Is it possible to encrypt a quantum state, such that a $BQP$ attacker who does not know the secret key cannot obtain any information about the original state, but a $BQP$ decryptor with the key can recover the original state?

Assume (classical) one-way functions that are hard for quantum attackers to break exist (as is believed to be the case).


1 Answer 1


One can encrypt an n-qubit state using a 2n-bit classical secret key. The idea is to use the key to select a random Pauli operator, and apply that operator to the secret as an encryption. (The inverse operator is applied to decrypt.)

The resulting scheme is perfectly secure -- if the key is selected uniformly at random, then even an attacker who know a state entangled with the plaintext cannot distinguish the true ciphertext from an independent random state.

This observation was first made in A Ambainis, M Mosca, A Tapp, R de Wolf. "Private quantum channels", FOCS 2000.

They also showed that 2n bits of key are necessary for encryption of entangled quantum states.

If you have a classical PRG secure against quantum distinguishers, then you can get away with a much shorter key (as short as seed length required by the generator to generate 2 n output bits).

  • $\begingroup$ Is this analogous to a one-time pad? $\endgroup$
    – Demi
    Apr 10, 2016 at 2:11
  • 2
    $\begingroup$ Yes, it is in fact called a quantum one-time pad. In the quantum case, you have to be a bit careful in how you define the security requirement for computational or statistical notions of secrecy. But for perfect secrecy, it is fairly straightforward: for every input state, the density matrix of the ciphertext should be the same when you average over keys. $\endgroup$
    – Adam Smith
    Apr 10, 2016 at 2:42
  • $\begingroup$ Ah okay. So the version with a CSPRNG is a quantum stream cipher. $\endgroup$
    – Demi
    Apr 10, 2016 at 2:43

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