Multi-Qubit Systems¶
Two qubits have four joint outcomes, and those outcomes can break a rule that independent qubits follow.
From One Qubit to Two¶
On the previous page, we could move one qubit's state anywhere and undo the move. We ended with two qubits side by side, each with its own state and nothing linking them.
What changes when there is more than one qubit?
Module 01 already showed a pair of qubits whose results were random alone and yet always agreed, and the CHSH game that tests such pairs. This page does not repeat the game. It works out the same pair from the joint outcomes, and shows how it breaks a rule that independent qubits follow.
Joint Outcomes¶
Two qubits, each read as 0 or 1, give four joint outcomes: 00, 01, 10, and 11.
Order convention
The first qubit, q₀, is written first, on the left. It is also the control of the CNOT gate below, and the top wire in a circuit. The left digit of an outcome belongs to q₀ and the right digit to q₁. Some software libraries write bit strings with qubit 0 on the right, so check the documentation of any tool before reading a bit string from it.
Why does 01 not mean the second qubit first?
On this page the left digit is q₀ and the right digit is q₁. Another tool may write them the other way round.
A joint probability table gives the probability of each joint outcome. Here q₀ is in |+⟩ and q₁ is in |0⟩. Each qubit's own probabilities are read off the table by adding along a row or a column:
The row totals are q₀'s own probabilities and the column totals are q₁'s.
Independent Qubits¶
Two qubits are independent when neither one's outcomes depend on the other. For independent qubits the joint probability is the product of the two separate probabilities:
Two examples. In the first, q₀ is in |+⟩ and q₁ is in |0⟩. In the second, both are in |+⟩:
The rule holds for independent qubits. We will use it as a test, and only in one direction: a joint table that breaks the rule cannot belong to independent qubits.
CNOT: A Conditional Flip¶
CNOT is a gate on two qubits. One qubit is the control and the other is the target:
- Control — the qubit whose value decides whether the flip happens. Here, q₀.
- Target — the qubit that is flipped when the control is 1. Here, q₁.
Its symbol is a dot on the control wire joined to a ⊕ on the target wire. Read the table down: when the control is 0, nothing happens, and when the control is 1, the target flips.
What if the control is in a superposition? CNOT acts on each part separately. The part of the state with the control at 0 is left alone, and the part with the control at 1 has its target flipped. Like every gate, CNOT can be undone.
Does CNOT look at the control?
No. It flips the target on the part of the state where the control is 1, and leaves the rest alone. It does not copy the control, and it does not read it.
Two-Qubit Explorer¶
Choose what each qubit starts in. Then switch on an H gate on q₀, and a CNOT gate after it, and watch the joint table. Switch on the product test to compare the table with what independent qubits would give.
1. Choose what each qubit starts in
q₀ starts in
q₁ starts in
2. Choose the steps (H first, then CNOT)
3. The steps
4. The joint outcomes
Intermediate readouts
5. The product test
The circuit, the bars, the amplitudes, and each qubit's own probabilities each sit in their own place. The bars are what the state would give, not a measurement result.
The Bell State¶
Start both qubits in |0⟩. Switch on H on q₀, and then CNOT. Here is the table, put through the product test next to the table for |+⟩ and |+⟩:
The pair that H then CNOT produces from |00⟩ is called the Bell state. It has two outcomes: 00 and 11, each 50% of the time. 01 and 10 never happen.
Now apply the product test. Each qubit alone is 50/50, so independent qubits would give 25% for every outcome. The Bell state gives 50%, 0%, 0%, and 50%. No pair of independent qubits can give this table.
Compare |+⟩ and |+⟩. Each qubit alone is also 50/50, and the joint table does match the product.
Both pairs have a 50/50 qubit on each side. What is different about the Bell state?
Its joint outcomes always agree, so the two qubits are not independent. Entanglement is not two independent superpositions.
Neither qubit of the Bell state has a state of its own to draw on the sphere. That is the limit the Bloch sphere page named: the picture works for one qubit. The pair has a state, but neither qubit alone does. And this is the pair from Module 01, random alone and always agreeing, now as a table.
Check your understanding. First, predict before the bars appear:
Then decide which tables could come from two independent qubits:
Now try one yourself, using the explorer above:
Make 01 and 10 each happen 50% of the time, with 00 and 11 never. Choose the starting states and steps, then check.
Beyond the Core¶
Tensor products and four-component vectors¶
The state of two qubits is four numbers, one per joint outcome:
The product of two states. If q₀ is in \(a_0|0\rangle + a_1|1\rangle\) and q₁ is in \(b_0|0\rangle + b_1|1\rangle\), the pair is their tensor product:
in the order 00, 01, 10, 11. The first factor is q₀. For example,
H on q₀. A gate on one qubit becomes a \(4\times 4\) matrix on the pair. H on q₀ is \(H\otimes I\), with the first factor for q₀:
Applied to \(|00\rangle = (1, 0, 0, 0)\), it gives \(\frac{1}{\sqrt 2}(1, 0, 1, 0)\), which is \(|+\rangle\otimes|0\rangle\) again. A circuit reads left to right, and a matrix product reads right to left, as on the previous page.
The explorer's matrix panel now shows this for each step you switch on.
CNOT as a matrix, the Bell state, and separability¶
In the order 00, 01, 10, 11, with q₀ as the control, CNOT is
On a general state, it exchanges the amplitudes of 10 and 11 and leaves 00 and 01 alone:
Deriving the Bell state. Start from \(|00\rangle\). H on q₀ gives \(\frac{|00\rangle + |10\rangle}{\sqrt 2}\). CNOT exchanges the amplitudes of 10 and 11, so the \(|10\rangle\) term becomes \(|11\rangle\):
The matrices are written right to left: \(H\otimes I\) acts first.
Separability. A two-qubit state is a product of two one-qubit states exactly when
For a product, \(\alpha\delta = a_0b_0\,a_1b_1 = a_0b_1\,a_1b_0 = \beta\gamma\). The explorer now shows \(\alpha\delta - \beta\gamma\). Three worked cases:
- The Bell state: \(\alpha = \delta = 1/\sqrt 2\) and \(\beta = \gamma = 0\), so \(\alpha\delta = \tfrac12\) and \(\beta\gamma = 0\). They differ, so it is entangled.
- \(|+\rangle\otimes|+\rangle\): every amplitude is \(\tfrac12\), so \(\alpha\delta = \beta\gamma = \tfrac14\). It is a product.
- The state with amplitudes \(\tfrac12, \tfrac12, \tfrac12, -\tfrac12\): it gives 25% for every outcome, so it passes the product test of the probabilities, yet \(\alpha\delta = -\tfrac14\) and \(\beta\gamma = \tfrac14\). It is entangled. A table that matches the product does not by itself prove the qubits are independent.
Naming the formalism¶
Any pure state of two qubits can be written as a sum of at most two product terms, with weights called Schmidt coefficients: the Schmidt decomposition. Numbers such as the entanglement entropy turn those weights into a measure of how entangled a state is. And CNOT is one case of a general controlled-U gate, which applies any one-qubit gate U to the target when the control is 1. Later material makes these ideas precise.
From Two Qubits to Circuits¶
Compare what you can do now with what is still hard.
Here are the Bell-state steps, drawn from the explorer:
The steps are described in words and tables. We need a standard way to draw them that anyone can read.
Takeaway¶
Two qubits have four joint outcomes. Independent qubits multiply their probabilities, and CNOT after H gives a pair whose outcomes do not, so entanglement is not two independent superpositions.
How can we write these steps down as a diagram?
Next: Quantum Circuits →