Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 59 (Quantum Information, Entanglement and Nonlocality)

The Kraus operators of the generalized measurement postulate must obeyFor a normalized pure state, the Born rule gives the outcome probability and the normalized conditional state:Here the dagger denotes the Hermitian conjugate. Completeness ensures ${\sum }_i{p}_i=1$. An outcome with $p_i=0$ never occurs for that input, so no conditional state is required there. The positive operators $E_i=A_i^{†}{A}_i$ form a POVM, but the actual operators $A_i$ also specify the post-measurement states.

Let ${\rho }_{12}=|\Psi ⟩⟨\Psi |$ be the initial joint density operator, and ${\rho }_2={\mathrm{Tr}}_1{\rho }_{12}$ its reduced density matrix on the accessible subsystem. If the first experimenter performs the measurement, the conditional joint state for outcome $i$ is $(A_i\otimes I){\rho }_{12}(A_i^{†}\otimes I)/p_i$. The second experimenter does not know that outcome. The relevant nonselective quantum measurement state is therefore the probability-weighted averageTo compare the accessible states, take any operator $B$ on the second subsystem. Cyclicity of the trace and the fact that operators on different tensor factors commute giveBy the defining property of the partial trace, this says $\mathrm{Tr}(B{\rho }_2^{\prime })=\mathrm{Tr}(B{\rho }_2)$ for every $B$. Matrix units, or arbitrary Hermitian test operators, distinguish any two different density matrices, soThis proves remote state invariance under a local trace-preserving operation directly from completeness.

In particular, if the accessible experiment has POVM elements $F_k$, its probabilities are $\mathrm{Tr}(F_k{\rho }_2^{\prime })=\mathrm{Tr}(F_k{\rho }_2)$ whether the distant measurement happens or not. Attaching an independent local ancilla, evolving locally, or performing a sequence of local measurements cannot help: these operations act on the identical input density operator and therefore give identical outcome distributions. No experiment confined to the second subsystem can distinguish whether the unannounced measurement occurred. This is a derivation of the no-communication theorem, not an appeal to a signalling postulate.

The qualification is that the outcome is not communicated. A selective quantum measurement can give different conditional remote states, as happens when measuring one half of a Bell state; their probability-weighted average is still ${\rho }_2$. An observer supplied with the distant outcome has extra classical information and is no longer restricted to the unannounced local experiment considered here.

Write $|0⟩=|\uparrow ⟩$, $|1⟩=|\downarrow ⟩$, and use $X_j,Y_j$ for the Pauli matrices on particle $j$. Their actions on the computational basis areSet $P_1=X_1{Y}_2{Y}_3$, $P_2=Y_1{X}_2{Y}_3$, $P_3=Y_1{Y}_2{X}_3$, and $P_4=X_1{X}_2{X}_3$. Any two distinct operators in this list differ by an $X$ versus $Y$ at exactly two sites. At each such site $XY=-YX$, while factors on different particles commute. Swapping the two full products therefore introduces two minus signs, givingThis is the Pauli-string commutation parity rule. Each product is Hermitian and squares to the identity, so its only eigenvalues are $\pm 1$.

For the negative-phase Greenberger–Horne–Zeilinger state, put $|G_{-}⟩=(|000⟩-|111⟩)/\sqrt{2}$. The two $Y$ factors in each of $P_1,P_2,P_3$ contribute $i^2=-1$ on $|000⟩$ and $(-i{)}^2=-1$ on $|111⟩$. Thus these operators take $|000⟩$ to $-|111⟩$ and $|111⟩$ to $-|000⟩$. By contrast, $P_4$ flips these two basis states without a phase. ConsequentlyThe GHZ state is therefore a simultaneous eigenstate of the four commuting products. As a sign check, direct multiplication givesThis is compatible with the displayed quantum eigenvalues.

Now apply the EPR criterion of reality together with locality for the spacelike separated particles. In the context with local settings $X_1,Y_2,Y_3$, the product of the three outcomes is certainly $+1$. Therefore the two remote $Y$ outcomes predict the first particle's $X$ outcome with certainty, without disturbing it under the locality assumption. The other perfect-correlation contexts similarly predict each particle's $X$ or $Y$ outcome from measurements performed only on the other particles. The EPR criterion of reality then assigns predetermined local values $x_j,y_j\in \{+1,-1\}$ to these six observables. Locality makes the value assigned at one site independent of which measurements were chosen at the other sites.

To reproduce the three certain mixed-setting quantum predictions, every such assignment must satisfyMultiplying these scalar equations cancels each squared $y_j$ and gives $x_1{x}_2{x}_3=+1$. But the all-$X$ quantum prediction, from the $P_4$ eigenvalue, is $x_1{x}_2{x}_3=-1$ with certainty. HenceThe perfect GHZ correlations contradict the simultaneous local elements of reality inferred by the EPR argument. No statistical Bell inequality is needed. The locality assumption is essential to that inference; the certainty criterion without the premise that distant choices cannot disturb the local element does not by itself imply a local hidden-variable model. Nor does the argument require simultaneous measurement of $X$ and $Y$ on a single particle: different experimental contexts establish the perfect predictions, and it is the EPR/locality interpretation that attempts to assign all their values at once.

Take the labels $r,s\in \{0,\dots ,n-1\}$, with basis indices read modulo $n$. There is a missing hypothesis in the printed claim: the root of unity must be primitive. For example, when $n=2$ and $\omega =1$, the two labels $(0,0)$ and $(1,0)$ produce the identical state $(|00⟩+|11⟩)/\sqrt{2}$, so they are not orthogonal. More generally, if the order of $\omega$ is $d<n$, changing $r$ to $r+d$ repeats a state. The orthonormality claim is false for a nonprimitive root.

For the intended result, choose a primitive root of unity, for example $\omega =e^{2\pi i/n}$. Since $|\omega |=1$, expansion in the orthonormal basis givesIf $r=r^{\prime }$, the sum is $n$. If $r\ne r^{\prime }$ modulo $n$, primitiveness gives $q={\omega }^{r^{\prime }-r}\ne 1$ while $q^n=1$, and the finite geometric series is $(1-q^n)/(1-q)=0$. ThusThere are $n^2$ vectors in an $n^2$-dimensional Hilbert space, so they form the generalized Bell basis. Each has equal Schmidt coefficients $1/\sqrt{n}$ and is a maximally entangled state.

For qudit teleportation, let $C$ be Alice's unknown input, write $|\chi {⟩}_C={\sum }_j{\alpha }_j|j{⟩}_C$, and let Alice and Bob share $|\Phi {⟩}_{AB}=n^{-1/2}{\sum }_k|k{⟩}_A|k{⟩}_B$. Alice holds $C,A$ and Bob holds $B$. Define the qudit shift and phase operatorsBoth are unitary. Alice makes a joint projective measurement of $C,A$ in the generalized Bell basis, with projectors $|{\psi }_{rs}⟩⟨{\psi }_{rs}|$. Completeness follows from the orthonormality just proved.

For outcome $(r,s)$, contracting the corresponding bra with the joint input and resource gives Bob's unnormalized state:The Born rule therefore gives $p_{rs}=1/n^2$, since $X^s{Z}^{-r}$ preserves the input norm. Conditional on this outcome, Bob holds $X^s{Z}^{-r}|\chi ⟩$. Alice sends him the classical label $(r,s)$, and he applies its inverse in the displayed order:The corrected receiver state is exactly the unknown input state for every outcome. Alice needs no knowledge of the amplitudes. The label has $n^2$ possibilities and can be encoded in $\lceil {\log }_2(n^2)\rceil$ classical bits; when $n$ is a power of two this is exactly $2{\log }_{2}n$. The protocol consumes the shared maximally entangled state and the original carrier is measured, so it does not retain an additional copy of the unknown state.

Equivalently, the complete joint quantum state has the teleportation expansionIt follows from the contraction calculation and completeness of the generalized Bell basis. After correction the branch operators are all $I/n$, so their sum defines the identity quantum channel: ${\sum }_{r,s}(I/n)\rho (I/n)=\rho$. Acting on the input part of a joint state with a reference gives the same identity, which proves teleportation as an identity channel on a reference and extends the protocol to mixed inputs.

Before receiving the label, Bob instead has the outcome-averaged stateTo check the equality, averaging over $r$ kills the off-diagonal matrix entries by the same root-of-unity sum; averaging the cyclic shifts $s$ makes every diagonal entry the total trace divided by $n$. This is the Heisenberg-Weyl twirling channel. Thus the classical message is necessary to recover the input, consistently with the no-communication theorem derived in Question 1.

Use $\oplus$ for addition modulo two, or exclusive or. The key is to write the known Boolean function as an XOR of products of locally computable bits. An explicit separated Boolean decomposition avoids any need to assume a particular circuit for the function. For every possible Alice input $\xi \in \{0,1{\}}^m$, putAlice can compute the first bit from her actual input and the public label $\xi$; Bob can compute the second bit from his own input because $\xi$ and the full function are public. Exactly one of the indicators is one, namely the one with $\xi =x$. Hence

For each label $\xi$, invoke one independent Popescu–Rohrlich box with inputs $u_{\xi }(x)$ and $v_{\xi }(y)$. Denote its outputs by $a_{\xi }$ and $b_{\xi }$. By the given correlation, $a_{\xi }\oplus b_{\xi }=u_{\xi }{v}_{\xi }$. Alice locally forms $A={⨁}_{\xi }{a}_{\xi }$, and Bob locally forms $B={⨁}_{\xi }{b}_{\xi }$. Taking the XOR of all box constraints yieldsThus the protocol isIt succeeds exactly for every input and every allowed sequence of box outputs. Randomness of the individual outputs introduces no error because each parity relation is exact. All box calls can be made before the message, with no extra classical communication. This proves one-bit computation using Popescu–Rohrlich boxes.

The truth-table construction uses $2^m$ PR boxes. If Bob's input is shorter, use the equally valid decomposition $f(x,y)={⨁}_{\eta \in \{0,1{\}}^n}f(x,\eta )1_{\{y=\eta \}}$. Alice computes the first factor, Bob the indicator, and Alice still sends her output parity. Therefore at most $2^{\min (m,n)}$ boxes are enough. This is a bound on box use, not on local computational work; the communication complexity remains one bit.

Another standard route is the algebraic normal form over the two-element field:The empty products are one. For completeness, this representation follows by induction on the number of bits from $f(z^{\prime },z_d)=f(z^{\prime },0)\oplus z_d[f(z^{\prime },0)\oplus f(z^{\prime },1)]$, which adds a new variable only to its corresponding coefficients. Each nonzero monomial supplies one local product to each party and can be evaluated in distributed parity form by one PR box. Terms depending on only one party can instead be included directly in that party's final parity; the constant may be assigned to either side. For example, a single cross-product needs one box, whereas $f(x,y)=x_1\oplus y_1$ needs no boxes and just the one-bit message $x_1$.

Finally, the scheme does not make the boxes themselves signalling devices. At any fixed box inputs, the two allowed output pairs give each party an individually uniform bit, independent of the distant input. Independence across box calls makes Bob's entire uncommunicated output string uniform, for any Alice input. Without $A$ he generally cannot identify the answer. The communicated parity combines with the nonlocal correlations to reveal it; it does not contradict the uniform local marginals of these no-signalling boxes.