + |1,i>), (Eq. 3a)

and

H(i)|1,i> = (|0,i> - |1,i>). (Eq. 3b)

I’ve left out a factor of 1 divided by the square root of 2, which is called a normalization constant. This does not affect the interconnections between the spaces on the game board.

On the game board, H links |ZERO> to |PLUS>. Thus, when H acts on |0>, the resulting state is |PLUS>. Then according to Eq. (3a),

|PLUS,i> = H(i)|0,i> = |0,i> + |1,i>. (Eq. 4)

Similarly, H links |ONE> to |MINUS> on the game board. According to Eq. (3b),

|MINUS,i> = H(i)|1,i> = |0,i> - |1,i>. (Eq. 5)

It may not be obvious that H is its own inverse, which means that when it acts twice in a row on a state, the original state is restored. For example, H(i)H(i)(|0,i>) = H(i)(|0,i> + |1,i>) = H(i)|0,i> + H(i)|1,i> = (|0,i> + |1,i>)+(|0,i> - |1,i>) = 2|0,i>. This is equivalent to state |0,i>; the factor of 2 can be ignored. (It wouldn’t have appeared if we’d used the normalization constant.) The fact that the operators are their own inverses is what allows an operator to move a spaceship either direction between two spaces. If the operators were not their own inverses, the lines connecting two spaces would have to be arrows pointing in only one direction.

All the lines connecting |ONE>, |ZERO>, |PLUS>, and |MINUS> are actually double lines, both blue and red. This indicates that an operator with either subscript (B or R) may effect the state transformation. Among the entangled states, some interconnections exist as a single line only. This indicates that the transition between those states can be accomplished only by an operator targeting a specific particle. For example, to follow a red dashed line that does not have a blue solid line beside it, we must apply an operator with an R subscript. Both spaceships move together when they are entangled, but the operator must be applied to the red state to move both spaceships along a red line.

The final operator is SWAP, which simply interchanges the B and R subscripts.

# Entangled States

The eight spaces arranged as an octagon are identified poetically as the Entanglion galaxy. To enter the Entanglion galaxy, one spaceship must be in |ONE> or |ZERO>, while the other spaceship is in |PLUS> or |MINUS>. From this configuration, the player in |ONE> or |ZERO> must play CNOT. Both spaceships then move to a single space within the Entanglion galaxy. The spaceship from |ONE> or |ZERO> moves to the right, and the spaceship from |PLUS> or |MINUS> moves up, until the two spaceships meet at the point of intersection.

For example, if CNOT is played by the player on |ONE> while the other spaceship is on |PLUS>, both spaceships move to |PSI PLUS>. This means that |PSI PLUS> = CNOT(i)(|1,i>|PLUS,j>). This becomes, through use of Eq. (4), CNOT(i)(|1,i>(|0,j> + |1,j>)), or CNOT(i)(|1,i>|0,j> + |1,i>|1,j>). Finally, applying Eqs. (2c) and (2d),

|PSI PLUS> = |1,i>|0,j> + |0,i>|1,j> = |1,blue>|0,red> + |0,blue>|1,red>. (Eq. 6)

The final expression recognizes that the result is the same whether (i) is (blue) or (red). Similar calculations yield

|PSI MINUS> = CNOT(i)(|1,i>|MINUS,j>) = |1,blue>|0,red> - |0,blue>|1,red>, (Eq. 7)

|PHI PLUS> = CNOT(i)(|0,i>|PLUS,j>) = |0,blue>|0,red> + |1,blue>|1,red>, (Eq. 8)

and

|PHI MINUS> = CNOT(i)(|0,i>|MINUS,j>) = |0,blue>|0,red> - |1,blue>|1,red>. (Eq. 9)

|PSI MINUS> is multiplied by -1 if it’s the red player who played CNOT, but Entanglion effectively ignores multiplication of a total state by -1. In general, we can ignore any factor that multiplies a total state.

Equations (6)-(9) are known as Bell states, named after John Bell, who famously proved that quantum physics is incompatible with common sense.

It may be not be obvious that CNOT precisely reverses itself to disentangle the particles. For example, if the spaceships are in |PSI PLUS> and the blue player plays CNOT, the blue spaceship moves to |ONE> and the red spaceship moves to |PLUS>. To confirm this, we evaluate CNOT(blue)|PSI PLUS> = CNOT(blue)(|1,blue>|0,red> + |0,blue>|1,red>) = |1,blue>|0,red> + |1,blue>|1,red> = |1,blue>|PLUS,red>, returning to a product state and using Eq. (4).

On the game board, the X operator links |PSI PLUS> and |PHI PLUS>. We can confirm this is correct, recalling that X acts on only one particle. If every term with a particular subscript is flipped (|0> to |1>, and |1> to |0>), the state is transformed from |PSI PLUS> to |PHI PLUS>, or from |PHI PLUS> to |PSI PLUS>. This is shown by comparing Eqs. (6) and (8). Similarly, the X operator connects |PSI MINUS> and |PHI MINUS>, as seen in Eqs. (7) and (9).

Only four states remain to be determined: |OMEGA ZERO>, |OMEGA ONE>, |OMEGA TWO>, and |OMEGA THREE>. There are multiple routes to each state. It is instructive to confirm the consistency of all possible paths to a state, though I develop only one path below.

A red dashed line, labeled H, connects |PSI PLUS> and |OMEGA ZERO>. Accordingly, |OMEGA ZERO> results from applying H(red) to |PSI PLUS>. On the game board, we see that H(blue) is not an option; there is no blue solid line connecting these spaces. We find that

|OMEGA ZERO> = H(red)|PSI PLUS> = H(red)(|1,blue>|0,red> + |0,blue>|1,red>)= |1,blue>(|0,red> + |1,red>) + |0,blue>(|0,red> - |1,red>). (Eq. 10)

Continuing clockwise around the board, we follow the red dashed line to arrive in |OMEGA ONE> by application of X(red):

|OMEGA ONE> = X(red)|OMEGA ZERO> = |1,blue>(|1,red> + |0,red>) + |0,blue>(|1,red> - |0,red>). (Eq. 11)

We can find |OMEGA THREE> by applying SWAP to |OMEGA ZERO>:

|OMEGA THREE> = SWAP|OMEGA ZERO> =|1,red>(|0,blue> + |1,blue>) + |0,red>(|0,blue> - |1,blue>). (Eq. 12)

Finally, we can operate on |OMEGA THREE> with X(red) to determine |OMEGA TWO>:

|OMEGA TWO> = X(red)|OMEGA THREE> =|0,red>(|0,blue> + |1,blue>) + |1,red>(|0,blue> - |1,blue>). (Eq. 13)

# Conclusions

The Entanglion game functions on multiple levels. At the most superficial level, it is a fun, cooperative activity. At an intermediate level, it familiarizes players with concepts relevant to a potentially important future technology. This is the level emphasized by the creators of the game, and their marketing targets are secondary schools and the general public. At the deepest level, Entanglion is a rigorous diagram of quantum operators and two-particle states.

Entanglion is even more instructive than its creators advertise. They note the technical accuracy as a small point of pride but do not focus on the resulting benefits. The game board vividly represents mathematical manipulations that are otherwise abstract or dull. These mathematical skills are at the heart of quantum computing and other investigations of quantum entanglement. I believe Entanglion can motivate us even in other areas of physics. Our access to the board’s hidden structure is an analogy to a still deeper truth: physics provides access to the hidden structure of the universe itself.

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## Qiskit

#### A community to discuss Qiskit, programming quantum computers, and anything else related to quantum computing.

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