July 11, 2026. Turns out you can’t send quantum information without first cloning and encrypting it.

Introduction

Famously, quantum information cannot be cloned, in the sense that there is no unitary operator $U$ which copies an arbitrary state $\vert\psi\rangle$ onto an ancilla,

\[U |\psi\rangle |\phi_0\rangle = |\psi\rangle|\psi\rangle.\]

This is true, but as Yamaguchi and Kempf (2025) argue, it is also unimaginative. They invent a scheme for cloning encrypted states, such that

\[U_E |\psi\rangle |\phi_0\rangle = |\psi\rangle E |\psi\rangle,\]

where $E$ is an operator that cryptographically obfuscates the state. Once you can copy once, you can copy multiple times, but they show that despite this, only one copy can ever be recovered! It’s a beautiful scheme. But the point of this post is to argue that, in some sense, the familiar quantum phenomenon of teleportation already has this form.

Teleportation

The basic idea is hopefully familiar: Alice wants to send a quantum state $\vert\psi\rangle$ to Bob, and they share a maximally entangled Bell pair,

\[|\Phi\rangle_{AB} = \frac{1}{\sqrt{2}}\big(|0\rangle_A |0\rangle_B + |1\rangle_A |1\rangle_B\big),\]

where I omit the tensor product for simplicity. The full system is initially $|\psi\rangle_A|\Phi\rangle_{AB}$. Alice then measures her qubits in the Bell basis, consisting of four states

\[|\Phi_{ab}\rangle = (I \otimes Z^b X^a) |\Phi\rangle = \frac{1}{\sqrt{2}}\bigg(|0\rangle_A |a\rangle_B + (-1)^b|1\rangle_A |\overline{a}\rangle_B\bigg),\]

for $a, b = 0, 1$, $\overline{a} = 1 - a$, and up to an overall minus sign we can ignore. This measurement couples Alice’s state $|\psi\rangle$ to her second qubit, which is entangled with Bob; when she measures, this has the effect of sending the state $\vert\psi\rangle$ through to Bob, with some corrections. More precisely, some algebra shows that

\[|\psi\rangle_A |\Phi\rangle_{AB} = \sum_{a, b} |\Phi_{ab}\rangle_A\big( X^a Z^b |\psi\rangle\big),\]

so after Alice measures $a, b$, Bob has state $X^a Z^b \vert \psi\rangle$. Alice sends Bob $a$ and $b$ over a classical channel, so he can undo $X^a Z^b$ and then voila! He has the state $\vert\psi\rangle$. If you want more detail on why it works, feel free to check out the explainer I wrote a while back. Admittedly, starting with the Coffman-Kundu-Wootters inequality may make things more confusing.

Encrypted cloning

The thing about entanglement is that it allows for “spooky action at a distance”, and in the case of teleportation, the state $X^a Z^b \vert \psi\rangle$ appears instantaneously in Bob’s system when Alice measures in the Bell basis. This is deeply disturbing, and in fact, gets even more disturbing: in a frame of reference moving relative to Alice and Bob, Bob receives the state $X^a Z^b \vert\psi\rangle$ before Alice measures in the Bell basis! So there are two copies of the state floating around. Formally, tracing out the middle qubit and working with density matrices, we have:

\[|\psi\rangle\langle\psi|_A \otimes \frac{1}{2}I_B \mapsto |\psi\rangle\langle\psi|_A \otimes X^aZ^b|\psi\rangle\langle\psi|_B Z^bX^a\]

since Bob’s qubit is maximally entangled, so when we trace out its partner, it’s in the maximally mixed state. This is encrypted cloning where the role of the encrypting operator is played by $E = X^aZ^b$. This encryption is essential: it prevents Alice and Bob both from cloning and from sending information faster than light, and in both cases it does so by completely obfuscating the identity of the state.

The proof of this is very neat. It turns out sums of products of the form $X^aZ^b$ provide a linear basis for density matrices (there are four such products and they are linearly independent), so

\[\rho = \sum_{a,b} c_{a,b} X^a Z^b, \quad c_{a,b} = \frac{1}{2}\mbox{tr}[\rho Z^bX^a].\]

From Bob’s perspective, the bits $a$ and $b$ are random, and any particular combination occurs with probability $p = 1/4$. So his density matrix is

\[\rho_B = p\sum_{ab} X^a Z^b |\psi\rangle\langle \psi| Z^b X^a = p\sum_{abcd} (-1)^{ad}(-1)^{bc}c_{cd} X^{c} Z^{d} = c_{00} I = \frac{1}{2}I_B,\]

using the following facts:

• $\sum_{a} (-1)^{ad} = 2\delta_{d,0}$ and similarly for $bc$;
• $X^a Z^b = (-1)^{ab} Z^b X^a$;
• and $c_{00} = \mbox{tr}[\rho]/2 = 1/2$.

Thus, Bob’s density matrix is maximally mixed and he knows absolutely nothing about the state! This means teleportation transfers no information until Alice supplies her Bell bits $a,b$, and there is no contradiction in cloning the state; it is hidden inaccessibly behind a random $X^a Z^b$.

Conclusion

What is genuinely different about Yamaguchi and Kempf’s construction is that multiple copies get encrypted and only one can be deciphered; efforts to generalize this to teleportation lead to rather different-looking techniques like port-based teleportation which are structurally distinct from the standard protocol, so I won’t go into them here. But it’s neat that what is perhaps the simplest quantum protocol is also the simplest example of quantum encrypted cloning!