Entanglement, Tensor Products & EPR Correlations

Part I — Composite Systems and the Tensor Product

When two independent quantum systems are combined, the mathematics requires a new construction. Let Alice's system have state space $\mathcal{H}_A$ of dimension $N_A$ and Bob's have $\mathcal{H}_B$ of dimension $N_B$. The combined state space is the tensor product:

A product state is written $|a\rangle \otimes |b\rangle$ (also $|a\rangle|b\rangle$ or $|ab\rangle$). For two spin-$\tfrac{1}{2}$ particles, the four-dimensional basis is:

where the first label belongs to Alice and the second to Bob. The most general two-spin state is:

with normalization $|\alpha_{uu}|^2+|\alpha_{ud}|^2+|\alpha_{du}|^2+|\alpha_{dd}|^2=1$. Operators on the combined space act as tensor products: $(A\otimes B)(|a\rangle\otimes|b\rangle) = (A|a\rangle)\otimes(B|b\rangle)$.

Part II — Entangled States: What They Are

A state $|\psi\rangle \in \mathcal{H}_A\otimes\mathcal{H}_B$ is a product state if it can be written as $|a\rangle\otimes|b\rangle$ for some $|a\rangle\in\mathcal{H}_A$ and $|b\rangle\in\mathcal{H}_B$. In this case Alice's and Bob's subsystems have independent states.

A state is entangled if it cannot be written as any product — the correlations between the two subsystems are non-factorizable. For a two-qubit state written with coefficient matrix $M$ where $(M)_{ij} = \alpha_{ij}$ (rows $=$ Alice's index, columns $=$ Bob's index), there is a simple test:

A two-qubit state is a product state if and only if $\det(M) = 0$. If $\det(M) \neq 0$, it is entangled.

Example — entangled: $\tfrac{1}{\sqrt{2}}(|ud\rangle-|du\rangle)$ has $M = \tfrac{1}{\sqrt{2}}\bigl(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\bigr)$, so $\det(M) = -\tfrac{1}{2} \neq 0$.

Example — product state: $\tfrac{1}{2}(|uu\rangle+|ud\rangle+|du\rangle+|dd\rangle) = |r\rangle_A\otimes|r\rangle_B$ has $M = \tfrac{1}{2}\bigl(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\bigr)$, so $\det(M)=0$.

Part III — The Singlet and Triplet States

The four two-spin basis states organize into angular momentum multiplets. Under total angular momentum classification:

Triplet states (symmetric, total spin $S=1$):

Singlet state (antisymmetric, total spin $S=0$):

The singlet has a remarkable property: it is rotationally invariant. In any basis defined by a unit vector $\hat{n}$, it takes the same form:

This follows because the singlet transforms as the $S=0$ representation of $SU(2)$ — the trivial representation that maps every rotation to the identity. The total spin satisfies $S_{\text{tot}}|\text{sing}\rangle = 0$.

Part IV — EPR Correlations

Alice and Bob separate and each holds one spin from a singlet pair. When Alice measures $\sigma_z$:

• If she gets $+1$, her spin collapses to $|u\rangle$, and Bob's spin instantly collapses to $|d\rangle$.
• If she gets $-1$, Bob's spin collapses to $|u\rangle$.
• Joint probabilities: $P(+1,+1)=0$, $P(+1,-1)=\tfrac{1}{2}$, $P(-1,+1)=\tfrac{1}{2}$, $P(-1,-1)=0$.

Perfect anti-correlation: whenever Alice measures $+1$, Bob measures $-1$, and vice versa. The two outcomes always disagree.

For measurements along general directions $\hat{n}_A$ and $\hat{n}_B$, the singlet gives the correlation function:

Same direction ($\hat{n}=\hat{m}$): $C=-1$ (perfect anti-correlation). Perpendicular ($\hat{n}\perp\hat{m}$): $C=0$ (no correlation). Opposite ($\hat{m}=-\hat{n}$): $C=+1$ (perfect correlation).

Part V — Why EPR Doesn't Violate Causality

Einstein, Podolsky, and Rosen (1935) found the instantaneous collapse unsettling — "spooky action at a distance." The crucial question: does Alice's measurement transmit information to Bob?

Consider Bob's perspective before and after Alice measures. Bob's reduced state is described by the density matrix $\rho_B = \text{Tr}_A(|\text{sing}\rangle\langle\text{sing}|) = \tfrac{1}{2}I$ — the maximally mixed state.

Regardless of what Alice does (or whether she measures at all), Bob's marginal statistics are:

If Alice gets $+1$ (probability $\tfrac{1}{2}$), Bob has $|d\rangle$. If Alice gets $-1$ (probability $\tfrac{1}{2}$), Bob has $|u\rangle$. But without knowing Alice's result, Bob's effective density matrix is still $\tfrac{1}{2}|u\rangle\langle u|+\tfrac{1}{2}|d\rangle\langle d|=\tfrac{1}{2}I$. Indistinguishable from before.

This is the no-communication theorem: quantum correlations alone cannot transmit information. A classical channel is required to compare results and reveal the correlations.

Part VI — Product vs Entangled States: Measurable Difference

For a product state $|a\rangle\otimes|b\rangle$, all joint probabilities factorize:

Alice's and Bob's outcomes are statistically independent — knowing one tells you nothing about the other.

For an entangled state, joint probabilities do not factorize in general. The correlation function $C(\hat{n}_A,\hat{n}_B)$ depends on both measurement axes simultaneously — this is only possible when the two subsystems share quantum correlations.

Could classical hidden variables explain the correlations? A classical system with shared randomness (a pre-agreed strategy) satisfies:

Quantum mechanics predicts violations of this bound — up to $2\sqrt{2}$ for the singlet (Tsirelson's bound). This is Bell's theorem, explored in Lecture 8.

Part VII — Wave Function Collapse in Entangled Systems

Before Alice measures, the composite state is $\tfrac{1}{\sqrt{2}}(|ud\rangle-|du\rangle)$ — a single entangled state of the joint system. There is no individual state for Alice's spin alone.

When Alice measures $\sigma_z$ and obtains $+1$:

The post-measurement state is a product state: the entanglement is destroyed. Bob's spin is now definitively $|d\rangle$. The collapse acts on the entire composite state, not just Alice's component.

Key points about the collapse:

Bob cannot tell which row he is in without classical communication from Alice. But once Alice communicates her result, Bob knows his spin with certainty.