Lecture 3.2 — Entanglement and the Bell States

Where We Left Off¶

Last lecture we built the machinery for two-qubit states: the tensor product $\otimes$, the amplitude table $C_{ij}$, and the $6 = 4 + 2$ DOF counting that told us correlations must exist.

Today we make that concrete. We’ll learn to test whether a state is entangled, meet the four Bell states — the maximally entangled two-qubit states that form the backbone of quantum information — and review the measurement formalism we’ll need to analyze them.

Part 1: Review and Practice¶

The General Two-Qubit State¶

The most general state of two qubits is:

Four complex amplitudes, satisfying $\sum|C_{ij}|^2 = 1$.

Independent Qubits: The Tensor Product¶

If qubit A is in state $|\psi\rangle = a|0\rangle + b|1\rangle$ and qubit B is in state $|\phi\rangle = c|0\rangle + d|1\rangle$, and they are independent, the joint state is:

The amplitudes are products: $\Psi_{ij} = u_i v_j$. This is the quantum version of $P_{AB}(i,j) = P_A(i) \cdot P_B(j)$ — independence means the joint state factors.

The answer is (C). With $a = 1$, $b = -1$, $c = 1$, $d = 1$:

The pattern makes sense: qubit A contributes a minus sign whenever it’s in state $|1\rangle$, because $b = -1$.

Part 2: Entanglement¶

Can Every State Be Factored?¶

Consider the state:

Can we write this as $|\psi\rangle \otimes |\phi\rangle = (a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle)$?

Let’s try. Matching coefficients to $ac|00\rangle + ad|01\rangle + bc|10\rangle + bd|11\rangle$:

From $ad = 0$: either $a = 0$ or $d = 0$. From $bc = 0$: either $b = 0$ or $c = 0$.

But if $a = 0$, then $ac = 0 \neq \frac{1}{\sqrt{2}}$. If $d = 0$, then $bd = 0 \neq \frac{1}{\sqrt{2}}$. Same problem with $b = 0$ or $c = 0$.

No assignment of $a, b, c, d$ works. This state cannot be written as a tensor product.

This means three things — all equivalent ways of saying the same thing:

1. The state cannot be written as a tensor product of single-qubit states.

2. The two qubits are not independent — they are correlated.

3. The state is entangled.

The Determinant Test¶

Trying to factor states by hand is tedious. We need a systematic test.

Recall from last lecture that we can arrange the four amplitudes of any two-qubit state into a $2 \times 2$ amplitude table:

For a product state, this matrix is an outer product of two vectors:

An outer product of two vectors is a rank-1 matrix — its rows (or columns) are proportional to each other. Rank-1 matrices have a simple property: their determinant is zero.

This gives us a clean test:

Why does this work? A $2 \times 2$ matrix can be written as an outer product of two vectors if and only if it has rank 1, which for a $2\times 2$ matrix is equivalent to having determinant zero. The determinant measures exactly the “entanglement content” — how far the amplitude table is from being factorable.

Practice: Entangled or Not?¶

Build the amplitude table and compute the determinant:

Not entangled. We can factor it:

(up to normalization). Both qubits are independently in the $|+\rangle$ state.

Not entangled. The factorization is:

Both qubits are in the $|-\rangle$ state.

Now try $|00\rangle + |11\rangle$:

Entangled. The amplitude table is the identity matrix — it has rank 2, not rank 1. No choice of single-qubit states can produce it. This is the state we tried (and failed) to factor by hand earlier.

The pattern is becoming clear: states where both diagonal entries are nonzero but the off-diagonal entries are zero (or vice versa) tend to be entangled — the two qubits are “doing different things together” in a way that can’t be separated.

Part 3: The Bell States¶

We’ve seen that $\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ is entangled. In fact, it’s one of four special entangled states that form a complete orthonormal basis for the two-qubit Hilbert space.

The Four Bell States¶

The naming convention: $\Phi$ states have same-outcome correlations ($|00\rangle$ and $|11\rangle$), $\Psi$ states have opposite-outcome correlations ($|01\rangle$ and $|10\rangle$). The $+/-$ superscript is the relative phase between the two terms.

$|\Psi^-\rangle$ is called the singlet state — it will play a starring role when we get to Bell’s theorem.

Properties¶

All four are maximally entangled. You can verify: each has an amplitude table with $|\det(C)| = 1/2$ (the maximum possible for a normalized state).

For example:

They form an orthonormal basis. Any two-qubit state can be expanded in the Bell basis instead of the computational basis $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$. The four Bell states are mutually orthogonal:

You can verify the first: $\langle\Phi^+|\Phi^-\rangle = \frac{1}{2}(\langle 00| + \langle 11|)(|00\rangle - |11\rangle) = \frac{1}{2}(1 - 1) = 0$.

Two complete bases for the same space:

The computational basis is made of product states — no correlations. The Bell basis is made entirely of maximally entangled states. Both are perfectly valid bases for the 4-dimensional two-qubit Hilbert space.

Part 4: Measurement Formalism¶

Before we can analyze what happens when you measure entangled states, we need to sharpen our tools. Let’s review the measurement formalism, first for a single qubit.

A Note on Labels and Eigenvalues¶

The computational basis labels $|0\rangle$ and $|1\rangle$ are names, not eigenvalues. The Pauli $Z$ operator has eigenvalues +1 and -1:

So $|0\rangle$ means “spin up” (eigenvalue +1) and $|1\rangle$ means “spin down” (eigenvalue -1). The X and Y eigenstates follow the same convention:

Throughout this course, measurement outcomes are always $\pm 1$, and basis states are labeled by their eigenvalue sign. When we get to Bell’s theorem, Alice measuring Z and getting $|0\rangle$ will correspond to outcome $A = +1$.

Projective Measurement: Single Qubit¶

Suppose we have a qubit in state $|\Psi\rangle = \alpha|0\rangle + \beta|1\rangle$ and we measure in the Z-basis.

Step 1: Build the projector. For the outcome $|0\rangle$ (eigenvalue +1), the projector is $|0\rangle\langle 0|$. As a matrix:

Step 2: Apply the projector. The (unnormalized) post-measurement state is:

The projector “kills” the $|1\rangle$ component and keeps the $|0\rangle$ component.

Step 3: Find the probability. The probability of this outcome is the squared norm of the projected state:

This is just Born’s rule: the probability equals the squared magnitude of the amplitude for that outcome. (We used the projector properties $(|0\rangle\langle 0|)^\dagger = |0\rangle\langle 0|$ and $(|0\rangle\langle 0|)^2 = |0\rangle\langle 0|$.)

Step 4: Normalize. The state after measurement is:

After measuring and getting $|0\rangle$, the qubit is in state $|0\rangle$ (up to global phase). The superposition has collapsed.

Measuring in a Different Basis¶

Now measure the same qubit in the X-basis instead. The eigenstates of $X$ are:

The projector onto $|+\rangle$ is:

Example: Suppose our qubit is already in $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. What happens when we measure X?

Probability: $|\langle+|+\rangle|^2 = 1$. The measurement gives $|+\rangle$ (eigenvalue +1) with certainty, and the state is unchanged. This makes sense: $|+\rangle$ is an eigenstate of X, so measuring X reveals its eigenvalue without disturbing it.

Contrast with Z-measurement: If we instead measured Z on the same state $|+\rangle$:

Now the outcome is random — 50/50. Measuring in a compatible basis (X) gives a definite result; measuring in an incompatible basis (Z) gives a random one. The state $|+\rangle$ has a definite X-value but no definite Z-value.

This isn’t an accident — it’s enforced by the mathematics. Let’s see why.

Why Incompatible Bases Give Random Outcomes¶

Why does $|0\rangle$ (definite Z) give completely random results when measured in X? Let’s see it two ways.

The quick way: expand in the other basis. Write $|0\rangle$ in the X-eigenbasis:

It’s an equal-weight superposition of $|+\rangle$ and $|-\rangle$. So a measurement of X gives +1 or -1 with equal probability $1/2$. The outcome is maximally uncertain — as random as a coin flip.

The same holds for Y. In the Y-eigenbasis:

(up to a relative phase that doesn’t affect probabilities). Again, 50/50.

The systematic way: use $\sigma_i^2 = I$. For any Pauli matrix, $\sigma_i^2 = I$, so $\langle\sigma_i^2\rangle = 1$ in any state. The variance is:

Maximum uncertainty is $\Delta\sigma_i = 1$, which happens when $\langle\sigma_i\rangle = 0$ — outcomes equally likely, completely random. In a Z-eigenstate:

Both orthogonal components are individually maximally uncertain. And the same argument holds cyclically: an eigenstate of X is maximally uncertain in Y and Z, and so on.

The Uncertainty Principle¶

The deeper reason is structural. The spin operators satisfy the commutation relations:

where $S_i = \frac{\hbar}{2}\sigma_i$. (In terms of the dimensionless Pauli matrices: $[\sigma_x, \sigma_y] = 2i\sigma_z$, etc.)

The generalized uncertainty relation says that for any two observables $A$ and $B$:

where $\Delta A = \sqrt{\langle A^2 \rangle - \langle A \rangle^2}$ is the standard deviation.

Applied to spin in state $|0\rangle$ (an eigenstate of $S_z$ with $\langle S_z\rangle = +\hbar/2$):

We already computed $\Delta S_x = \Delta S_y = \hbar/2$, so $\Delta S_x \cdot \Delta S_y = \hbar^2/4$. The bound is saturated. The uncertainty isn’t just bounded from below — it’s as large as the algebra allows.

This also explains the “$6 = 4 + 2$” DOF counting from last lecture in a new light. A single qubit on the Bloch sphere has a definite value along one axis, but that forces uncertainty along the other two. You can’t have definite values for all three components simultaneously — the information budget is 2 real numbers (one axis direction), not 6.

Next lecture, we’ll extend the measurement formalism to two qubits — measuring just one qubit of an entangled pair — and discover that the choice of measurement basis for one qubit determines what happens to the other.

Summary¶

1. Entanglement means a two-qubit state cannot be written as $|\psi\rangle_A \otimes |\phi\rangle_B$. Equivalently: the qubits are not independent, and the amplitude table $C$ is not rank 1.

2. The determinant test: arrange amplitudes into a $2\times 2$ matrix. If $\det(C) = 0$, the state is a product (not entangled). If $\det(C) \neq 0$, the state is entangled. The determinant measures how far the state is from being factorable.

3. The four Bell states $\{|\Phi^\pm\rangle, |\Psi^\pm\rangle\}$ are maximally entangled and form an orthonormal basis — an alternative to the computational basis for two qubits. The singlet $|\Psi^-\rangle$ will be central to Bell’s theorem.

4. Projective measurement uses projectors $|m\rangle\langle m|$ to compute probabilities ($p_m = |\langle m|\Psi\rangle|^2$) and post-measurement states ($|m\rangle\langle m|\Psi\rangle / \sqrt{p_m}$). Measuring in a compatible basis gives a definite result; measuring in an incompatible basis gives a random one.

5. The uncertainty principle for spin-$\frac{1}{2}$: an eigenstate of $S_x$, $S_y$, or $S_z$ is definite along one axis and maximally uncertain along the other two (50/50 outcomes). This follows from $[S_x, S_y] = i\hbar S_z$ and the generalized uncertainty relation $\Delta A \cdot \Delta B \geq \frac{1}{2}|\langle[A,B]\rangle|$.

Homework¶

Problem 1: Bell State Verification¶

(a) Write the amplitude table for each of the four Bell states and verify that $|\det(C)| = 1/2$ for all of them.

(b) Verify orthogonality: compute $\langle\Phi^+|\Phi^-\rangle$, $\langle\Phi^+|\Psi^+\rangle$, and $\langle\Psi^+|\Psi^-\rangle$.

(c) Show that $|\Phi^+\rangle$ can be written as: $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|{+}{+}\rangle + |{-}{-}\rangle)$, where $|{+}{+}\rangle = |+\rangle \otimes |+\rangle$ and $|{-}{-}\rangle = |-\rangle \otimes |-\rangle$. What does this say about correlations in the X-basis?

Problem 2: Projective Measurement¶

A qubit is in state $|\Psi\rangle = \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac{2}{3}}|1\rangle$.

(a) Measure in the Z-basis. What are the probabilities for each outcome? What is the state after each outcome?

(b) Measure in the X-basis. Construct the projectors $|+\rangle\langle+|$ and $|-\rangle\langle-|$. What are the probabilities? What is the state after each outcome?

(c) Verify that the probabilities sum to 1 and the projectors satisfy $|+\rangle\langle+| + |-\rangle\langle-| = I$.

Problem 3: The Uncertainty Principle for Spin¶

(a) Verify the commutation relation $[\sigma_x, \sigma_y] = 2i\sigma_z$ by explicit matrix multiplication, where:

(b) Show that $\sigma_i^2 = I$ for each Pauli matrix. Conclude that $(\Delta\sigma_i)^2 = 1 - \langle\sigma_i\rangle^2$ in any state.

(c) For the state $|0\rangle$: compute $\langle\sigma_x\rangle$ and $\langle\sigma_y\rangle$. Use part (b) to find $\Delta\sigma_x$ and $\Delta\sigma_y$. Are they maximal?

(d) Verify the same result by expanding $|0\rangle$ in the X-eigenbasis. What are the probabilities for getting +1 and -1?

(e) For the state $|+\rangle$: which component is definite? Compute $\Delta\sigma_y$ and $\Delta\sigma_z$. Verify $\Delta\sigma_y \cdot \Delta\sigma_z \geq |\langle\sigma_x\rangle|$.

(f) Can a state have $\Delta\sigma_x = \Delta\sigma_y = \Delta\sigma_z = 0$ (definite values for all three components simultaneously)? Explain using the commutation relations.

Looking Ahead¶

We now have all the ingredients: entangled Bell states, the determinant test, and the projective measurement formalism. Next lecture, we put them together. What happens when Alice measures one qubit of a Bell pair? How does her measurement basis affect what Bob sees? The answers lead directly to EPR, Bell’s theorem, and the deepest question in quantum foundations: are quantum correlations “just” classical correlations in disguise?