QuantumCore

Lesson 23 · 80 min · core

Entanglement & the Bell States

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The circuit CNOT(H⊗I)\mathrm{CNOT}(H\otimes I) turns the four computational basis states into the four Bell states ∣Φ±⟩=12(∣00⟩±∣11⟩)\lvert\Phi^\pm\rangle=\tfrac1{\sqrt2}(\lvert00\rangle\pm\lvert11\rangle) and ∣Ψ±⟩=12(∣01⟩±∣10⟩)\lvert\Psi^\pm\rangle=\tfrac1{\sqrt2}(\lvert01\rangle\pm\lvert10\rangle). This lesson computes each of them line by line, proves that they form an orthonormal basis, proves that none of them is a product state (via the criterion c00c11=c01c10c_{00}c_{11}=c_{01}c_{10}), shows that each qubit on its own is maximally mixed, proves that the singlet ∣Ψ−⟩\lvert\Psi^-\rangle is invariant under U⊗UU\otimes U up to the phase det⁡U\det U, and evaluates the CHSH expression: S=22S=2\sqrt2 for ∣Φ+⟩\lvert\Phi^+\rangle, beyond the bound ∣S∣≤2\lvert S\rvert\le2 obeyed by every local hidden-variable model.

Prerequisites: Controlled Gates: CNOT, CZ, SWAP, Partial Trace & Reduced States

Objectives and key results

  • Compute CNOT(H⊗I)∣ab⟩=12(∣0,b⟩+(−1)a∣1,bˉ⟩)\mathrm{CNOT}(H\otimes I)\lvert ab\rangle=\tfrac1{\sqrt2}\big(\lvert0,b\rangle+(-1)^a\lvert1,\bar b\rangle\big) for all four inputs and name the resulting Bell states.

  • Prove that the Bell states are orthonormal by computing their 4×44\times4 Gram matrix, and recognise CNOT(H⊗I)\mathrm{CNOT}(H\otimes I) as the unitary change of basis from the computational basis to the Bell basis.

  • State and prove the product criterion c00c11=c01c10c_{00}c_{11}=c_{01}c_{10} for two-qubit pure states and use it to show that every Bell state is entangled.

  • Compute ρA=TrB∣β⟩⟨β∣=12I\rho_A=\mathrm{Tr}_B\lvert\beta\rangle\langle\beta\rvert=\tfrac12I and ρB=12I\rho_B=\tfrac12I for each Bell state ∣β⟩\lvert\beta\rangle.

  • Prove (U⊗U)∣Ψ−⟩=det⁡(U)∣Ψ−⟩(U\otimes U)\lvert\Psi^-\rangle=\det(U)\lvert\Psi^-\rangle with the antisymmetric symbol εab\varepsilon_{ab} and deduce perfect anti-correlation along every axis.

  • Prove ⟨Φ+∣P⊗Q∣Φ+⟩=12Tr(PTQ)\langle\Phi^+\vert P\otimes Q\vert\Phi^+\rangle=\tfrac12\mathrm{Tr}(P^TQ), evaluate the CHSH value S=22S=2\sqrt2, and prove the local hidden-variable bound ∣S∣≤2\lvert S\rvert\le2 and the Tsirelson bound ∣S∣≤22\lvert S\rvert\le2\sqrt2.

definition · Bell states
∣Φ±⟩=12(∣00⟩±∣11⟩),∣Ψ±⟩=12(∣01⟩±∣10⟩)\lvert\Phi^\pm\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle\pm\lvert11\rangle\big),\qquad\lvert\Psi^\pm\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle\pm\lvert10\rangle\big)
formula · Bell circuit
CNOT0→1(H⊗I)∣a,b⟩=12(∣0,b⟩+(−1)a∣1,bˉ⟩)\mathrm{CNOT}_{0\to1}(H\otimes I)\lvert a,b\rangle=\tfrac1{\sqrt2}\big(\lvert0,b\rangle+(-1)^a\lvert1,\bar b\rangle\big)
theorem · Product criterion
∑a,bcab∣ab⟩ is a product state  ⟺  c00c11−c01c10=0\textstyle\sum_{a,b}c_{ab}\lvert ab\rangle\ \text{is a product state}\iff c_{00}c_{11}-c_{01}c_{10}=0
identity · Maximally mixed marginals
TrB∣β⟩⟨β∣=TrA∣β⟩⟨β∣=12Ifor ∣β⟩∈{∣Φ±⟩,∣Ψ±⟩}\mathrm{Tr}_B\lvert\beta\rangle\langle\beta\rvert=\mathrm{Tr}_A\lvert\beta\rangle\langle\beta\rvert=\tfrac12I\quad\text{for }\lvert\beta\rangle\in\{\lvert\Phi^\pm\rangle,\lvert\Psi^\pm\rangle\}
identity · Singlet invariance
(U⊗U)∣Ψ−⟩=det⁡(U) ∣Ψ−⟩for every U∈U(2)(U\otimes U)\lvert\Psi^-\rangle=\det(U)\,\lvert\Psi^-\rangle\quad\text{for every }U\in U(2)
theorem · CHSH: classical, quantum and Tsirelson values
S=E00+E01+E10−E11,∣SLHV∣≤2,SΦ+=22,∣SQM∣≤22S=E_{00}+E_{01}+E_{10}-E_{11},\qquad\lvert S_{\mathrm{LHV}}\rvert\le2,\qquad S_{\Phi^+}=2\sqrt2,\qquad\lvert S_{\mathrm{QM}}\rvert\le2\sqrt2

The Bell circuit#

Definition

Bell states

The four Bell states are the two-qubit vectors ∣Φ+⟩=12(∣00⟩+∣11⟩),∣Φ−⟩=12(∣00⟩−∣11⟩),∣Ψ+⟩=12(∣01⟩+∣10⟩),∣Ψ−⟩=12(∣01⟩−∣10⟩).\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big),\quad\lvert\Phi^-\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle-\lvert11\rangle\big),\quad\lvert\Psi^+\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle+\lvert10\rangle\big),\quad\lvert\Psi^-\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert10\rangle\big). In the Φ\Phi states the two qubits always give equal computational-basis outcomes; in the Ψ\Psi states they always differ. The superscript is the relative sign. ∣Ψ−⟩\lvert\Psi^-\rangle is called the singlet.

Derivation

The general formula CNOT(H⊗I)∣a,b⟩\mathrm{CNOT}(H\otimes I)\lvert a,b\rangle

  1. Step 1:
    H∣a⟩=12(∣0⟩+(−1)a∣1⟩),a∈{0,1}H\lvert a\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle+(-1)^a\lvert1\rangle\big),\qquad a\in\{0,1\}
    For a=0a=0 this is the first column of HH, 12(1,1)T\tfrac1{\sqrt2}(1,1)^T; for a=1a=1 it is the second column, 12(1,−1)T\tfrac1{\sqrt2}(1,-1)^T.
  2. Step 2:
    (H⊗I)∣a,b⟩=H∣a⟩⊗∣b⟩=12(∣0,b⟩+(−1)a∣1,b⟩)(H\otimes I)\lvert a,b\rangle=H\lvert a\rangle\otimes\lvert b\rangle=\tfrac1{\sqrt2}\big(\lvert0,b\rangle+(-1)^a\lvert1,b\rangle\big)
    (A⊗B)(∣u⟩⊗∣v⟩)=A∣u⟩⊗B∣v⟩(A\otimes B)(\lvert u\rangle\otimes\lvert v\rangle)=A\lvert u\rangle\otimes B\lvert v\rangle with B=IB=I, then distribute the tensor product over the sum.
  3. Step 3:
    CNOT∣0,b⟩=∣0,b⊕0⟩=∣0,b⟩,CNOT∣1,b⟩=∣1,b⊕1⟩=∣1,bˉ⟩\mathrm{CNOT}\lvert0,b\rangle=\lvert0,b\oplus0\rangle=\lvert0,b\rangle,\qquad\mathrm{CNOT}\lvert1,b\rangle=\lvert1,b\oplus1\rangle=\lvert1,\bar b\rangle
    The CNOT rule ∣c,t⟩↦∣c,t⊕c⟩\lvert c,t\rangle\mapsto\lvert c,t\oplus c\rangle on the two basis states that occur.
  4. Step 4:
    CNOT(H⊗I)∣a,b⟩=12(∣0,b⟩+(−1)a∣1,bˉ⟩)\mathrm{CNOT}(H\otimes I)\lvert a,b\rangle=\tfrac1{\sqrt2}\big(\lvert0,b\rangle+(-1)^a\lvert1,\bar b\rangle\big)
    CNOT is linear, so it acts term by term on the superposition of the previous step.

Result

CNOT(H⊗I)∣a,b⟩=12(∣0,b⟩+(−1)a∣1,bˉ⟩)\boxed{\mathrm{CNOT}(H\otimes I)\lvert a,b\rangle=\tfrac1{\sqrt2}\big(\lvert0,b\rangle+(-1)^a\lvert1,\bar b\rangle\big)}
Derivation

The four inputs, line by line

  1. Step 1:
    (H⊗I)∣00⟩=12(∣0⟩+∣1⟩)⊗∣0⟩=12(∣00⟩+∣10⟩)(H\otimes I)\lvert00\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle+\lvert1\rangle\big)\otimes\lvert0\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert10\rangle\big)
    Input ∣00⟩\lvert00\rangle: H∣0⟩=∣+⟩H\lvert0\rangle=\lvert+\rangle on qubit 00; qubit 11 is untouched.
  2. Step 2:
    CNOT 12(∣00⟩+∣10⟩)=12(∣00⟩+∣11⟩)=∣Φ+⟩=12(1,0,0,1)T\mathrm{CNOT}\,\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert10\rangle\big)=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\lvert\Phi^+\rangle=\tfrac1{\sqrt2}(1,0,0,1)^T
    ∣00⟩\lvert00\rangle has control 00 and is unchanged; ∣10⟩\lvert10\rangle has control 11, so its target flips: ∣10⟩↦∣11⟩\lvert10\rangle\mapsto\lvert11\rangle.
  3. Step 3:
    (H⊗I)∣01⟩=12(∣0⟩+∣1⟩)⊗∣1⟩=12(∣01⟩+∣11⟩)(H\otimes I)\lvert01\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle+\lvert1\rangle\big)\otimes\lvert1\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle+\lvert11\rangle\big)
    Input ∣01⟩\lvert01\rangle: the same Hadamard step, now with qubit 11 in ∣1⟩\lvert1\rangle.
  4. Step 4:
    CNOT 12(∣01⟩+∣11⟩)=12(∣01⟩+∣10⟩)=∣Ψ+⟩=12(0,1,1,0)T\mathrm{CNOT}\,\tfrac1{\sqrt2}\big(\lvert01\rangle+\lvert11\rangle\big)=\tfrac1{\sqrt2}\big(\lvert01\rangle+\lvert10\rangle\big)=\lvert\Psi^+\rangle=\tfrac1{\sqrt2}(0,1,1,0)^T
    ∣01⟩\lvert01\rangle is unchanged (control 00); ∣11⟩↦∣1,1⊕1⟩=∣10⟩\lvert11\rangle\mapsto\lvert1,1\oplus1\rangle=\lvert10\rangle.
  5. Step 5:
    (H⊗I)∣10⟩=12(∣0⟩−∣1⟩)⊗∣0⟩=12(∣00⟩−∣10⟩)(H\otimes I)\lvert10\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle-\lvert1\rangle\big)\otimes\lvert0\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle-\lvert10\rangle\big)
    Input ∣10⟩\lvert10\rangle: H∣1⟩=∣−⟩H\lvert1\rangle=\lvert-\rangle carries the minus sign onto the ∣1⟩\lvert1\rangle branch of qubit 00.
  6. Step 6:
    CNOT 12(∣00⟩−∣10⟩)=12(∣00⟩−∣11⟩)=∣Φ−⟩=12(1,0,0,−1)T\mathrm{CNOT}\,\tfrac1{\sqrt2}\big(\lvert00\rangle-\lvert10\rangle\big)=\tfrac1{\sqrt2}\big(\lvert00\rangle-\lvert11\rangle\big)=\lvert\Phi^-\rangle=\tfrac1{\sqrt2}(1,0,0,-1)^T
    ∣00⟩\lvert00\rangle is unchanged and −∣10⟩↦−∣11⟩-\lvert10\rangle\mapsto-\lvert11\rangle; linearity keeps the coefficient −1-1.
  7. Step 7:
    (H⊗I)∣11⟩=12(∣0⟩−∣1⟩)⊗∣1⟩=12(∣01⟩−∣11⟩)(H\otimes I)\lvert11\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle-\lvert1\rangle\big)\otimes\lvert1\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert11\rangle\big)
    Input ∣11⟩\lvert11\rangle: H∣1⟩=∣−⟩H\lvert1\rangle=\lvert-\rangle on qubit 00, qubit 11 stays ∣1⟩\lvert1\rangle.
  8. Step 8:
    CNOT 12(∣01⟩−∣11⟩)=12(∣01⟩−∣10⟩)=∣Ψ−⟩=12(0,1,−1,0)T\mathrm{CNOT}\,\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert11\rangle\big)=\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert10\rangle\big)=\lvert\Psi^-\rangle=\tfrac1{\sqrt2}(0,1,-1,0)^T
    ∣01⟩\lvert01\rangle is unchanged and −∣11⟩↦−∣10⟩-\lvert11\rangle\mapsto-\lvert10\rangle.

Result

∣00⟩↦∣Φ+⟩,∣01⟩↦∣Ψ+⟩,∣10⟩↦∣Φ−⟩,∣11⟩↦∣Ψ−⟩\boxed{\lvert00\rangle\mapsto\lvert\Phi^+\rangle,\qquad\lvert01\rangle\mapsto\lvert\Psi^+\rangle,\qquad\lvert10\rangle\mapsto\lvert\Phi^-\rangle,\qquad\lvert11\rangle\mapsto\lvert\Psi^-\rangle}

Writing the four outputs as columns gives the matrix of the Bell circuit. Column 2a+b2a+b is B∣ab⟩B\lvert ab\rangle: B=CNOT(H⊗I)=(1000010000010010)12(1010010110−10010−1)=12(10100101010−110−10).B=\mathrm{CNOT}(H\otimes I)=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}\frac1{\sqrt2}\begin{pmatrix}1&0&1&0\\0&1&0&1\\1&0&-1&0\\0&1&0&-1\end{pmatrix}=\frac1{\sqrt2}\begin{pmatrix}1&0&1&0\\0&1&0&1\\0&1&0&-1\\1&0&-1&0\end{pmatrix}. The CNOT matrix only swaps rows 22 and 33 (indices ∣10⟩\lvert10\rangle and ∣11⟩\lvert11\rangle) of H⊗IH\otimes I. Because HH and CNOT are both self-inverse, B−1=B†=(H⊗I) CNOTB^{-1}=B^\dagger=(H\otimes I)\,\mathrm{CNOT}: running the circuit backwards maps each Bell state to its basis label, which is how a Bell measurement is performed with computational-basis readout.

B=12(10100101010−110−10)=(∣Φ+⟩  ∣Ψ+⟩  ∣Φ−⟩  ∣Ψ−⟩)B=\frac1{\sqrt2}\begin{pmatrix}1&0&1&0\\0&1&0&1\\0&1&0&-1\\1&0&-1&0\end{pmatrix}=\Big(\lvert\Phi^+\rangle\ \ \lvert\Psi^+\rangle\ \ \lvert\Phi^-\rangle\ \ \lvert\Psi^-\rangle\Big)
Bell circuit B = CNOT(H ⊗ I)4×4

Unitary verified: U†U=IU^\dagger U = I (max deviation 2.2e-16)

Circuit · 2 qubits

Quantum circuit on 2 qubitsH on q0; X on q1 controlled by q0q0q1H

Bell preparation CNOT(H⊗I)∣00⟩=∣Φ+⟩\mathrm{CNOT}(H\otimes I)\lvert00\rangle=\lvert\Phi^+\rangle: the statevector has amplitude 12≈0.7071\tfrac1{\sqrt2}\approx0.7071 on ∣00⟩\lvert00\rangle and ∣11⟩\lvert11\rangle, and the histogram shows outcomes 0000 and 1111 with probability 12\tfrac12 each, never 0101 or 1010.

∣ψ⟩=12∣00⟩+12∣11⟩\lvert\psi\rangle = \tfrac{1}{\sqrt{2}}\lvert 00\rangle + \tfrac{1}{\sqrt{2}}\lvert 11\rangle
  • |00⟩0.5
  • |01⟩0
  • |10⟩0
  • |11⟩0.5

Circuit · 2 qubits

Quantum circuit on 2 qubitsX on q1; H on q0; X on q1 controlled by q0q0q1XH

An XX on qubit 11 first changes the input to ∣01⟩\lvert01\rangle, and the same circuit produces ∣Ψ+⟩=12(∣01⟩+∣10⟩)\lvert\Psi^+\rangle=\tfrac1{\sqrt2}(\lvert01\rangle+\lvert10\rangle): the outcomes are now always different.

∣ψ⟩=12∣01⟩+12∣10⟩\lvert\psi\rangle = \tfrac{1}{\sqrt{2}}\lvert 01\rangle + \tfrac{1}{\sqrt{2}}\lvert 10\rangle
  • |00⟩0
  • |01⟩0.5
  • |10⟩0.5
  • |11⟩0

Quick check

Which Bell state does CNOT(H⊗I)\mathrm{CNOT}(H\otimes I) produce from the input ∣10⟩\lvert10\rangle?

Select one answer

The Bell basis is orthonormal#

Derivation

The Gram matrix of (∣Φ+⟩,∣Ψ+⟩,∣Φ−⟩,∣Ψ−⟩)(\lvert\Phi^+\rangle,\lvert\Psi^+\rangle,\lvert\Phi^-\rangle,\lvert\Psi^-\rangle)

  1. Step 1:
    Gjk=⟨βj∣βk⟩,(β0,β1,β2,β3)=(Φ+,Ψ+,Φ−,Ψ−)G_{jk}=\langle\beta_j\vert\beta_k\rangle,\qquad(\beta_0,\beta_1,\beta_2,\beta_3)=(\Phi^+,\Psi^+,\Phi^-,\Psi^-)
    The Gram matrix collects all pairwise inner products; a family is orthonormal exactly when G=IG=I. All amplitudes are real, so complex conjugation does not change them.
  2. Step 2:
    ⟨Φ±∣Φ±⟩=12(1⋅1+(±1)(±1))=1,⟨Ψ±∣Ψ±⟩=12(1⋅1+(±1)(±1))=1\langle\Phi^\pm\vert\Phi^\pm\rangle=\tfrac12\big(1\cdot1+(\pm1)(\pm1)\big)=1,\qquad\langle\Psi^\pm\vert\Psi^\pm\rangle=\tfrac12\big(1\cdot1+(\pm1)(\pm1)\big)=1
    Each state has two non-zero amplitudes of modulus 12\tfrac1{\sqrt2}, so the norm squared is 12+12\tfrac12+\tfrac12.
  3. Step 3:
    ⟨Φ+∣Φ−⟩=12(1⋅1+1⋅(−1))=0,⟨Ψ+∣Ψ−⟩=12(1⋅1+1⋅(−1))=0\langle\Phi^+\vert\Phi^-\rangle=\tfrac12\big(1\cdot1+1\cdot(-1)\big)=0,\qquad\langle\Psi^+\vert\Psi^-\rangle=\tfrac12\big(1\cdot1+1\cdot(-1)\big)=0
    The two Φ\Phi states share the support {∣00⟩,∣11⟩}\{\lvert00\rangle,\lvert11\rangle\} and differ by the sign of one amplitude, so the two contributions cancel; the same holds for the Ψ\Psi pair on {∣01⟩,∣10⟩}\{\lvert01\rangle,\lvert10\rangle\}.
  4. Step 4:
    ⟨Φ±∣Ψ±⟩=⟨Φ±∣Ψ∓⟩=0\langle\Phi^\pm\vert\Psi^\pm\rangle=\langle\Phi^\pm\vert\Psi^\mp\rangle=0
    A Φ\Phi state lives on {∣00⟩,∣11⟩}\{\lvert00\rangle,\lvert11\rangle\} and a Ψ\Psi state on {∣01⟩,∣10⟩}\{\lvert01\rangle,\lvert10\rangle\}; the supports are disjoint, so every term of the inner product contains a zero factor.
  5. Step 5:
    G=(1000010000100001)=I4G=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}=I_4
    Collect the sixteen entries: the diagonal entries are 11, every off-diagonal entry is 00, and GG is Hermitian, so Gkj=Gjk‾G_{kj}=\overline{G_{jk}} fills the lower triangle.
  6. Step 6:
    G=B†B=I4G=B^\dagger B=I_4
    Independent check: the columns of BB are the Bell vectors, so (B†B)jk=⟨βj∣βk⟩=Gjk(B^\dagger B)_{jk}=\langle\beta_j\vert\beta_k\rangle=G_{jk}, and BB is unitary as a product of the unitaries CNOT\mathrm{CNOT} and H⊗IH\otimes I.

Result

⟨βj∣βk⟩=δjk⟹{∣Φ+⟩,∣Ψ+⟩,∣Φ−⟩,∣Ψ−⟩} is an orthonormal basis of C2⊗C2\boxed{\langle\beta_j\vert\beta_k\rangle=\delta_{jk}\quad\Longrightarrow\quad\{\lvert\Phi^+\rangle,\lvert\Psi^+\rangle,\lvert\Phi^-\rangle,\lvert\Psi^-\rangle\}\ \text{is an orthonormal basis of}\ \mathbb C^2\otimes\mathbb C^2}
G=(⟨Φ+∣Φ+⟩⟨Φ+∣Ψ+⟩⟨Φ+∣Φ−⟩⟨Φ+∣Ψ−⟩⟨Ψ+∣Φ+⟩⟨Ψ+∣Ψ+⟩⟨Ψ+∣Φ−⟩⟨Ψ+∣Ψ−⟩⟨Φ−∣Φ+⟩⟨Φ−∣Ψ+⟩⟨Φ−∣Φ−⟩⟨Φ−∣Ψ−⟩⟨Ψ−∣Φ+⟩⟨Ψ−∣Ψ+⟩⟨Ψ−∣Φ−⟩⟨Ψ−∣Ψ−⟩)=(1000010000100001)G=\begin{pmatrix}\langle\Phi^+\vert\Phi^+\rangle&\langle\Phi^+\vert\Psi^+\rangle&\langle\Phi^+\vert\Phi^-\rangle&\langle\Phi^+\vert\Psi^-\rangle\\\langle\Psi^+\vert\Phi^+\rangle&\langle\Psi^+\vert\Psi^+\rangle&\langle\Psi^+\vert\Phi^-\rangle&\langle\Psi^+\vert\Psi^-\rangle\\\langle\Phi^-\vert\Phi^+\rangle&\langle\Phi^-\vert\Psi^+\rangle&\langle\Phi^-\vert\Phi^-\rangle&\langle\Phi^-\vert\Psi^-\rangle\\\langle\Psi^-\vert\Phi^+\rangle&\langle\Psi^-\vert\Psi^+\rangle&\langle\Psi^-\vert\Phi^-\rangle&\langle\Psi^-\vert\Psi^-\rangle\end{pmatrix}=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}
Gram matrix of the Bell basis4×4

Unitary verified: U†U=IU^\dagger U = I (max deviation 0.0e+0)

The Bell states are entangled#

Definition

Product state, entangled state

A two-qubit pure state ∣ψ⟩∈C2⊗C2\lvert\psi\rangle\in\mathbb C^2\otimes\mathbb C^2 is a product state if ∣ψ⟩=∣u⟩⊗∣v⟩\lvert\psi\rangle=\lvert u\rangle\otimes\lvert v\rangle for some single-qubit vectors ∣u⟩,∣v⟩\lvert u\rangle,\lvert v\rangle. Otherwise it is entangled. We write ∣ψ⟩=∑a,b∈{0,1}cab∣ab⟩\lvert\psi\rangle=\sum_{a,b\in\{0,1\}}c_{ab}\lvert ab\rangle and collect the amplitudes in the coefficient matrix C=(c00c01c10c11)C=\begin{pmatrix}c_{00}&c_{01}\\c_{10}&c_{11}\end{pmatrix}.

Theorem

Product criterion for two qubits

A non-zero vector ∣ψ⟩=∑a,bcab∣ab⟩\lvert\psi\rangle=\sum_{a,b}c_{ab}\lvert ab\rangle is a product state if and only if det⁡C=c00c11−c01c10=0\det C=c_{00}c_{11}-c_{01}c_{10}=0.

Proof (7 steps)
  1. Step 1:
    ∣ψ⟩=(u0∣0⟩+u1∣1⟩)⊗(v0∣0⟩+v1∣1⟩)=∑a,buavb∣ab⟩\lvert\psi\rangle=\big(u_0\lvert0\rangle+u_1\lvert1\rangle\big)\otimes\big(v_0\lvert0\rangle+v_1\lvert1\rangle\big)=\textstyle\sum_{a,b}u_av_b\lvert ab\rangle
    (⇒\Rightarrow) Assume ∣ψ⟩=∣u⟩⊗∣v⟩\lvert\psi\rangle=\lvert u\rangle\otimes\lvert v\rangle and expand the tensor product bilinearly.
  2. Step 2:
    cab=uavb ⇒ c00c11−c01c10=u0v0u1v1−u0v1u1v0=0c_{ab}=u_av_b\ \Rightarrow\ c_{00}c_{11}-c_{01}c_{10}=u_0v_0u_1v_1-u_0v_1u_1v_0=0
    Compare coefficients in the basis {∣ab⟩}\{\lvert ab\rangle\}; both products contain the same four factors u0u1v0v1u_0u_1v_0v_1.
  3. Step 3:
    Case (c00,c01)=(0,0):∣ψ⟩=∣1⟩⊗(c10∣0⟩+c11∣1⟩)\text{Case }(c_{00},c_{01})=(0,0):\quad\lvert\psi\rangle=\lvert1\rangle\otimes\big(c_{10}\lvert0\rangle+c_{11}\lvert1\rangle\big)
    (⇐\Leftarrow) Assume det⁡C=0\det C=0. If the first row of CC vanishes, only the terms with a=1a=1 survive and they factor directly.
  4. Step 4:
    Case c00≠0:λ:=c10c00,c11=c01c10c00=λc01\text{Case }c_{00}\neq0:\quad\lambda:=\tfrac{c_{10}}{c_{00}},\qquad c_{11}=\tfrac{c_{01}c_{10}}{c_{00}}=\lambda c_{01}
    Divide c00c11=c01c10c_{00}c_{11}=c_{01}c_{10} by c00c_{00}; trivially also c10=λc00c_{10}=\lambda c_{00}, so the second row of CC is λ\lambda times the first.
  5. Step 5:
    Case c00=0, c01≠0:c01c10=c00c11=0⇒c10=0,λ:=c11c01\text{Case }c_{00}=0,\ c_{01}\neq0:\quad c_{01}c_{10}=c_{00}c_{11}=0\Rightarrow c_{10}=0,\qquad\lambda:=\tfrac{c_{11}}{c_{01}}
    Then (c10,c11)=(0,λc01)=λ (c00,c01)(c_{10},c_{11})=(0,\lambda c_{01})=\lambda\,(c_{00},c_{01}): again the second row is λ\lambda times the first.
  6. Step 6:
    ∣ψ⟩=∑bc0b∣0b⟩+∑bλc0b∣1b⟩=(∣0⟩+λ∣1⟩)⊗(c00∣0⟩+c01∣1⟩)\lvert\psi\rangle=\textstyle\sum_b c_{0b}\lvert0b\rangle+\sum_b\lambda c_{0b}\lvert1b\rangle=\big(\lvert0\rangle+\lambda\lvert1\rangle\big)\otimes\big(c_{00}\lvert0\rangle+c_{01}\lvert1\rangle\big)
    In both remaining cases c1b=λc0bc_{1b}=\lambda c_{0b}; factor ∑bc0b∣b⟩\sum_bc_{0b}\lvert b\rangle out of both rows. The three cases are exhaustive, so det⁡C=0\det C=0 implies a product state.
  7. Step 7:
    ∣ψ⟩=∣0⟩+λ∣1⟩1+∣λ∣2⊗1+∣λ∣2 (c00∣0⟩+c01∣1⟩)\lvert\psi\rangle=\frac{\lvert0\rangle+\lambda\lvert1\rangle}{\sqrt{1+\lvert\lambda\rvert^2}}\otimes\sqrt{1+\lvert\lambda\rvert^2}\,\big(c_{00}\lvert0\rangle+c_{01}\lvert1\rangle\big)
    Scalars move freely between the factors of a tensor product, so the factors can be normalised individually when ∣ψ⟩\lvert\psi\rangle is a unit vector: the product of their norms is ∥ψ∥=1\lVert\psi\rVert=1.

∎

Derivation

det⁡C=±12\det C=\pm\tfrac12 for each Bell state

  1. Step 1:
    ∣Φ±⟩:C=12(100±1),c00c11−c01c10=12⋅±12−0⋅0=±12\lvert\Phi^\pm\rangle:\quad C=\frac1{\sqrt2}\begin{pmatrix}1&0\\0&\pm1\end{pmatrix},\qquad c_{00}c_{11}-c_{01}c_{10}=\tfrac1{\sqrt2}\cdot\tfrac{\pm1}{\sqrt2}-0\cdot0=\pm\tfrac12
    ∣Φ±⟩\lvert\Phi^\pm\rangle has c00=12c_{00}=\tfrac1{\sqrt2}, c11=±12c_{11}=\pm\tfrac1{\sqrt2} and c01=c10=0c_{01}=c_{10}=0.
  2. Step 2:
    ∣Ψ±⟩:C=12(01±10),c00c11−c01c10=0⋅0−12⋅±12=∓12\lvert\Psi^\pm\rangle:\quad C=\frac1{\sqrt2}\begin{pmatrix}0&1\\\pm1&0\end{pmatrix},\qquad c_{00}c_{11}-c_{01}c_{10}=0\cdot0-\tfrac1{\sqrt2}\cdot\tfrac{\pm1}{\sqrt2}=\mp\tfrac12
    ∣Ψ±⟩\lvert\Psi^\pm\rangle has c01=12c_{01}=\tfrac1{\sqrt2}, c10=±12c_{10}=\pm\tfrac1{\sqrt2} and c00=c11=0c_{00}=c_{11}=0. For the singlet the value is +12+\tfrac12.
  3. Step 3:
    det⁡C≠0 ⟹ ∣Φ±⟩,∣Ψ±⟩ are entangled\det C\neq0\ \Longrightarrow\ \lvert\Phi^\pm\rangle,\lvert\Psi^\pm\rangle\ \text{are entangled}
    Apply the product criterion: a product state would have det⁡C=0\det C=0.
  4. Step 4:
    C(ψ)=2 ∣c00c11−c01c10∣=2⋅12=1\mathcal C(\psi)=2\,\lvert c_{00}c_{11}-c_{01}c_{10}\rvert=2\cdot\tfrac12=1
    The concurrence C=2∣det⁡C∣\mathcal C=2\lvert\det C\rvert measures how far a pure two-qubit state is from the product condition. For a unit vector, ∣det⁡C∣\lvert\det C\rvert is the product of the two singular values s1,s2s_1,s_2 of CC with s12+s22=1s_1^2+s_2^2=1, so ∣det⁡C∣≤12\lvert\det C\rvert\le\tfrac12 by the AM–GM inequality: every Bell state attains the maximum C=1\mathcal C=1.

Result

c00c11−c01c10={+12Φ+, Ψ−−12Φ−, Ψ+⟹all four Bell states are maximally entangled\boxed{c_{00}c_{11}-c_{01}c_{10}=\begin{cases}+\tfrac12&\Phi^+,\ \Psi^-\\-\tfrac12&\Phi^-,\ \Psi^+\end{cases}\qquad\Longrightarrow\qquad\text{all four Bell states are maximally entangled}}

Quick check

Compute c00c11−c01c10c_{00}c_{11}-c_{01}c_{10} for the singlet ∣Ψ−⟩=12(∣01⟩−∣10⟩)\lvert\Psi^-\rangle=\tfrac1{\sqrt2}(\lvert01\rangle-\lvert10\rangle).

tolerance ±0.000001

Each half of a Bell pair is maximally mixed#

The partial trace over BB acts on operator products as TrB(∣a⟩⟨a′∣⊗∣b⟩⟨b′∣)=∣a⟩⟨a′∣ ⟨b′∣b⟩\mathrm{Tr}_B\big(\lvert a\rangle\langle a'\rvert\otimes\lvert b\rangle\langle b'\rvert\big)=\lvert a\rangle\langle a'\rvert\,\langle b'\vert b\rangle, and over AA as TrA(∣a⟩⟨a′∣⊗∣b⟩⟨b′∣)=⟨a′∣a⟩ ∣b⟩⟨b′∣\mathrm{Tr}_A\big(\lvert a\rangle\langle a'\rvert\otimes\lvert b\rangle\langle b'\rvert\big)=\langle a'\vert a\rangle\,\lvert b\rangle\langle b'\rvert; both are extended by linearity. For two qubits ∣ab⟩⟨a′b′∣=∣a⟩⟨a′∣⊗∣b⟩⟨b′∣\lvert ab\rangle\langle a'b'\rvert=\lvert a\rangle\langle a'\rvert\otimes\lvert b\rangle\langle b'\rvert.

Derivation

ρA\rho_A and ρB\rho_B for the singlet ∣Ψ−⟩\lvert\Psi^-\rangle

  1. Step 1:
    ρ=∣Ψ−⟩⟨Ψ−∣=12(∣01⟩−∣10⟩)(⟨01∣−⟨10∣)\rho=\lvert\Psi^-\rangle\langle\Psi^-\rvert=\tfrac12\big(\lvert01\rangle-\lvert10\rangle\big)\big(\langle01\rvert-\langle10\rvert\big)
    The bra of 12(∣01⟩−∣10⟩)\tfrac1{\sqrt2}(\lvert01\rangle-\lvert10\rangle) is 12(⟨01∣−⟨10∣)\tfrac1{\sqrt2}(\langle01\rvert-\langle10\rvert) (real coefficients); the two factors 12\tfrac1{\sqrt2} multiply to 12\tfrac12.
  2. Step 2:
    =12(∣01⟩⟨01∣−∣01⟩⟨10∣−∣10⟩⟨01∣+∣10⟩⟨10∣)=\tfrac12\big(\lvert01\rangle\langle01\rvert-\lvert01\rangle\langle10\rvert-\lvert10\rangle\langle01\rvert+\lvert10\rangle\langle10\rvert\big)
    Expand the product of the two binomials into four outer products.
  3. Step 3:
    TrB∣01⟩⟨01∣=∣0⟩⟨0∣⟨1∣1⟩=∣0⟩⟨0∣,TrB∣10⟩⟨10∣=∣1⟩⟨1∣⟨0∣0⟩=∣1⟩⟨1∣\mathrm{Tr}_B\lvert01\rangle\langle01\rvert=\lvert0\rangle\langle0\rvert\langle1\vert1\rangle=\lvert0\rangle\langle0\rvert,\qquad\mathrm{Tr}_B\lvert10\rangle\langle10\rvert=\lvert1\rangle\langle1\rvert\langle0\vert0\rangle=\lvert1\rangle\langle1\rvert
    Diagonal terms: the qubit-11 labels of ket and bra agree, so ⟨b′∣b⟩=1\langle b'\vert b\rangle=1.
  4. Step 4:
    TrB∣01⟩⟨10∣=∣0⟩⟨1∣⟨0∣1⟩=0,TrB∣10⟩⟨01∣=∣1⟩⟨0∣⟨1∣0⟩=0\mathrm{Tr}_B\lvert01\rangle\langle10\rvert=\lvert0\rangle\langle1\rvert\langle0\vert1\rangle=0,\qquad\mathrm{Tr}_B\lvert10\rangle\langle01\rvert=\lvert1\rangle\langle0\rvert\langle1\vert0\rangle=0
    Cross terms: the qubit-11 labels differ, so ⟨b′∣b⟩=0\langle b'\vert b\rangle=0 and the coherence between ∣0⟩A\lvert0\rangle_A and ∣1⟩A\lvert1\rangle_A is erased.
  5. Step 5:
    ρA=TrB ρ=12(∣0⟩⟨0∣−0−0+∣1⟩⟨1∣)=12(1001)=12I\rho_A=\mathrm{Tr}_B\,\rho=\tfrac12\big(\lvert0\rangle\langle0\rvert-0-0+\lvert1\rangle\langle1\rvert\big)=\tfrac12\begin{pmatrix}1&0\\0&1\end{pmatrix}=\tfrac12I
    Linearity of TrB\mathrm{Tr}_B applied to the four terms.
  6. Step 6:
    ρB=TrA ρ=12(⟨0∣0⟩∣1⟩⟨1∣−⟨1∣0⟩∣1⟩⟨0∣−⟨0∣1⟩∣0⟩⟨1∣+⟨1∣1⟩∣0⟩⟨0∣)=12I\rho_B=\mathrm{Tr}_A\,\rho=\tfrac12\big(\langle0\vert0\rangle\lvert1\rangle\langle1\rvert-\langle1\vert0\rangle\lvert1\rangle\langle0\rvert-\langle0\vert1\rangle\lvert0\rangle\langle1\rvert+\langle1\vert1\rangle\lvert0\rangle\langle0\rvert\big)=\tfrac12I
    The same four terms traced over qubit 00: the cross terms carry ⟨1∣0⟩=⟨0∣1⟩=0\langle1\vert0\rangle=\langle0\vert1\rangle=0, the diagonal terms leave ∣1⟩⟨1∣\lvert1\rangle\langle1\rvert and ∣0⟩⟨0∣\lvert0\rangle\langle0\rvert.
  7. Step 7:
    r⃗A=(Tr(XρA),Tr(YρA),Tr(ZρA))=12(TrX,TrY,TrZ)=(0,0,0),TrρA2=14Tr I=12\vec r_A=\big(\mathrm{Tr}(X\rho_A),\mathrm{Tr}(Y\rho_A),\mathrm{Tr}(Z\rho_A)\big)=\tfrac12\big(\mathrm{Tr}X,\mathrm{Tr}Y,\mathrm{Tr}Z\big)=(0,0,0),\qquad\mathrm{Tr}\rho_A^2=\tfrac14\mathrm{Tr}\,I=\tfrac12
    The Pauli matrices are traceless, so the reduced Bloch vector sits at the centre of the ball; the purity 12\tfrac12 is the minimum for a qubit and the von Neumann entropy is S(ρA)=−2⋅12log⁡212=1S(\rho_A)=-2\cdot\tfrac12\log_2\tfrac12=1 bit.

Result

TrB∣Ψ−⟩⟨Ψ−∣=TrA∣Ψ−⟩⟨Ψ−∣=12I\boxed{\mathrm{Tr}_B\lvert\Psi^-\rangle\langle\Psi^-\rvert=\mathrm{Tr}_A\lvert\Psi^-\rangle\langle\Psi^-\rvert=\tfrac12I}
Derivation

The same pattern for ∣Φ±⟩\lvert\Phi^\pm\rangle and ∣Ψ+⟩\lvert\Psi^+\rangle

  1. Step 1:
    ∣βab⟩=12(∣0,b⟩+s∣1,bˉ⟩),s=(−1)a\lvert\beta_{ab}\rangle=\tfrac1{\sqrt2}\big(\lvert0,b\rangle+s\lvert1,\bar b\rangle\big),\qquad s=(-1)^a
    Every Bell state has this form by the Bell-circuit formula; the two terms differ on both qubits.
  2. Step 2:
    ∣βab⟩⟨βab∣=12(∣0b⟩⟨0b∣+s∣0b⟩⟨1bˉ∣+s∣1bˉ⟩⟨0b∣+∣1bˉ⟩⟨1bˉ∣)\lvert\beta_{ab}\rangle\langle\beta_{ab}\rvert=\tfrac12\big(\lvert0b\rangle\langle0b\rvert+s\lvert0b\rangle\langle1\bar b\rvert+s\lvert1\bar b\rangle\langle0b\rvert+\lvert1\bar b\rangle\langle1\bar b\rvert\big)
    ss is real with s2=1s^2=1, so the diagonal terms have coefficient 11 and the cross terms coefficient ss.
  3. Step 3:
    TrB:  ⟨bˉ∣b⟩=0 ⇒ ρA=12(∣0⟩⟨0∣+∣1⟩⟨1∣)=12I\mathrm{Tr}_B:\ \ \langle\bar b\vert b\rangle=0\ \Rightarrow\ \rho_A=\tfrac12\big(\lvert0\rangle\langle0\rvert+\lvert1\rangle\langle1\rvert\big)=\tfrac12I
    The cross terms contain ⟨bˉ∣b⟩=0\langle\bar b\vert b\rangle=0 or ⟨b∣bˉ⟩=0\langle b\vert\bar b\rangle=0 on qubit 11; the diagonal terms contribute ∣0⟩⟨0∣⟨b∣b⟩\lvert0\rangle\langle0\rvert\langle b\vert b\rangle and ∣1⟩⟨1∣⟨bˉ∣bˉ⟩\lvert1\rangle\langle1\rvert\langle\bar b\vert\bar b\rangle.
  4. Step 4:
    TrA:  ⟨1∣0⟩=0 ⇒ ρB=12(∣b⟩⟨b∣+∣bˉ⟩⟨bˉ∣)=12I\mathrm{Tr}_A:\ \ \langle1\vert0\rangle=0\ \Rightarrow\ \rho_B=\tfrac12\big(\lvert b\rangle\langle b\rvert+\lvert\bar b\rangle\langle\bar b\rvert\big)=\tfrac12I
    The cross terms carry ⟨1∣0⟩\langle1\vert0\rangle or ⟨0∣1⟩\langle0\vert1\rangle on qubit 00; {∣b⟩,∣bˉ⟩}={∣0⟩,∣1⟩}\{\lvert b\rangle,\lvert\bar b\rangle\}=\{\lvert0\rangle,\lvert1\rangle\} for either value of bb.
  5. Step 5:
    ∣Φ+⟩ (a,b)=(0,0),∣Ψ+⟩ (0,1),∣Φ−⟩ (1,0):ρA=ρB=12I\lvert\Phi^+\rangle\ (a,b)=(0,0),\quad\lvert\Psi^+\rangle\ (0,1),\quad\lvert\Phi^-\rangle\ (1,0):\qquad\rho_A=\rho_B=\tfrac12I
    The sign ss only ever multiplies the cross terms, which the partial trace removes, so all four Bell states have the same marginals.

Result

ρA=ρB=12Ifor all four Bell states\boxed{\rho_A=\rho_B=\tfrac12I\quad\text{for all four Bell states}}

Quick check

What is the purity Tr ρA2\mathrm{Tr}\,\rho_A^2 of qubit AA when the pair is in ∣Φ−⟩\lvert\Phi^-\rangle?

tolerance ±0.000001

The singlet is invariant under U ⊗ U#

Lemma

The antisymmetric symbol and the determinant

Let εab\varepsilon_{ab} for a,b∈{0,1}a,b\in\{0,1\} be the antisymmetric symbol ε01=1\varepsilon_{01}=1, ε10=−1\varepsilon_{10}=-1, ε00=ε11=0\varepsilon_{00}=\varepsilon_{11}=0. For every complex 2×22\times2 matrix UU and all i,j∈{0,1}i,j\in\{0,1\}, ∑a,bUiaUjbεab=det⁡(U) εij\sum_{a,b}U_{ia}U_{jb}\varepsilon_{ab}=\det(U)\,\varepsilon_{ij}.

Proof (4 steps)
  1. Step 1:
    ∑a,bUiaUjbεab=Ui0Uj1ε01+Ui1Uj0ε10=Ui0Uj1−Ui1Uj0\textstyle\sum_{a,b}U_{ia}U_{jb}\varepsilon_{ab}=U_{i0}U_{j1}\varepsilon_{01}+U_{i1}U_{j0}\varepsilon_{10}=U_{i0}U_{j1}-U_{i1}U_{j0}
    Only the two terms with a≠ba\neq b survive, because ε00=ε11=0\varepsilon_{00}=\varepsilon_{11}=0.
  2. Step 2:
    i=j:Ui0Ui1−Ui1Ui0=0=det⁡(U) εiii=j:\quad U_{i0}U_{i1}-U_{i1}U_{i0}=0=\det(U)\,\varepsilon_{ii}
    The two products coincide, and εii=0\varepsilon_{ii}=0.
  3. Step 3:
    (i,j)=(0,1):U00U11−U01U10=det⁡U=det⁡(U) ε01(i,j)=(0,1):\quad U_{00}U_{11}-U_{01}U_{10}=\det U=\det(U)\,\varepsilon_{01}
    This is the definition of the 2×22\times2 determinant, and ε01=1\varepsilon_{01}=1.
  4. Step 4:
    (i,j)=(1,0):U10U01−U11U00=−det⁡U=det⁡(U) ε10(i,j)=(1,0):\quad U_{10}U_{01}-U_{11}U_{00}=-\det U=\det(U)\,\varepsilon_{10}
    The same two products with the opposite sign, and ε10=−1\varepsilon_{10}=-1. The four cases exhaust i,j∈{0,1}i,j\in\{0,1\}.

∎

Theorem

Rotational invariance of the singlet

For every U∈U(2)U\in U(2), (U⊗U)∣Ψ−⟩=det⁡(U) ∣Ψ−⟩(U\otimes U)\lvert\Psi^-\rangle=\det(U)\,\lvert\Psi^-\rangle. Since ∣det⁡U∣=1\lvert\det U\rvert=1, the singlet is unchanged up to a global phase, and it is exactly invariant for U∈SU(2)U\in SU(2).

Proof (5 steps)
  1. Step 1:
    ∣Ψ−⟩=12(∣01⟩−∣10⟩)=12∑a,bεab∣a⟩⊗∣b⟩\lvert\Psi^-\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert10\rangle\big)=\tfrac1{\sqrt2}\textstyle\sum_{a,b}\varepsilon_{ab}\lvert a\rangle\otimes\lvert b\rangle
    The sum has the terms ε01∣01⟩=∣01⟩\varepsilon_{01}\lvert01\rangle=\lvert01\rangle and ε10∣10⟩=−∣10⟩\varepsilon_{10}\lvert10\rangle=-\lvert10\rangle; the others vanish.
  2. Step 2:
    (U⊗U)∣Ψ−⟩=12∑a,bεab U∣a⟩⊗U∣b⟩(U\otimes U)\lvert\Psi^-\rangle=\tfrac1{\sqrt2}\textstyle\sum_{a,b}\varepsilon_{ab}\,U\lvert a\rangle\otimes U\lvert b\rangle
    Linearity and (U⊗U)(∣a⟩⊗∣b⟩)=U∣a⟩⊗U∣b⟩(U\otimes U)(\lvert a\rangle\otimes\lvert b\rangle)=U\lvert a\rangle\otimes U\lvert b\rangle.
  3. Step 3:
    U∣a⟩=∑iUia∣i⟩ ⇒ (U⊗U)∣Ψ−⟩=12∑i,j(∑a,bUiaUjbεab)∣ij⟩U\lvert a\rangle=\textstyle\sum_iU_{ia}\lvert i\rangle\ \Rightarrow\ (U\otimes U)\lvert\Psi^-\rangle=\tfrac1{\sqrt2}\sum_{i,j}\Big(\sum_{a,b}U_{ia}U_{jb}\varepsilon_{ab}\Big)\lvert ij\rangle
    Column aa of UU is U∣a⟩U\lvert a\rangle; expand both factors and exchange the finite sums.
  4. Step 4:
    =12∑i,jdet⁡(U) εij∣ij⟩=det⁡(U) ∣Ψ−⟩=\tfrac1{\sqrt2}\textstyle\sum_{i,j}\det(U)\,\varepsilon_{ij}\lvert ij\rangle=\det(U)\,\lvert\Psi^-\rangle
    Apply the lemma to the bracket, then recognise the first line with (a,b)(a,b) renamed (i,j)(i,j).
  5. Step 5:
    U†U=I ⇒ ∣det⁡U∣2=det⁡(U†)det⁡(U)=det⁡(U†U)=1U^\dagger U=I\ \Rightarrow\ \lvert\det U\rvert^2=\det(U^\dagger)\det(U)=\det(U^\dagger U)=1
    The determinant is multiplicative and det⁡U†=det⁡U‾\det U^\dagger=\overline{\det U}, so det⁡U=eiγ\det U=e^{i\gamma} is a pure phase, equal to 11 on SU(2)SU(2).

∎

Derivation

Perfect anti-correlation along every axis

  1. Step 1:
    n^⋅σ⃗=∣n+⟩⟨n+∣−∣n−⟩⟨n−∣,V:=∣n+⟩⟨0∣+∣n−⟩⟨1∣\hat n\cdot\vec\sigma=\lvert n_+\rangle\langle n_+\rvert-\lvert n_-\rangle\langle n_-\rvert,\qquad V:=\lvert n_+\rangle\langle0\rvert+\lvert n_-\rangle\langle1\rvert
    For a unit vector n^\hat n, n^⋅σ⃗\hat n\cdot\vec\sigma is Hermitian with (n^⋅σ⃗)2=I(\hat n\cdot\vec\sigma)^2=I and trace 00, so its eigenvalues are +1+1 and −1-1 with orthonormal eigenvectors ∣n±⟩\lvert n_\pm\rangle. VV maps an orthonormal basis to an orthonormal basis, hence is unitary.
  2. Step 2:
    VZV†=V(∣0⟩⟨0∣−∣1⟩⟨1∣)V†=∣n+⟩⟨n+∣−∣n−⟩⟨n−∣=n^⋅σ⃗VZV^\dagger=V\big(\lvert0\rangle\langle0\rvert-\lvert1\rangle\langle1\rvert\big)V^\dagger=\lvert n_+\rangle\langle n_+\rvert-\lvert n_-\rangle\langle n_-\rvert=\hat n\cdot\vec\sigma
    V∣0⟩=∣n+⟩V\lvert0\rangle=\lvert n_+\rangle, V∣1⟩=∣n−⟩V\lvert1\rangle=\lvert n_-\rangle and ⟨k∣V†=(V∣k⟩)†\langle k\rvert V^\dagger=(V\lvert k\rangle)^\dagger.
  3. Step 3:
    (V†⊗V†)∣Ψ−⟩=det⁡(V†)∣Ψ−⟩,⟨Ψ−∣(V⊗V)=det⁡(V)⟨Ψ−∣(V^\dagger\otimes V^\dagger)\lvert\Psi^-\rangle=\det(V^\dagger)\lvert\Psi^-\rangle,\qquad\langle\Psi^-\rvert(V\otimes V)=\det(V)\langle\Psi^-\rvert
    Apply the theorem to the unitary V†V^\dagger; the second identity is the adjoint of the first, since det⁡V†‾=det⁡V\overline{\det V^\dagger}=\det V.
  4. Step 4:
    ⟨Ψ−∣(n^⋅σ⃗)⊗(n^⋅σ⃗)∣Ψ−⟩=⟨Ψ−∣(V⊗V)(Z⊗Z)(V†⊗V†)∣Ψ−⟩=∣det⁡V∣2⟨Ψ−∣Z⊗Z∣Ψ−⟩\langle\Psi^-\vert(\hat n\cdot\vec\sigma)\otimes(\hat n\cdot\vec\sigma)\vert\Psi^-\rangle=\langle\Psi^-\rvert(V\otimes V)(Z\otimes Z)(V^\dagger\otimes V^\dagger)\lvert\Psi^-\rangle=\lvert\det V\rvert^2\langle\Psi^-\vert Z\otimes Z\vert\Psi^-\rangle
    Use (VZV†)⊗(VZV†)=(V⊗V)(Z⊗Z)(V†⊗V†)(VZV^\dagger)\otimes(VZV^\dagger)=(V\otimes V)(Z\otimes Z)(V^\dagger\otimes V^\dagger) and the previous line.
  5. Step 5:
    ⟨Ψ−∣Z⊗Z∣Ψ−⟩=12(⟨01∣Z⊗Z∣01⟩+⟨10∣Z⊗Z∣10⟩)=12((1)(−1)+(−1)(1))=−1\langle\Psi^-\vert Z\otimes Z\vert\Psi^-\rangle=\tfrac12\big(\langle01\vert Z\otimes Z\vert01\rangle+\langle10\vert Z\otimes Z\vert10\rangle\big)=\tfrac12\big((1)(-1)+(-1)(1)\big)=-1
    Z⊗ZZ\otimes Z is diagonal with entry (−1)a(−1)b(-1)^{a}(-1)^{b} on ∣ab⟩\lvert ab\rangle, so the cross terms vanish; ∣det⁡V∣2=1\lvert\det V\rvert^2=1.

Result

⟨Ψ−∣(n^⋅σ⃗)⊗(n^⋅σ⃗)∣Ψ−⟩=−1for every unit vector n^\boxed{\langle\Psi^-\vert(\hat n\cdot\vec\sigma)\otimes(\hat n\cdot\vec\sigma)\vert\Psi^-\rangle=-1\quad\text{for every unit vector }\hat n}

A correlator equal to −1-1 for observables with eigenvalues ±1\pm1 means the two outcomes are always opposite: ⟨O⊗O⟩=P(equal)−P(different)=−1\langle O\otimes O\rangle=P(\text{equal})-P(\text{different})=-1 forces P(different)=1P(\text{different})=1. The singlet is the same anti-correlated state whichever axis the two parties choose, as long as they choose the same one. The other three Bell states lack this symmetry: for example ⟨Φ+∣Y⊗Y∣Φ+⟩=−1\langle\Phi^+\vert Y\otimes Y\vert\Phi^+\rangle=-1 while ⟨Φ+∣Z⊗Z∣Φ+⟩=+1\langle\Phi^+\vert Z\otimes Z\vert\Phi^+\rangle=+1.

Quick check

Alice and Bob share ∣Ψ−⟩\lvert\Psi^-\rangle and both measure XX. What is ⟨Ψ−∣X⊗X∣Ψ−⟩\langle\Psi^-\vert X\otimes X\vert\Psi^-\rangle?

Select one answer

The CHSH inequality#

In a CHSH experiment Alice chooses one of two measurement settings x∈{0,1}x\in\{0,1\} and Bob one of two settings y∈{0,1}y\in\{0,1\}; every measurement has outcomes ±1\pm1. From many runs they estimate the four correlators Exy=⟨Ax⊗By⟩E_{xy}=\langle A_x\otimes B_y\rangle, the average of the product of their outcomes, and form S=E00+E01+E10−E11.S=E_{00}+E_{01}+E_{10}-E_{11}. Quantum mechanically the settings are Hermitian observables with eigenvalues ±1\pm1. We use A0=Z,A1=X,B0=12(Z+X),B1=12(Z−X),A_0=Z,\qquad A_1=X,\qquad B_0=\tfrac1{\sqrt2}(Z+X),\qquad B_1=\tfrac1{\sqrt2}(Z-X), where B0,B1B_0,B_1 are the Pauli observables along the unit vectors 12(1,0,±1)\tfrac1{\sqrt2}(1,0,\pm1) in the xzxz plane, so By2=IB_y^2=I.

Lemma

Correlations in the state Φ+

For all complex 2×22\times2 matrices PP and QQ, ⟨Φ+∣P⊗Q∣Φ+⟩=12Tr(PTQ)\langle\Phi^+\vert P\otimes Q\vert\Phi^+\rangle=\tfrac12\mathrm{Tr}\big(P^TQ\big), where PP acts on qubit 00 and QQ on qubit 11.

Proof (4 steps)
  1. Step 1:
    ∣Φ+⟩=12∑k∈{0,1}∣k⟩⊗∣k⟩\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\textstyle\sum_{k\in\{0,1\}}\lvert k\rangle\otimes\lvert k\rangle
    The two terms are ∣00⟩\lvert00\rangle and ∣11⟩\lvert11\rangle.
  2. Step 2:
    ⟨Φ+∣P⊗Q∣Φ+⟩=12∑k,l(⟨k∣⊗⟨k∣)(P⊗Q)(∣l⟩⊗∣l⟩)\langle\Phi^+\vert P\otimes Q\vert\Phi^+\rangle=\tfrac12\textstyle\sum_{k,l}\big(\langle k\rvert\otimes\langle k\rvert\big)(P\otimes Q)\big(\lvert l\rangle\otimes\lvert l\rangle\big)
    Expand bra and ket; the prefactor is 12⋅12\tfrac1{\sqrt2}\cdot\tfrac1{\sqrt2}.
  3. Step 3:
    =12∑k,l⟨k∣P∣l⟩ ⟨k∣Q∣l⟩=12∑k,lPkl Qkl=\tfrac12\textstyle\sum_{k,l}\langle k\vert P\vert l\rangle\,\langle k\vert Q\vert l\rangle=\tfrac12\sum_{k,l}P_{kl}\,Q_{kl}
    (⟨u∣⊗⟨v∣)(P⊗Q)(∣w⟩⊗∣z⟩)=⟨u∣P∣w⟩⟨v∣Q∣z⟩(\langle u\rvert\otimes\langle v\rvert)(P\otimes Q)(\lvert w\rangle\otimes\lvert z\rangle)=\langle u\vert P\vert w\rangle\langle v\vert Q\vert z\rangle, and ⟨k∣P∣l⟩=Pkl\langle k\vert P\vert l\rangle=P_{kl}.
  4. Step 4:
    =12∑l∑k(PT)lk Qkl=12∑l(PTQ)ll=12Tr(PTQ)=\tfrac12\textstyle\sum_{l}\sum_{k}(P^T)_{lk}\,Q_{kl}=\tfrac12\sum_l\big(P^TQ\big)_{ll}=\tfrac12\mathrm{Tr}\big(P^TQ\big)
    Write Pkl=(PT)lkP_{kl}=(P^T)_{lk}; the inner sum over kk is the (l,l)(l,l) entry of the product PTQP^TQ, and summing the diagonal gives the trace.

∎

Derivation

The four correlators and S=22S=2\sqrt2

  1. Step 1:
    ZT=Z,XT=X,Tr(ZZ)=Tr(XX)=Tr I=2,Tr(ZX)=Tr(XZ)=Tr(01−10)=0Z^T=Z,\quad X^T=X,\qquad\mathrm{Tr}(ZZ)=\mathrm{Tr}(XX)=\mathrm{Tr}\,I=2,\qquad\mathrm{Tr}(ZX)=\mathrm{Tr}(XZ)=\mathrm{Tr}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=0
    ZZ and XX are real symmetric; Z2=X2=IZ^2=X^2=I; ZX=(01−10)=iYZX=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=iY and XZ=−iYXZ=-iY are traceless.
  2. Step 2:
    E00=12Tr(Z Z+X2)=122(Tr(ZZ)+Tr(ZX))=122(2+0)=12E_{00}=\tfrac12\mathrm{Tr}\Big(Z\,\tfrac{Z+X}{\sqrt2}\Big)=\tfrac1{2\sqrt2}\big(\mathrm{Tr}(ZZ)+\mathrm{Tr}(ZX)\big)=\tfrac1{2\sqrt2}(2+0)=\tfrac1{\sqrt2}
    Lemma with P=A0=ZP=A_0=Z, Q=B0Q=B_0, and linearity of the trace.
  3. Step 3:
    E01=12Tr(Z Z−X2)=122(Tr(ZZ)−Tr(ZX))=122(2−0)=12E_{01}=\tfrac12\mathrm{Tr}\Big(Z\,\tfrac{Z-X}{\sqrt2}\Big)=\tfrac1{2\sqrt2}\big(\mathrm{Tr}(ZZ)-\mathrm{Tr}(ZX)\big)=\tfrac1{2\sqrt2}(2-0)=\tfrac1{\sqrt2}
    P=A0=ZP=A_0=Z, Q=B1Q=B_1.
  4. Step 4:
    E10=12Tr(X Z+X2)=122(Tr(XZ)+Tr(XX))=122(0+2)=12E_{10}=\tfrac12\mathrm{Tr}\Big(X\,\tfrac{Z+X}{\sqrt2}\Big)=\tfrac1{2\sqrt2}\big(\mathrm{Tr}(XZ)+\mathrm{Tr}(XX)\big)=\tfrac1{2\sqrt2}(0+2)=\tfrac1{\sqrt2}
    P=A1=XP=A_1=X, Q=B0Q=B_0.
  5. Step 5:
    E11=12Tr(X Z−X2)=122(Tr(XZ)−Tr(XX))=122(0−2)=−12E_{11}=\tfrac12\mathrm{Tr}\Big(X\,\tfrac{Z-X}{\sqrt2}\Big)=\tfrac1{2\sqrt2}\big(\mathrm{Tr}(XZ)-\mathrm{Tr}(XX)\big)=\tfrac1{2\sqrt2}(0-2)=-\tfrac1{\sqrt2}
    P=A1=XP=A_1=X, Q=B1Q=B_1: the only negative correlator.
  6. Step 6:
    S=E00+E01+E10−E11=12+12+12+12=42=22≈2.828S=E_{00}+E_{01}+E_{10}-E_{11}=\tfrac1{\sqrt2}+\tfrac1{\sqrt2}+\tfrac1{\sqrt2}+\tfrac1{\sqrt2}=\tfrac4{\sqrt2}=2\sqrt2\approx2.828
    The minus sign in front of E11E_{11} turns the negative correlator into a positive contribution; 42=422\tfrac4{\sqrt2}=\tfrac{4\sqrt2}{2}.

Result

E00=E01=E10=12,E11=−12,SΦ+=22\boxed{E_{00}=E_{01}=E_{10}=\tfrac1{\sqrt2},\quad E_{11}=-\tfrac1{\sqrt2},\qquad S_{\Phi^+}=2\sqrt2}

Circuit · 2 qubits

Quantum circuit on 2 qubitsH on q0; X on q1 controlled by q0; RY(-π/4) on q1q0q1HRY(-π/4)

Measuring the pair (A0,B0)(A_0,B_0) with computational-basis readout: RY(−π4)R_Y(-\tfrac\pi4) on qubit 11 rotates B0B_0 into ZZ, since RY(π4) Z RY(π4)†=cos⁡π4 Z+sin⁡π4 X=B0R_Y(\tfrac\pi4)\,Z\,R_Y(\tfrac\pi4)^\dagger=\cos\tfrac\pi4\,Z+\sin\tfrac\pi4\,X=B_0. The output is 12(c∣00⟩−s∣01⟩+s∣10⟩+c∣11⟩)\tfrac1{\sqrt2}\big(c\lvert00\rangle-s\lvert01\rangle+s\lvert10\rangle+c\lvert11\rangle\big) with c=cos⁡π8c=\cos\tfrac\pi8, s=sin⁡π8s=\sin\tfrac\pi8, so equal outcomes occur with probability c2=2+24≈0.854c^2=\tfrac{2+\sqrt2}4\approx0.854 and E00=c2−s2=cos⁡π4=12E_{00}=c^2-s^2=\cos\tfrac\pi4=\tfrac1{\sqrt2}.

∣ψ⟩=0.6533∣00⟩−0.2706∣01⟩+0.2706∣10⟩+0.6533∣11⟩\lvert\psi\rangle = 0.6533\lvert 00\rangle - 0.2706\lvert 01\rangle + 0.2706\lvert 10\rangle + 0.6533\lvert 11\rangle
  • |00⟩0.4268
  • |01⟩0.0732
  • |10⟩0.0732
  • |11⟩0.4268

Definition

Local hidden-variable model

A local hidden-variable (LHV) model assigns to each run a hidden variable λ\lambda drawn from a probability distribution p(λ)p(\lambda) that does not depend on the settings, and outcome functions Ax(λ)∈{±1}A_x(\lambda)\in\{\pm1\}, By(λ)∈{±1}B_y(\lambda)\in\{\pm1\}: Alice's outcome depends only on her own setting xx and on λ\lambda, Bob's only on yy and λ\lambda (locality). The model predicts Exy=∫p(λ) Ax(λ)By(λ) dλE_{xy}=\int p(\lambda)\,A_x(\lambda)B_y(\lambda)\,d\lambda. A deterministic strategy is the special case of a fixed λ\lambda: four numbers a0,a1,b0,b1∈{±1}a_0,a_1,b_0,b_1\in\{\pm1\}.

Theorem

CHSH inequality (Clauser–Horne–Shimony–Holt)

Every local hidden-variable model satisfies ∣S∣=∣E00+E01+E10−E11∣≤2\lvert S\rvert=\lvert E_{00}+E_{01}+E_{10}-E_{11}\rvert\le2. Consequently no LHV model reproduces the quantum prediction S=22S=2\sqrt2 for ∣Φ+⟩\lvert\Phi^+\rangle.

Proof (7 steps)
  1. Step 1:
    S(λ)=a0b0+a0b1+a1b0−a1b1=a0(b0+b1)+a1(b0−b1)S(\lambda)=a_0b_0+a_0b_1+a_1b_0-a_1b_1=a_0(b_0+b_1)+a_1(b_0-b_1)
    For one deterministic strategy ax=Ax(λ)a_x=A_x(\lambda), by=By(λ)b_y=B_y(\lambda) each correlator is the product axbya_xb_y; group the terms by Alice's outcome.
  2. Step 2:
    b0=b1 ⇒ b0−b1=0,  b0+b1=±2;b0=−b1 ⇒ b0+b1=0,  b0−b1=±2b_0=b_1\ \Rightarrow\ b_0-b_1=0,\ \ b_0+b_1=\pm2;\qquad b_0=-b_1\ \Rightarrow\ b_0+b_1=0,\ \ b_0-b_1=\pm2
    Since b0,b1∈{±1}b_0,b_1\in\{\pm1\}, they are either equal or opposite: exactly one bracket vanishes and the other is ±2\pm2.
  3. Step 3:
    S(λ)=±2a0  or  S(λ)=±2a1 ⇒ S(λ)∈{−2,+2}S(\lambda)=\pm2a_0\ \ \text{or}\ \ S(\lambda)=\pm2a_1\ \Rightarrow\ S(\lambda)\in\{-2,+2\}
    ∣ax∣=1\lvert a_x\rvert=1, so the surviving term has modulus 22. All 24=162^4=16 deterministic strategies therefore give ∣S∣=2\lvert S\rvert=2 exactly.
  4. Step 4:
    S=∫p(λ)[A0B0+A0B1+A1B0−A1B1](λ) dλ=∫p(λ) S(λ) dλS=\textstyle\int p(\lambda)\big[A_0B_0+A_0B_1+A_1B_0-A_1B_1\big](\lambda)\,d\lambda=\int p(\lambda)\,S(\lambda)\,d\lambda
    For a general LHV model, each ExyE_{xy} is an average over the same distribution p(λ)p(\lambda), which does not depend on (x,y)(x,y), so the four averages combine into one average of S(λ)S(\lambda).
  5. Step 5:
    ∣S∣≤∫p(λ) ∣S(λ)∣ dλ=∫p(λ)⋅2 dλ=2\lvert S\rvert\le\textstyle\int p(\lambda)\,\lvert S(\lambda)\rvert\,d\lambda=\int p(\lambda)\cdot2\,d\lambda=2
    Triangle inequality for integrals, p(λ)≥0p(\lambda)\ge0 and ∫p(λ) dλ=1\int p(\lambda)\,d\lambda=1: a convex combination of numbers in [−2,2][-2,2] stays in [−2,2][-2,2].
  6. Step 6:
    Ax(λ,μ)={+1μ<P(+1∣x,λ)−1otherwise,μ∼Uniform[0,1)A_x(\lambda,\mu)=\begin{cases}+1&\mu<P(+1\mid x,\lambda)\\-1&\text{otherwise}\end{cases},\qquad\mu\sim\mathrm{Uniform}[0,1)
    Models with random local responses P(±1∣x,λ)P(\pm1\mid x,\lambda) are covered too: adjoin independent uniform variables μ\mu (Alice) and ν\nu (Bob) to the hidden variable. The pair (λ,μ,ν)(\lambda,\mu,\nu) is a new hidden variable with deterministic outcome functions and the same correlators, so the bound applies.
  7. Step 7:
    SΦ+=22>2S_{\Phi^+}=2\sqrt2>2
    The quantum value computed above violates the bound, so quantum correlations of the Bell state cannot be explained by any LHV model.

∎

Theorem

Tsirelson bound

For every quantum state ρ\rho on HA⊗HB\mathcal H_A\otimes\mathcal H_B and Hermitian observables with Ax2=IA_x^2=I and By2=IB_y^2=I, the CHSH operator C=A0⊗(B0+B1)+A1⊗(B0−B1)\mathcal C=A_0\otimes(B_0+B_1)+A_1\otimes(B_0-B_1) satisfies ∣Tr(ρ C)∣≤22\lvert\mathrm{Tr}(\rho\,\mathcal C)\rvert\le2\sqrt2. The settings above attain the bound on ∣Φ+⟩\lvert\Phi^+\rangle.

Proof (8 steps)
  1. Step 1:
    (B0±B1)2=2I±(B0B1+B1B0),(B0+B1)(B0−B1)=−[B0,B1],(B0−B1)(B0+B1)=[B0,B1](B_0\pm B_1)^2=2I\pm(B_0B_1+B_1B_0),\qquad(B_0+B_1)(B_0-B_1)=-[B_0,B_1],\qquad(B_0-B_1)(B_0+B_1)=[B_0,B_1]
    Expand the products using B02=B12=IB_0^2=B_1^2=I; for instance (B0+B1)(B0−B1)=I−B0B1+B1B0−I(B_0+B_1)(B_0-B_1)=I-B_0B_1+B_1B_0-I.
  2. Step 2:
    C2=A02⊗(B0+B1)2+A12⊗(B0−B1)2+A0A1⊗(B0+B1)(B0−B1)+A1A0⊗(B0−B1)(B0+B1)\mathcal C^2=A_0^2\otimes(B_0+B_1)^2+A_1^2\otimes(B_0-B_1)^2+A_0A_1\otimes(B_0+B_1)(B_0-B_1)+A_1A_0\otimes(B_0-B_1)(B_0+B_1)
    Square C\mathcal C with (P⊗Q)(P′⊗Q′)=PP′⊗QQ′(P\otimes Q)(P'\otimes Q')=PP'\otimes QQ'.
  3. Step 3:
    C2=I⊗4I−A0A1⊗[B0,B1]+A1A0⊗[B0,B1]=4I−[A0,A1]⊗[B0,B1]\mathcal C^2=I\otimes4I-A_0A_1\otimes[B_0,B_1]+A_1A_0\otimes[B_0,B_1]=4I-[A_0,A_1]\otimes[B_0,B_1]
    Ax2=IA_x^2=I; the anticommutator terms ±(B0B1+B1B0)\pm(B_0B_1+B_1B_0) cancel between the first two terms, which sum to 4I4I.
  4. Step 4:
    ∥[A0,A1]∥≤∥A0A1∥+∥A1A0∥≤2,∥[B0,B1]∥≤2\lVert[A_0,A_1]\rVert\le\lVert A_0A_1\rVert+\lVert A_1A_0\rVert\le2,\qquad\lVert[B_0,B_1]\rVert\le2
    In the operator norm, ∥Ax∥=1\lVert A_x\rVert=1 because the eigenvalues of AxA_x are ±1\pm1, and ∥PQ∥≤∥P∥∥Q∥\lVert PQ\rVert\le\lVert P\rVert\lVert Q\rVert.
  5. Step 5:
    ∥C∥2=∥C2∥≤4+∥[A0,A1]∥ ∥[B0,B1]∥≤4+4=8\lVert\mathcal C\rVert^2=\lVert\mathcal C^2\rVert\le4+\lVert[A_0,A_1]\rVert\,\lVert[B_0,B_1]\rVert\le4+4=8
    C\mathcal C is Hermitian, so ∥C2∥=∥C∥2\lVert\mathcal C^2\rVert=\lVert\mathcal C\rVert^2; also ∥P⊗Q∥=∥P∥∥Q∥\lVert P\otimes Q\rVert=\lVert P\rVert\lVert Q\rVert and the triangle inequality.
  6. Step 6:
    ∣Tr(ρ C)∣≤∥C∥≤22\lvert\mathrm{Tr}(\rho\,\mathcal C)\rvert\le\lVert\mathcal C\rVert\le2\sqrt2
    ρ=∑ipi∣ψi⟩⟨ψi∣\rho=\sum_ip_i\lvert\psi_i\rangle\langle\psi_i\rvert with pi≥0p_i\ge0, ∑pi=1\sum p_i=1, and ∣⟨ψ∣C∣ψ⟩∣≤∥C∥\lvert\langle\psi\vert\mathcal C\vert\psi\rangle\rvert\le\lVert\mathcal C\rVert for unit vectors.
  7. Step 7:
    [Z,X]=2iY,[B0,B1]=12[Z+X,Z−X]=−[Z,X]=−2iY ⇒ C2=4I−4 Y⊗Y[Z,X]=2iY,\quad[B_0,B_1]=\tfrac12[Z+X,Z-X]=-[Z,X]=-2iY\ \Rightarrow\ \mathcal C^2=4I-4\,Y\otimes Y
    For the settings of this lesson; [Z+X,Z−X]=−[Z,X]+[X,Z]=−2[Z,X][Z+X,Z-X]=-[Z,X]+[X,Z]=-2[Z,X].
  8. Step 8:
    Y⊗Y∣Φ+⟩=12(i⋅i∣11⟩+(−i)(−i)∣00⟩)=−∣Φ+⟩ ⇒ C2∣Φ+⟩=8∣Φ+⟩Y\otimes Y\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\big(i\cdot i\lvert11\rangle+(-i)(-i)\lvert00\rangle\big)=-\lvert\Phi^+\rangle\ \Rightarrow\ \mathcal C^2\lvert\Phi^+\rangle=8\lvert\Phi^+\rangle
    Y∣0⟩=i∣1⟩Y\lvert0\rangle=i\lvert1\rangle and Y∣1⟩=−i∣0⟩Y\lvert1\rangle=-i\lvert0\rangle. Since Y⊗YY\otimes Y has eigenvalues ±1\pm1, C2=4I−4 Y⊗Y\mathcal C^2=4I-4\,Y\otimes Y has eigenvalues 00 and 88, so ∥C∥=22\lVert\mathcal C\rVert=2\sqrt2 exactly for these settings. The value ⟨Φ+∣C∣Φ+⟩=S=22=∥C∥\langle\Phi^+\vert\mathcal C\vert\Phi^+\rangle=S=2\sqrt2=\lVert\mathcal C\rVert computed above therefore attains the bound; equality in ⟨ψ∣C∣ψ⟩≤∥C∥\langle\psi\vert\mathcal C\vert\psi\rangle\le\lVert\mathcal C\rVert for Hermitian C\mathcal C forces C∣Φ+⟩=22 ∣Φ+⟩\mathcal C\lvert\Phi^+\rangle=2\sqrt2\,\lvert\Phi^+\rangle.

∎

Quick check

To three decimal places, what CHSH value SS do the settings A0=ZA_0=Z, A1=XA_1=X, B0,1=12(Z±X)B_{0,1}=\tfrac1{\sqrt2}(Z\pm X) give on ∣Φ+⟩\lvert\Phi^+\rangle?

tolerance ±0.001
TypeScriptrunnable
import { Statevector, reducedStateFromPure, purity, concurrence, expectationObservable, matrix, PAULI_X, PAULI_Z, u3Matrix } from '@/lib/quantum-simulator';

// The Bell circuit CNOT(H x I) on the four basis inputs |ab> (qubit 0 = a = most significant bit)
const fmt = (sv) => sv.amplitudes().map((a) => a.re.toFixed(4)).join(', ');
for (const ab of ['00', '01', '10', '11']) {
  const bell = Statevector.fromBitstring(ab).applyGate('H', [0]).applyGate('X', [1], [], [0]);
  const rhoA = reducedStateFromPure(bell, [0]);
  console.log('|' + ab + '> -> [' + fmt(bell) + ']  purity(rho_A) = ' + purity(rhoA).toFixed(4) + '  concurrence = ' + concurrence(bell).toFixed(4));
}

// CHSH on |Phi+> with A0 = Z, A1 = X, B0 = (Z + X)/sqrt2, B1 = (Z - X)/sqrt2
const phiPlus = Statevector.fromBitstring('00').applyGate('H', [0]).applyGate('X', [1], [], [0]);
const A = [PAULI_Z, PAULI_X];
const B = [matrix.scale(matrix.add(PAULI_Z, PAULI_X), Math.SQRT1_2), matrix.scale(matrix.sub(PAULI_Z, PAULI_X), Math.SQRT1_2)];
const E = (x, y) => expectationObservable(phiPlus, matrix.kron(A[x], B[y]), [0, 1]);
console.log('E00 E01 E10 E11 =', E(0, 0).toFixed(4), E(0, 1).toFixed(4), E(1, 0).toFixed(4), E(1, 1).toFixed(4));
console.log('S =', (E(0, 0) + E(0, 1) + E(1, 0) - E(1, 1)).toFixed(6), ' 2*sqrt(2) =', (2 * Math.SQRT2).toFixed(6));

// Local hidden variables: all 16 deterministic strategies (a0, a1, b0, b1)
let best = 0;
for (let m = 0; m < 16; m++) {
  const [a0, a1, b0, b1] = [0, 1, 2, 3].map((k) => ((m >> k) & 1 ? -1 : 1));
  best = Math.max(best, Math.abs(a0 * (b0 + b1) + a1 * (b0 - b1)));
}
console.log('max |S| over deterministic strategies =', best);

// Singlet invariance: <Psi-|(U x U)|Psi-> = det U = e^{i(phi + lambda)} for U = U3(0.4, 1.3, -0.8)
const U = u3Matrix(0.4, 1.3, -0.8);
const singlet = Statevector.fromBitstring('11').applyGate('H', [0]).applyGate('X', [1], [], [0]);
const overlap = singlet.inner(singlet.clone().applyMatrix(U, [0]).applyMatrix(U, [1]));
console.log('<Psi-|(U x U)|Psi-> =', overlap.re.toFixed(4), overlap.im.toFixed(4), ' e^{0.5 i} =', Math.cos(0.5).toFixed(4), Math.sin(0.5).toFixed(4));

Expected output

|00> -> [0.7071, 0.0000, 0.0000, 0.7071]  purity(rho_A) = 0.5000  concurrence = 1.0000
|01> -> [0.0000, 0.7071, 0.7071, 0.0000]  purity(rho_A) = 0.5000  concurrence = 1.0000
|10> -> [0.7071, 0.0000, 0.0000, -0.7071]  purity(rho_A) = 0.5000  concurrence = 1.0000
|11> -> [0.0000, 0.7071, -0.7071, 0.0000]  purity(rho_A) = 0.5000  concurrence = 1.0000
E00 E01 E10 E11 = 0.7071 0.7071 0.7071 -0.7071
S = 2.828427  2*sqrt(2) = 2.828427
max |S| over deterministic strategies = 2
<Psi-|(U x U)|Psi-> = 0.8776 0.4794  e^{0.5 i} = 0.8776 0.4794

References

  1. M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), §1.3.6 (Bell states), §2.4.3 (partial trace), §2.6 (EPR and the Bell inequality).doi:10.1017/CBO9780511976667
  2. J. S. Bell, On the Einstein Podolsky Rosen paradox, Physics Physique Fizika 1, 195–200 (1964).doi:10.1103/PhysicsPhysiqueFizika.1.195
  3. J. F. Clauser, M. A. Horne, A. Shimony and R. A. Holt, Proposed experiment to test local hidden-variable theories, Physical Review Letters 23, 880–884 (1969).doi:10.1103/PhysRevLett.23.880
  4. B. S. Cirel'son (Tsirelson), Quantum generalizations of Bell's inequality, Letters in Mathematical Physics 4, 93–100 (1980).doi:10.1007/BF00417500

Hands-on labs

  • Prepare the singlet state

    1 step · auto-graded

    Starting from ∣00⟩\lvert00\rangle, prepare the singlet ∣Ψ−⟩=12(∣01⟩−∣10⟩)\lvert\Psi^-\rangle=\tfrac1{\sqrt2}(\lvert01\rangle-\lvert10\rangle) with at most four gates from {H,X,Z,CNOT}\{H,X,Z,\mathrm{CNOT}\}. The grader checks the state fidelity and, separately, the perfectly anti-correlated measurement statistics, so it can tell a wrong relative sign from a wrong pair of basis states.

    Open lab