Nature of the one-way function

A classical one-way function as said above is based on a classical infeasible mathematical task, whereas a quantum one-way function exploits the uncertainty principle which makes it impossible even for a quantum computer to compute the inverse. This is done by providing a quantum output state, with whom one cannot learn enough about the input string to reproduce it. In case of the first group of schemes this is shown by Holevo's theorem, which says, that from a given n-qubit quantum state one cannot extract more than n classical bits of information.^[3] One possibility to ensure that the scheme uses less qubits for a bit string of a certain length is by using nearly orthogonal states

${\displaystyle |\langle f_{k}|f_{k}'\rangle |<\delta \qquad {\text{ for }}k\neq k'\land 0\leq \delta \leq 1}$

That gives us the possibility to induce a basis with more than two states.^[1] So to describe an information of ${\displaystyle 2^{n}}$ bits, we can use less than n qubits. An example with a 3 qubit basis

• ${\displaystyle |0\rangle }$
• ${\displaystyle |1\rangle }$
• ${\displaystyle {\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )}$

Only m qubits are needed to describe n classical bits when ${\displaystyle 3^{m}=2^{n}}$ holds.

Because of Holevo's theorem and the fact, that m can be much smaller than n, we can only get m bits out of the n bits message. More general, if one gets T copies of the public key he can extract at most Tm bits of the private key. If ${\displaystyle \delta }$ is big ${\displaystyle n-Tm}$ becomes very large, which makes it impossible for a dishonest person to guess the sign key.

Note: You cannot distinguish between non-orthogonal states, if you only have a small amount of identical states. That's how the quantum one-way functions works.
Nevertheless ${\displaystyle |f_{k}\rangle }$ leaks information about the private key, in contrast to the classical public key, which forces one to get nothing or all about the private key.

Copying the public key

In the classical case we create a classical public key out of a classical sign key, thus it is easy to provide every potential recipient with a copy of the public key. The public key can be freely distributed. This becomes more difficult in the quantum case, because copying a quantum state is forbidden by the no cloning theorem, as long as the state itself is unknown.^[4] So public keys can only be created and distributed by a person who knows the exact quantum state he wants to create, thus who knows the sign key (This can be the sender or in more general a trustful institution). Nevertheless, in contrast to the classical public key there is an upper bound for the number of public quantum keys T which can be created, without enabling one to guess the sign key and thus endangering the security of the scheme (${\displaystyle n-Tm}$ has to be big)

Public Key should be the same for every recipient (Swap Test)

To make sure that every recipient gets identical results when testing the authenticity of a message, public keys distributed have to be the same. This is straightforward in the classical case, because one can easily compare two classical bit strings and see if those match. Nevertheless, in the quantum state it is more complicated. To test, if two public quantum states are the same one has to compare the following

${\displaystyle |f_{k}'\rangle =|f_{k}\rangle }$

Swap test for qubits

This is done with the following quantum circuit which uses one Fredkin gate F, one Hadamard gate H and an ancilla qubit a. First of all the ancilla qubit is set to a symmetric state ${\displaystyle |a\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )}$.

Right after the ancilla qubit is used as a control on the targets ${\displaystyle |f_{k}\rangle \ }$ and ${\displaystyle \ |f_{k}'\rangle }$ in a Fredkin Gate.

Furthermore, a Hadamard gate is applied on the ancilla qubit and finally the first qubit gets measured. If both states are the same, the result ${\displaystyle |0\rangle }$ is measured. If both states are nearly orthogonal, the result can be either ${\displaystyle |0\rangle \ }$ or ${\displaystyle \ |1\rangle }$.^[1]

The calculation of the swap test in more detail:

The overall state

${\displaystyle |\psi _{0}\rangle =|a\rangle |f_{k}\rangle |f_{k}'\rangle }$

${\displaystyle |\psi _{0}\rangle ={\frac {1}{\sqrt {2}}}{\bigg (}|0\rangle +|1\rangle {\bigg )}|f_{k}\rangle |f_{k}'\rangle }$

${\displaystyle |\psi _{0}\rangle ={\frac {1}{\sqrt {2}}}{\bigg (}|0\rangle |f_{k}\rangle |f_{k}'\rangle +|1\rangle |f_{k}\rangle |f_{k}'\rangle {\bigg )}}$

After the Fredkin gate is applied

${\displaystyle \Rightarrow {\frac {1}{\sqrt {2}}}{\bigg (}|0\rangle |f_{k}\rangle |f_{k}'\rangle +|1\rangle \mathbf {|f_{k}'\rangle |f_{k}\rangle } {\bigg )}}$

After the Hadamard gate is applied on the first qubit

${\displaystyle \Rightarrow {\frac {1}{2}}{\bigg [}{\bigg (}|0\rangle +|1\rangle {\bigg )}|f_{k}\rangle |f_{k}'\rangle +{\bigg (}|0\rangle -|1\rangle {\bigg )}|f_{k}'\rangle |f_{k}\rangle {\bigg ]}}$

After sorting for ${\displaystyle |0\rangle {\text{ and }}|1\rangle }$

${\displaystyle \Rightarrow |\psi \rangle ={\frac {1}{2}}{\bigg [}|0\rangle {\bigg (}|f_{k}\rangle |f_{k}'\rangle +|f_{k}'\rangle |f_{k}\rangle {\bigg )}+|1\rangle {\bigg (}|f_{k}\rangle |f_{k}'\rangle -|f_{k}'\rangle |f_{k}\rangle {\bigg )}{\bigg ]}}$

Now it is easy to see, if the states ${\displaystyle |f_{k}\rangle =|f_{k}'\rangle \ }$ then ${\displaystyle \ |\psi \rangle =|0\rangle |f_{k}\rangle |f_{k}\rangle }$, which gives us a 0 whenever it is measured.