Lesson 26 · 60 min · core

Superdense Coding

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\lvert\Phi^+\rangle

Prerequisites: Entanglement & the Bell States

Objectives and key results

$(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle$
$Prove that the four encoded states are orthonormal by computing their Gram matrix, and explain why orthonormality means perfect distinguishability.$
$(H\otimes I)\,\mathrm{CNOT}=B^\dagger$
$11$
$\chi\le S(\rho)\le\log_2d$
$1\,\text{ebit}+1\,\text{qubit}\succeq2\,\text{cbits}$

formula · Encoding: $(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\big(\lvert0,b_2\rangle+(-1)^{b_1}\lvert1,\bar b_2\rangle\big)=B\lvert b_1b_2\rangle$
identity · Decoding: $(H\otimes I)\,\mathrm{CNOT}\,(Z^{b_1}X^{b_2}\otimes I)\,\mathrm{CNOT}\,(H\otimes I)\lvert00\rangle=\lvert b_1b_2\rangle$
theorem · Holevo bound: $I(X{:}Y)\le\chi=S\Big(\sum_xp_x\rho_x\Big)-\sum_xp_xS(\rho_x)\le S(\rho)\le\log_2d$
theorem · Resource inequalities: $1\,\text{ebit}+1\,\text{qubit}\succeq2\,\text{cbits},\qquad1\,\text{ebit}+2\,\text{cbits}\succeq1\,\text{qubit}$

The protocol and the four encodings#

Definition

Superdense coding

b_1b_2\in\{00,01,10,11\}

Convention

0

Derivation

The general encoding $(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle=B\lvert b_1b_2\rangle$

Step 1:
$\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\textstyle\sum_{c\in\{0,1\}}\lvert c\rangle\otimes\lvert c\rangle$
$\lvert\Phi^+\rangle=\tfrac1{\sqrt2}(\lvert00\rangle+\lvert11\rangle)$ : both qubits carry the same bit $c$ .
Step 2:
$X^{b_2}\lvert c\rangle=\lvert c\oplus b_2\rangle,\qquad Z^{b_1}\lvert u\rangle=(-1)^{b_1u}\lvert u\rangle$
$X$ flips a bit and $X^0=I$ ; $Z$ multiplies $\lvert1\rangle$ by $-1$ , so $Z^{b_1}$ contributes the sign $-1$ exactly when $b_1=1$ and $u=1$ .
Step 3:
$(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\textstyle\sum_{c}(-1)^{b_1(c\oplus b_2)}\lvert c\oplus b_2\rangle\otimes\lvert c\rangle$
Apply $X^{b_2}$ and then $Z^{b_1}$ to the first factor of each term; the second factor is untouched by $I$ .
Step 4:
$=\tfrac1{\sqrt2}\textstyle\sum_{u}(-1)^{b_1u}\lvert u\rangle\otimes\lvert u\oplus b_2\rangle$
Relabel the sum by Alice's bit $u=c\oplus b_2$ ; since $\oplus$ is its own inverse, $c=u\oplus b_2$ , and $u$ runs over $\{0,1\}$ exactly when $c$ does.
Step 5:
$=\tfrac1{\sqrt2}\big(\lvert0,b_2\rangle+(-1)^{b_1}\lvert1,\bar b_2\rangle\big)=B\lvert b_1b_2\rangle$
Write out $u=0$ (sign $+1$ , Bob's bit $b_2$ ) and $u=1$ (sign $(-1)^{b_1}$ , Bob's bit $1\oplus b_2=\bar b_2$ ), and compare with the Bell-circuit formula with $a=b_1$ , $b=b_2$ .

Result

\boxed{(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle=B\lvert b_1b_2\rangle\qquad\text{(no extra sign for any message)}}

Derivation

The four encoded states, one by one

Step 1:
$b=00:\quad(I\otimes I)\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\lvert\Phi^+\rangle=\tfrac1{\sqrt2}(1,0,0,1)^T$
$Z^0X^0=I$ : Alice does nothing.
Step 2:
$b=01:\quad(X\otimes I)\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\tfrac1{\sqrt2}\big(\lvert10\rangle+\lvert01\rangle\big)=\lvert\Psi^+\rangle=\tfrac1{\sqrt2}(0,1,1,0)^T$
$X$ on qubit $0$ maps $\lvert00\rangle\mapsto\lvert10\rangle$ and $\lvert11\rangle\mapsto\lvert01\rangle$ ; the two qubits now disagree.
Step 3:
$b=10:\quad(Z\otimes I)\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\tfrac1{\sqrt2}\big(\lvert00\rangle-\lvert11\rangle\big)=\lvert\Phi^-\rangle=\tfrac1{\sqrt2}(1,0,0,-1)^T$
$Z$ on qubit $0$ leaves $\lvert00\rangle$ alone and sends $\lvert11\rangle\mapsto-\lvert11\rangle$ , because qubit $0$ is $\lvert1\rangle$ there.
Step 4:
$b=11,\ \text{after }X:\quad(X\otimes I)\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\big(\lvert10\rangle+\lvert01\rangle\big)$
The rightmost factor of $ZX$ acts first; this is the $b=01$ state.
Step 5:
$b=11,\ \text{then }Z:\quad(Z\otimes I)\tfrac1{\sqrt2}\big(\lvert10\rangle+\lvert01\rangle\big)=\tfrac1{\sqrt2}\big(-\lvert10\rangle+\lvert01\rangle\big)=+\lvert\Psi^-\rangle=\tfrac1{\sqrt2}(0,1,-1,0)^T$
Only the term with qubit $0$ in $\lvert1\rangle$ , namely $\lvert10\rangle$ , picks up the sign; reordering gives $\tfrac1{\sqrt2}(\lvert01\rangle-\lvert10\rangle)$ , which is $\lvert\Psi^-\rangle$ with sign $+1$ .
Step 6:
$ZX=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=iY,\qquad XZ=\begin{pmatrix}0&-1\\1&0\end{pmatrix}=-ZX$
Matrix check of the sign. With $Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}$ we get $iY=\begin{pmatrix}0&1\\-1&0\end{pmatrix}=ZX$ . The opposite order $XZ$ would give $(XZ\otimes I)\lvert\Phi^+\rangle=-\lvert\Psi^-\rangle$ , which differs only by a global phase and decodes to the same outcome.

Result

\boxed{00\mapsto\lvert\Phi^+\rangle,\qquad01\mapsto\lvert\Psi^+\rangle,\qquad10\mapsto\lvert\Phi^-\rangle,\qquad11\mapsto+\lvert\Psi^-\rangle}

b_2

Quick check

\lvert\Phi^+\rangle

The encoded states are perfectly distinguishable#

Proposition

The four encoded states are orthonormal

\lvert\beta_{b}\rangle=(P_b\otimes I)\lvert\Phi^+\rangle

Proof (6 steps)

Step 1:
$\langle\Phi^+\vert(A\otimes I)\vert\Phi^+\rangle=\tfrac12\textstyle\sum_{c,c'}\langle c\vert A\vert c'\rangle\langle c\vert c'\rangle=\tfrac12\sum_c A_{cc}=\tfrac12\mathrm{Tr}A$
Expand both copies of $\lvert\Phi^+\rangle=\tfrac1{\sqrt2}\sum_c\lvert cc\rangle$ ; Bob's factor gives $\langle c\vert c'\rangle=\delta_{cc'}$ . This is the case $Q=I$ of $\langle\Phi^+\vert P\otimes Q\vert\Phi^+\rangle=\tfrac12\mathrm{Tr}(P^TQ)$ from the Bell-states lesson.
Step 2:
$\langle\beta_b\vert\beta_{b'}\rangle=\langle\Phi^+\vert(P_b^\dagger P_{b'}\otimes I)\vert\Phi^+\rangle=\tfrac12\mathrm{Tr}\big(P_b^\dagger P_{b'}\big)$
$\langle\beta_b\rvert=\langle\Phi^+\rvert(P_b^\dagger\otimes I)$ and $(P_b^\dagger\otimes I)(P_{b'}\otimes I)=P_b^\dagger P_{b'}\otimes I$ ; then apply the previous line with $A=P_b^\dagger P_{b'}$ .
Step 3:
$b=b':\quad P_b^\dagger P_b=I\ \Rightarrow\ \tfrac12\mathrm{Tr}\,I=1$
Each $P_b$ is unitary ( $ZX=iY$ is a phase times a Pauli), so every encoded state is normalised.
Step 4:
$\mathrm{Tr}X=\mathrm{Tr}Z=0,\qquad\mathrm{Tr}(ZX)=\mathrm{Tr}\begin{pmatrix}0&1\\-1&0\end{pmatrix}=0$
Pairs involving $P_{00}=I$ : $P_{00}^\dagger P_{b'}=P_{b'}$ , and the three non-identity operators are traceless.
Step 5:
$X^\dagger Z=XZ=-iY,\qquad X^\dagger ZX=XZX=-Z,\qquad Z^\dagger ZX=X$
The remaining pairs $(01,10)$ , $(01,11)$ , $(10,11)$ , using $X^\dagger=X$ , $Z^\dagger=Z$ , $XZ=-ZX$ and $X^2=Z^2=I$ ; all three products are traceless. The pairs with $b>b'$ are complex conjugates of these, $\langle\beta_{b'}\vert\beta_b\rangle=\overline{\langle\beta_b\vert\beta_{b'}\rangle}=0$ .
Step 6:
$G_{bb'}=\langle\beta_b\vert\beta_{b'}\rangle=\delta_{bb'}\quad\Longrightarrow\quad G=I_4$
Collect the diagonal ( $=1$ ) and off-diagonal ( $=0$ ) entries.

∎

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 encoded states (rows and columns 00, 01, 10, 11)4×4

\lvert\beta_b\rangle=B\lvert b\rangle

Decoding with the inverse Bell circuit#

Theorem

The decoder returns the message

B=\mathrm{CNOT}(H\otimes I)

Proof (4 steps)

Step 1:
$B^\dagger=\big(\mathrm{CNOT}(H\otimes I)\big)^\dagger=(H\otimes I)^\dagger\,\mathrm{CNOT}^\dagger$
$(UV)^\dagger=V^\dagger U^\dagger$ : the adjoint reverses the order of a product.
Step 2:
$H^\dagger=H,\quad(H\otimes I)^\dagger=H^\dagger\otimes I^\dagger=H\otimes I,\quad\mathrm{CNOT}^\dagger=\mathrm{CNOT}$
$H=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ is real and symmetric; $\mathrm{CNOT}$ is a real permutation matrix that swaps $\lvert10\rangle$ and $\lvert11\rangle$ , hence symmetric.
Step 3:
$B^\dagger=(H\otimes I)\,\mathrm{CNOT}=D,\qquad DB=B^\dagger B=I_4$
Substitute; $B$ is a product of unitaries, hence unitary, so $B^\dagger B=I$ .
Step 4:
$D\,(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle=D\,B\lvert b_1b_2\rangle=\lvert b_1b_2\rangle$
Use the encoding formula $(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle=B\lvert b_1b_2\rangle$ derived above, then $DB=I$ .

∎

H\otimes I

D=(H\otimes I)\,\mathrm{CNOT}=\frac1{\sqrt2}\begin{pmatrix}1&0&0&1\\0&1&1&0\\1&0&0&-1\\0&1&-1&0\end{pmatrix}=\begin{pmatrix}\langle\Phi^+\rvert\\\langle\Psi^+\rvert\\\langle\Phi^-\rvert\\\langle\Psi^-\rvert\end{pmatrix}

Decoder D = (H ⊗ I) CNOT4×4

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

Derivation

Decoding each encoded state gate by gate

Step 1:
$\mathrm{CNOT}\lvert\Phi^\pm\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle\pm\lvert10\rangle\big)=\lvert\pm\rangle\otimes\lvert0\rangle$
$\mathrm{CNOT}\lvert c,t\rangle=\lvert c,t\oplus c\rangle$ : $\lvert00\rangle$ is fixed and $\lvert11\rangle\mapsto\lvert10\rangle$ ; factor out $\lvert0\rangle$ on qubit $1$ with $\lvert\pm\rangle=\tfrac1{\sqrt2}(\lvert0\rangle\pm\lvert1\rangle)$ .
Step 2:
$\mathrm{CNOT}\lvert\Psi^\pm\rangle=\tfrac1{\sqrt2}\big(\lvert01\rangle\pm\lvert11\rangle\big)=\lvert\pm\rangle\otimes\lvert1\rangle$
$\lvert01\rangle$ is fixed and $\lvert10\rangle\mapsto\lvert11\rangle$ . The CNOT has disentangled the pair: qubit $1$ now holds the parity bit $b_2$ and qubit $0$ holds the sign as $\lvert\pm\rangle$ .
Step 3:
$H\lvert+\rangle=\lvert0\rangle,\qquad H\lvert-\rangle=\lvert1\rangle$
$H$ is self-inverse and $H\lvert0\rangle=\lvert+\rangle$ , $H\lvert1\rangle=\lvert-\rangle$ .
Step 4:
$(H\otimes I)\big(\lvert\pm\rangle\otimes\lvert b_2\rangle\big)=\lvert b_1\rangle\otimes\lvert b_2\rangle,\qquad b_1=\begin{cases}0&\text{sign }+\\1&\text{sign }-\end{cases}$
The Hadamard converts the sign into the bit $b_1$ .

Result

\boxed{\lvert\Phi^+\rangle\mapsto\lvert00\rangle,\qquad\lvert\Psi^+\rangle\mapsto\lvert01\rangle,\qquad\lvert\Phi^-\rangle\mapsto\lvert10\rangle,\qquad\lvert\Psi^-\rangle\mapsto\lvert11\rangle}

Truth tableTruthTable

Applies a circuit to every computational-basis input and lists the output basis state (or superposition) for each, exposing reversible-logic behaviour.

Not classical: creates superpositionsAt least one basis input is mapped to a superposition of several basis states.

U|b⟩ for every basis input (qubit 0 leftmost)
Input \|b⟩	Output U\|b⟩	Inspect
\|00⟩	(0.7071)\|00⟩(0.7071)\|10⟩
\|01⟩	(0.7071)\|01⟩(0.7071)\|11⟩
\|10⟩	(0.7071)\|01⟩(−0.7071)\|11⟩
\|11⟩	(0.7071)\|00⟩(−0.7071)\|10⟩

Quick check

\mathrm{CNOT}_{0\to1}

Worked example: the message 11#

Derivation

Every amplitude vector for $b_1b_2=11$

Step 1:
$\lvert00\rangle=(1,0,0,0)^T$
Initial register; qubit $0$ is Alice's, qubit $1$ is Bob's.
Step 2:
$(H\otimes I)\lvert00\rangle=\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert10\rangle\big)=\tfrac1{\sqrt2}(1,0,1,0)^T$
$H\lvert0\rangle=\lvert+\rangle$ on qubit $0$ (start of the sharing step).
Step 3:
$\mathrm{CNOT}\to\tfrac1{\sqrt2}\big(\lvert00\rangle+\lvert11\rangle\big)=\lvert\Phi^+\rangle=\tfrac1{\sqrt2}(1,0,0,1)^T$
$\lvert10\rangle\mapsto\lvert11\rangle$ . The pair is now shared; Bob takes qubit $1$ .
Step 4:
$(X\otimes I)\to\tfrac1{\sqrt2}\big(\lvert10\rangle+\lvert01\rangle\big)=\lvert\Psi^+\rangle=\tfrac1{\sqrt2}(0,1,1,0)^T$
Alice's first encoding gate ( $b_2=1$ ): $\lvert00\rangle\mapsto\lvert10\rangle$ , $\lvert11\rangle\mapsto\lvert01\rangle$ .
Step 5:
$(Z\otimes I)\to\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert10\rangle\big)=\lvert\Psi^-\rangle=\tfrac1{\sqrt2}(0,1,-1,0)^T$
Alice's second encoding gate ( $b_1=1$ ): only $\lvert10\rangle$ has qubit $0$ in $\lvert1\rangle$ , so only it changes sign. Hence $ZX\lvert\Phi^+\rangle=\tfrac1{\sqrt2}(\lvert01\rangle-\lvert10\rangle)$ . Alice now sends qubit $0$ to Bob.
Step 6:
$\mathrm{CNOT}\to\tfrac1{\sqrt2}\big(\lvert01\rangle-\lvert11\rangle\big)=\lvert-\rangle\otimes\lvert1\rangle=\tfrac1{\sqrt2}(0,1,0,-1)^T$
Bob's CNOT: $\lvert01\rangle$ is fixed (control $0$ ), $-\lvert10\rangle\mapsto-\lvert11\rangle$ .
Step 7:
$(H\otimes I)\to\tfrac1{\sqrt2}\Big(\tfrac{\lvert0\rangle+\lvert1\rangle}{\sqrt2}-\tfrac{\lvert0\rangle-\lvert1\rangle}{\sqrt2}\Big)\otimes\lvert1\rangle=\lvert1\rangle\otimes\lvert1\rangle=(0,0,0,1)^T$
$H\lvert0\rangle=\lvert+\rangle$ and $H\lvert1\rangle=\lvert-\rangle$ ; the $\lvert0\rangle$ components cancel and the $\lvert1\rangle$ components add to $\tfrac{2}{2}=1$ .

Result

\boxed{(H\otimes I)\,\mathrm{CNOT}\,(Z\otimes I)(X\otimes I)\,\mathrm{CNOT}\,(H\otimes I)\lvert00\rangle=\lvert11\rangle}

Circuit · 2 qubits

11

\lvert\psi\rangle = 1\lvert 11\rangle

|00⟩0
|01⟩0
|10⟩0
|11⟩1

TypeScriptrunnable

import { Statevector, reducedStateFromPure, vonNeumannEntropy, DensityMatrix, matrix } from '@/lib/quantum-simulator';

// Superdense coding for all four messages b1 b2 (qubit 0 = Alice = most significant bit, qubit 1 = Bob)
const fmt = (sv) => sv.amplitudes().map((a) => a.re.toFixed(4)).join(', ');
let average = matrix.zeros(4, 4);
for (const msg of ['00', '01', '10', '11']) {
  // Share |Phi+> = CNOT (H x I)|00>
  const sv = Statevector.zero(2).applyGate('H', [0]).applyGate('X', [1], [], [0]);
  // Alice encodes with Z^b1 X^b2 on her qubit: X first (b2), then Z (b1)
  if (msg[1] === '1') sv.applyGate('X', [0]);
  if (msg[0] === '1') sv.applyGate('Z', [0]);
  const encoded = fmt(sv);
  const sAlice = vonNeumannEntropy(reducedStateFromPure(sv, [0]));
  average = matrix.add(average, matrix.scale(DensityMatrix.fromStatevector(sv).toMatrix(), 0.25));
  // Bob decodes with (H x I) CNOT: CNOT(0 -> 1), then H on qubit 0
  sv.applyGate('X', [1], [], [0]).applyGate('H', [0]);
  const p = Array.from(sv.probabilities());
  const k = p.indexOf(Math.max(...p));
  console.log('message ' + msg + ': encoded [' + encoded + ']  S(rho_Alice) = ' + sAlice.toFixed(4) + '  decoded ' + k.toString(2).padStart(2, '0') + ' with P = ' + p[k].toFixed(4));
}
// Holevo quantity of the ensemble Bob receives: chi = S(average) - (1/4) sum S(pure) = S(I/4)
console.log('chi (two qubits at Bob) = ' + vonNeumannEntropy(average).toFixed(4) + ' bits');

Expected output

message 00: encoded [0.7071, 0.0000, 0.0000, 0.7071]  S(rho_Alice) = 1.0000  decoded 00 with P = 1.0000
message 01: encoded [0.0000, 0.7071, 0.7071, 0.0000]  S(rho_Alice) = 1.0000  decoded 01 with P = 1.0000
message 10: encoded [0.7071, 0.0000, 0.0000, -0.7071]  S(rho_Alice) = 1.0000  decoded 10 with P = 1.0000
message 11: encoded [0.0000, 0.7071, -0.7071, 0.0000]  S(rho_Alice) = 1.0000  decoded 11 with P = 1.0000
chi (two qubits at Bob) = 2.0000 bits

Why entanglement is essential: the Holevo bound#

Definition

Holevo quantity

x

Proposition

$\chi\le S(\rho)\le\log_2d$

\{p_x,\rho_x\}

Proof (5 steps)

Step 1:
$S(\rho_x)=-\textstyle\sum_k\mu_k\log_2\mu_k\ge0$
$\mu_k\in[0,1]$ are the eigenvalues of $\rho_x$ , so each term $-\mu_k\log_2\mu_k\ge0$ (with $0\log0:=0$ ).
Step 2:
$\chi=S(\rho)-\textstyle\sum_xp_xS(\rho_x)\le S(\rho)$
Subtracting a non-negative quantity ( $p_x\ge0$ , $S(\rho_x)\ge0$ ).
Step 3:
$S(\rho)=\textstyle\sum_{k:\lambda_k>0}\lambda_k\log_2\frac1{\lambda_k}\le\log_2\Big(\sum_{k:\lambda_k>0}\lambda_k\cdot\frac1{\lambda_k}\Big)$
$\lambda_k$ are the eigenvalues of $\rho$ , a probability distribution. $\log_2$ is concave, so Jensen's inequality $\sum_k\lambda_kf(t_k)\le f\big(\sum_k\lambda_kt_k\big)$ applies with $t_k=1/\lambda_k$ .
Step 4:
$=\log_2\#\{k:\lambda_k>0\}=\log_2\mathrm{rank}\,\rho\le\log_2d$
Each positive eigenvalue contributes $\lambda_k\cdot\tfrac1{\lambda_k}=1$ , and a $d\times d$ matrix has at most $d$ non-zero eigenvalues. Equality holds iff $\rho=I/d$ .
Step 5:
$d=2:\quad I(X{:}Y)\le\chi\le S(\rho)\le\log_22=1$
Combine with the Holevo bound $I(X{:}Y)\le\chi$ .

∎

d=4

Derivation

The travelling qubit alone carries no information

Step 1:
$\rho_A^{(b)}=\mathrm{Tr}_B\lvert\beta_b\rangle\langle\beta_b\rvert,\qquad\lvert\beta_b\rangle=\tfrac1{\sqrt2}\big(\lvert0\rangle\lvert b_2\rangle+(-1)^{b_1}\lvert1\rangle\lvert\bar b_2\rangle\big)$
An eavesdropper Eve who intercepts only the qubit in transit holds Alice's qubit $A$ (qubit $0$ ), whose state is the partial trace over Bob's qubit $B$ .
Step 2:
$\lvert\beta_b\rangle\langle\beta_b\rvert=\tfrac12\Big(\lvert0\rangle\langle0\rvert\otimes\lvert b_2\rangle\langle b_2\rvert+(-1)^{b_1}\lvert0\rangle\langle1\rvert\otimes\lvert b_2\rangle\langle\bar b_2\rvert+(-1)^{b_1}\lvert1\rangle\langle0\rvert\otimes\lvert\bar b_2\rangle\langle b_2\rvert+\lvert1\rangle\langle1\rvert\otimes\lvert\bar b_2\rangle\langle\bar b_2\rvert\Big)$
Expand the outer product of the two-term vector; the sign appears once in each cross term and $((-1)^{b_1})^2=1$ in the second diagonal term.
Step 3:
$\mathrm{Tr}_B\big(\lvert a\rangle\langle a'\rvert\otimes\lvert u\rangle\langle v\rvert\big)=\langle v\vert u\rangle\,\lvert a\rangle\langle a'\rvert,\qquad\langle b_2\vert b_2\rangle=\langle\bar b_2\vert\bar b_2\rangle=1,\quad\langle\bar b_2\vert b_2\rangle=0$
Definition of the partial trace on product operators, and orthonormality of $\lvert0\rangle,\lvert1\rangle$ ( $b_2\neq\bar b_2$ ).
Step 4:
$\rho_A^{(b)}=\tfrac12\big(\lvert0\rangle\langle0\rvert+\lvert1\rangle\langle1\rvert\big)=\tfrac12I\qquad\text{for all four messages }b$
The cross terms vanish because Bob's bits in the two branches are different; the diagonal terms survive with weight $\tfrac12$ each.
Step 5:
$\chi_{\mathrm{Eve}}=S\Big(\tfrac14\textstyle\sum_b\tfrac12I\Big)-\tfrac14\sum_bS\big(\tfrac12I\big)=1-1=0\ \Longrightarrow\ I(X{:}Y_{\mathrm{Eve}})=0$
Every message produces the same state $\tfrac12I$ (entropy $\log_22=1$ ), so the Holevo quantity of Eve's ensemble is $0$ and no measurement on the intercepted qubit reveals anything about the message.

Result

\boxed{\mathrm{Tr}_B\big[(Z^{b_1}X^{b_2}\otimes I)\lvert\Phi^+\rangle\langle\Phi^+\rvert(Z^{b_1}X^{b_2}\otimes I)^\dagger\big]=\tfrac12I\quad\text{for every }b_1b_2}

\mathrm{Tr}_A\lvert\beta_b\rangle\langle\beta_b\rvert=\tfrac12I

Proposition

Two bits per qubit is the most entanglement can give

\rho_{QB}

Proof (4 steps)

Step 1:
$\rho_B^{(x)}=\mathrm{Tr}_Q\,\rho_{QB}^{(x)}=\rho_B\quad\text{for every }x$
An operation on $Q$ alone does not change the reduced state of $B$ (the trace over $Q$ absorbs any trace-preserving map on $Q$ ): this is the no-signalling property already seen above.
Step 2:
$S\big(\textstyle\sum_xp_x\rho_{QB}^{(x)}\big)\le S(\rho_Q)+S(\rho_B)\le\log_2d_Q+S(\rho_B)$
Subadditivity $S(QB)\le S(Q)+S(B)$ for the average state, whose $B$ -marginal is $\rho_B$ by the previous line, and $S(\rho_Q)\le\log_2d_Q$ from the proposition above.
Step 3:
$S\big(\rho_{QB}^{(x)}\big)\ge S(\rho_B)-S\big(\rho_Q^{(x)}\big)\ge S(\rho_B)-\log_2d_Q$
Araki–Lieb triangle inequality $S(QB)\ge\lvert S(B)-S(Q)\rvert$ , again with $S(\rho_Q^{(x)})\le\log_2d_Q$ .
Step 4:
$\chi\le\log_2d_Q+S(\rho_B)-\textstyle\sum_xp_x\big(S(\rho_B)-\log_2d_Q\big)=2\log_2d_Q$
Insert both bounds into $\chi=S(\sum_xp_x\rho^{(x)}_{QB})-\sum_xp_xS(\rho^{(x)}_{QB})$ and use $\sum_xp_x=1$ .

∎

Quick check

\chi_{\mathrm{Eve}}

Quick check

\chi

Duality with teleportation#

Definition

Resource inequality

1\,\text{ebit}

\underbrace{1\,\text{ebit}+1\,\text{qubit}\ \succeq\ 2\,\text{cbits}}_{\text{superdense coding}}\qquad\qquad\underbrace{1\,\text{ebit}+2\,\text{cbits}\ \succeq\ 1\,\text{qubit}}_{\text{teleportation}}

References

C. H. Bennett and S. J. Wiesner, Communication via one- and two-particle operators on Einstein-Podolsky-Rosen states, Physical Review Letters 69, 2881–2884 (1992).doi:10.1103/PhysRevLett.69.2881
M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), §2.3 (superdense coding), §12.1.1 (the Holevo bound).doi:10.1017/CBO9780511976667
A. S. Holevo, Bounds for the quantity of information transmitted by a quantum communication channel, Problemy Peredachi Informatsii 9(3), 3–11 (1973); English translation: Problems of Information Transmission 9, 177–183 (1973).

Hands-on labs

Send the message 11 with one qubit
1 step · auto-graded
$0$
Open lab

Input \|b⟩	Output U\|b⟩	Inspect
\|00⟩	(0.7071)\|00⟩(0.7071)\|10⟩
\|01⟩	(0.7071)\|01⟩(0.7071)\|11⟩
\|10⟩	(0.7071)\|01⟩(−0.7071)\|11⟩
\|11⟩	(0.7071)\|00⟩(−0.7071)\|10⟩

The protocol and the four encodings#

Superdense coding

The encoded states are perfectly distinguishable#

Decoding with the inverse Bell circuit#

Worked example: the message 11#

Why entanglement is essential: the Holevo bound#

Holevo quantity

Duality with teleportation#

Resource inequality

References

Hands-on labs

Send the message 11 with one qubit