On the previous page we saw that some multi-qubit states can be factored into independent single-qubit states (product states) and some cannot (entangled states). But why should you care?
Entanglement is the resource that makes quantum computing and quantum communication fundamentally different from classical. Here's the key insight:
If two qubits are in the entangled state \(\frac{1}{\sqrt{2}}(\ket{00} + \ket{11})\), then measuring one qubit instantly determines the other — even if they're on opposite sides of the universe. Measure the first qubit and get 0? The second qubit is guaranteed to be 0. Get 1? The second is guaranteed to be 1.
This isn't because the qubits "agreed in advance" which value to show. Bell's theorem (1964) proved that no such pre-agreement (local hidden variable theory) can reproduce quantum predictions. The correlations are genuinely nonclassical.
Entanglement is the engine behind:
Recall: a two-qubit state \(\ket{\phi}\) is a product state if \(\ket{\phi} = \ket{\phi_A} \otimes \ket{\phi_B}\) for some single-qubit states. It is entangled if no such factorization exists.
Write the state as \(a\ket{00} + b\ket{01} + c\ket{10} + d\ket{11}\). Then:
$$\text{Entangled} \iff ad \neq bc$$
This is the determinant of the coefficient matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\). If the determinant is nonzero, the state is entangled.
The Bell states (named after physicist John Stewart Bell) are the four maximally entangled two-qubit states. They form an orthonormal basis for \(\mathbb{C}^4\) — meaning any two-qubit state can be written as a linear combination of Bell states. They are the "entangled basis," just as \(\{\ket{00}, \ket{01}, \ket{10}, \ket{11}\}\) is the "computational basis."
$$\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11}) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \end{pmatrix}$$
$$\ket{\Phi^-} = \frac{1}{\sqrt{2}}(\ket{00} - \ket{11}) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix}$$
$$\ket{\Psi^+} = \frac{1}{\sqrt{2}}(\ket{01} + \ket{10}) = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$$
$$\ket{\Psi^-} = \frac{1}{\sqrt{2}}(\ket{01} - \ket{10}) = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ -1 \\ 0 \end{pmatrix}$$
The naming follows a simple pattern:
Use the \(ad \neq bc\) test. For \(\ket{\Phi^+} = \frac{1}{\sqrt{2}}\ket{00} + 0\ket{01} + 0\ket{10} + \frac{1}{\sqrt{2}}\ket{11}\):
$$ad = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \qquad bc = 0 \cdot 0 = 0 \qquad ad \neq bc \implies \text{entangled } \checkmark$$
The same check works for all four Bell states.
Each Bell state has norm 1 (both amplitudes are \(\frac{1}{\sqrt{2}}\), so \(|\frac{1}{\sqrt{2}}|^2 + |\frac{1}{\sqrt{2}}|^2 = 1\)). And they are mutually orthogonal — you can verify by computing inner products. For example:
$$\braket{\Phi^+}{\Phi^-} = \frac{1}{2}(1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 + 1 \cdot (-1)) = \frac{1}{2}(1 - 1) = 0 \quad \checkmark$$
How do you actually make an entangled state? You start with a product state and apply gates that create correlations between the qubits. The standard recipe uses just two gates: Hadamard (H) on the first qubit, then CNOT.
CNOT (Controlled-NOT) is a two-qubit gate. It flips the second qubit if and only if the first qubit is \(\ket{1}\):
$$\text{CNOT}\ket{00} = \ket{00} \qquad \text{CNOT}\ket{01} = \ket{01}$$
$$\text{CNOT}\ket{10} = \ket{11} \qquad \text{CNOT}\ket{11} = \ket{10}$$
Matrix: \(\text{CNOT} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\)
Full treatment of CNOT comes in W4 (multi-qubit gates). For now, just use the table above.
Step 1: Apply H to the first qubit (and I to the second):
$$(H \otimes I)\ket{00} = H\ket{0} \otimes I\ket{0} = \ket{+} \otimes \ket{0} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1}) \otimes \ket{0} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{10})$$
Step 2: Apply CNOT:
$$\text{CNOT}\left(\frac{1}{\sqrt{2}}(\ket{00} + \ket{10})\right) = \frac{1}{\sqrt{2}}(\text{CNOT}\ket{00} + \text{CNOT}\ket{10}) = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11}) = \ket{\Phi^+}$$
H created a superposition on the first qubit: \(\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})\). Then CNOT created a correlation: "if the first qubit is \(\ket{1}\), flip the second." The result is that the second qubit's value is now tied to the first qubit's value. That's entanglement.
Before CNOT: \(\frac{1}{\sqrt{2}}(\ket{00} + \ket{10})\) — this is a product state (\(\ket{+}\ket{0}\)).
After CNOT: \(\frac{1}{\sqrt{2}}(\ket{00} + \ket{11})\) — this is entangled (can't factor it).
The same circuit (H then CNOT) produces all four Bell states depending on the input:
| Input | After H⊗I | After CNOT | Bell State |
|---|---|---|---|
| \(\ket{00}\) | \(\frac{1}{\sqrt{2}}(\ket{00} + \ket{10})\) | \(\frac{1}{\sqrt{2}}(\ket{00} + \ket{11})\) | \(\ket{\Phi^+}\) |
| \(\ket{01}\) | \(\frac{1}{\sqrt{2}}(\ket{01} + \ket{11})\) | \(\frac{1}{\sqrt{2}}(\ket{01} + \ket{10})\) | \(\ket{\Psi^+}\) |
| \(\ket{10}\) | \(\frac{1}{\sqrt{2}}(\ket{00} - \ket{10})\) | \(\frac{1}{\sqrt{2}}(\ket{00} - \ket{11})\) | \(\ket{\Phi^-}\) |
| \(\ket{11}\) | \(\frac{1}{\sqrt{2}}(\ket{01} - \ket{11})\) | \(\frac{1}{\sqrt{2}}(\ket{01} - \ket{10})\) | \(\ket{\Psi^-}\) |
Let's trace through \(\ket{10}\) as a second worked example:
Step 1: \((H \otimes I)\ket{10} = H\ket{1} \otimes \ket{0} = \ket{-}\ket{0} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1})\ket{0} = \frac{1}{\sqrt{2}}(\ket{00} - \ket{10})\)
Step 2: \(\text{CNOT}\left(\frac{1}{\sqrt{2}}(\ket{00} - \ket{10})\right) = \frac{1}{\sqrt{2}}(\ket{00} - \ket{11}) = \ket{\Phi^-}\)
The minus sign came from \(H\ket{1} = \ket{-}\), which has a minus between its terms. This propagated through CNOT to create the \(-\) variant of the Bell state.
Since all four Bell states are related by simple single-qubit operations, you can transform between them using local gates:
Starting from \(\ket{\Phi^+} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})\):
We need to transform \(\frac{1}{\sqrt{2}}(\ket{00} + \ket{11})\) into \(\frac{1}{\sqrt{2}}(\ket{01} - \ket{10})\). Compare term by term:
Try \(Z \otimes X\):
$$(Z \otimes X)\frac{1}{\sqrt{2}}(\ket{00} + \ket{11}) = \frac{1}{\sqrt{2}}(Z\ket{0} \otimes X\ket{0} + Z\ket{1} \otimes X\ket{1})$$
$$= \frac{1}{\sqrt{2}}(\ket{0}\ket{1} + (-\ket{1})\ket{0}) = \frac{1}{\sqrt{2}}(\ket{01} - \ket{10}) = \ket{\Psi^-} \quad\checkmark$$
This used the mixed-product rule: \((Z \otimes X)(\ket{0} \otimes \ket{0}) = Z\ket{0} \otimes X\ket{0}\), applied to each term in the superposition by linearity.
What happens when you measure one qubit of an entangled state? The measurement "collapses" the entire system.
State: \(\ket{\Phi^+} = \frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}\)
Measure the first qubit in the computational basis:
Outcome 0: probability \(|\frac{1}{\sqrt{2}}|^2 = \frac{1}{2}\). Post-measurement state: \(\ket{00}\).
Outcome 1: probability \(|\frac{1}{\sqrt{2}}|^2 = \frac{1}{2}\). Post-measurement state: \(\ket{11}\).
In both cases, measuring one qubit completely determines the other. If the first qubit measured 0, the second qubit is guaranteed to be 0. If the first measured 1, the second is guaranteed to be 1. This is the essence of entanglement — perfect correlation between measurement outcomes.
For the product state \(\ket{+0} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{10})\), measuring the first qubit gives 0 or 1 with equal probability. But the second qubit is \(\ket{0}\) regardless — the measurement didn't tell you anything new about the second qubit. In an entangled state, it does.
Full details of multi-qubit measurement (including partial measurement in arbitrary bases) are on the next page.