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Two initial maximally entangled pairs $A{B}_1$ and $B_{2}C$ can be turned into an entangled pair $AC$ by a generalized Bell basis measurement on $B_1{B}_2$, followed by a local correction at an endpoint using its classical label. This is qudit teleportation of $B_1$, including its correlations with $A$, to $C$. The initial two pairs are consumed. Without the classical label, averaging the conditional endpoint states need not leave entanglement.

The root $\omega$ must be a primitive $n$th root of unity, for example $e^{2\pi i/n}$. The printed statement merely says $n$th root; that is insufficient. With $n=2$ and $\omega =1$, the states with $r=0$ and $r=1$ coincide. More generally, a root of order $d<n$ repeats the $r$ labels modulo $d$. For $n=1$ the construction is trivial.

For a primitive root of unity, the generalized Bell basis obeysThe last equality follows from the finite geometric series: a nonzero exponent difference modulo $n$ has sum zero, while zero difference has sum $n$. There are $n^2$ orthonormal vectors in an $n^2$-dimensional Hilbert space, so they form a complete basis.

For qudit teleportation, Alice's input is $|\chi {⟩}_U={\sum }_j{c}_j|j⟩$. Alice and Bob share $|{\psi }_{00}{⟩}_{aB}=n^{-1/2}{\sum }_k|kk⟩$. Define the qudit shift and phase operators by $X|j⟩=|j+1⟩$ and $Z|j⟩={\omega }^j|j⟩$. Alice measures $Ua$ in the generalized Bell basis. On outcome $(r,s)$, Bob's unnormalized state isAll outcomes have probability $1/n^2$. Alice communicates $(r,s)$ and Bob applieswhich restores $|\chi ⟩$ exactly. The protocol consumes one maximally entangled pair of $n$-level systems, a local $n^2$-outcome measurement, and a classical message with $n^2$ possibilities. For a fixed-length binary encoding, $\lceil {\log }_2{n}^2\rceil$ bits suffice. No measurement depends on the unknown amplitudes.

Bob uses qudit teleportation to teleport his carrier $B_1$ to Clare through the pair $B_{2}C$. This transfers Bob's entanglement with Alice to Clare. Explicitly, Bob performs a generalized Bell basis measurement on $B_1{B}_2$. Contraction of the two initial pairs givesThe probability is $1/n^2$ for each outcome. Bob sends $(r,s)$ to Clare, who applies $Z_C^r{X}_C^{-s}$. Alice and Clare then share $|{\psi }_{00}{⟩}_{AC}$ with certainty. This is entanglement swapping; neither an additional entangled pair nor a quantum transmission during the protocol is required.

The two initial pairs are consumed, and Bob's measured carriers cease to be entangled with $AC$. Without Bob's classical record, averaging the possible generalized Bell states leaves $AC$ maximally mixed. Thus the conditional entanglement swapping does not supply faster-than-light communication.

For an $n$-level qudit and primitive root of unity $\omega$, these unitary operators obey $X^n=Z^n=I$ and $ZX=\omega XZ$. The inverse of $X^s{Z}^{-r}$ is $Z^r{X}^{-s}$, with the order reversed. These operators implement the receiver corrections in qudit teleportation.