Quantum Computing Stack Exchange is a question and answer site for engineers, scientists, programmers, and computing professionals interested in quantum computing. It only takes a minute to sign up.
Sign up to join this community
In my textbook, it's said
the unitarity constraint is the only constraint on quantum gates. Any unitary matrix specifies a valid quantum gate!
Why do quantum gates have to have to be unitary? How do we know that the unitarity constraint is the only constraint?
Time Evolution Postulate: A pure state $|\psi(t_0)\rangle$ in a Hilbert Space $ \mathbb{H}$ evolves to another state $|\psi(t)\rangle$ is given by the time evolution operator $U(t,t_0)$ $$ |\psi(t) \rangle = U(t,t_0)|\psi(t_0)\rangle $$ where $U(t,t_0)$ is the solution of the initial value problem $$ i \dfrac{d}{dt} U(t,t_0) = H(t)U(t,t_0) $$ $$ U(t_0, t_0) = \boldsymbol{I} $$ here $H(t)$ is a self adjoint operator also known as the Hamiltonian.
If you think about quantum circuits, then you will see that the circuit start at some initial state, usually $|0\rangle^{\otimes n}$, then you evolve this state to another state. This then must follow the above postulate. That is, the quantum gate, says $V$, that takes the state $|\psi (t_0)\rangle$ to $|\psi(t) \rangle$ must be $V = U(t,t_0)$.
Now, it turns out that from the way $U$ is defined, it is unitary. This can be shown as follow:
For any $|\psi\rangle \in \mathbb{H}$
\begin{align} \dfrac{d}{dt} \|U(t,t_0) \psi\|^2 &= \dfrac{d}{dt} \langle U(t,t_0) \psi| U(t,t_0) \psi \rangle\\ &= \langle \dfrac{d}{dt} U(t,t_0) \psi| U(t,t_0) \psi \rangle + \langle U(t,t_0)\psi| \dfrac{d}{dt} U(t,t_0) \psi \rangle \hspace{1 cm} \textrm{product rule}\\ &= \langle -iH(t)U(t,t_0)\psi | U(t,t_0) \psi \rangle + \langle U(t,t_0) \psi | -iH(t)U(t,t_0) \psi \rangle \\ &= i\bigg(\langle H(t) U(t,t_0) \psi | U(t,t_0)\psi \rangle - \langle U(t,t_0)\psi | H(t) U(t,t_0) \psi \rangle \bigg) \hspace{0.5 cm} \textrm{since} \ \ \langle c \psi | \phi \rangle = c^*\langle \psi | \phi\rangle \\ &= i\bigg(\langle H(t)^* U(t,t_0) \psi | U(t,t_0)\psi \rangle - \langle U(t,t_0)\psi | H(t) U(t,t_0) \psi \rangle \bigg) \\ &= 0 \hspace{0.75 cm} \textrm{since}\hspace{0.5 cm} \langle A^* \psi | \phi \rangle = \langle \psi | A \phi \rangle \end{align}
This tells us that $ \|U(t,t_0) |\psi \rangle \|$ must be a constant. And since $U(t_0, t_0) = \boldsymbol{I}$ so $\|U(t_0,t_0) \psi \| = \|\psi\|$ we must have $\| U(t,t_0) \psi \| = \|\psi\|$ as well.
Thus, $U(t,t_0)$ preserves the inner product. Hence, it is a unitary operator.
In summary, quantum gates are built from operator $U(t,t_0)$ generate from certain Hamiltonian $H(t)$ and $t$ by the time evolution postulate. And because of how $U(t,t_0)$ is formulated, it must be unitary. Thus, quantum gates are neccesary to be unitary.
Reference: Mathematics of Quantum Computing
The unitary , after you applied the gate on a state. the new state should keep the property if you add all the squared components of the state vector, this should be equal, to 1.In another way since the state vector is a vector that contains the probabilities, and it should keep that property adding all the amplitudes square(norm square, they are complex numbers) it always must give 1, so after you applied a gate on the state vector it will remain that property on the state vector.
A Quantum Gate is a frame to describe the quantum process. For a closed quantum system, opposites to an open quantum system, the operator describes the evolution of the system can be described with $e^{-i\int Hdt/\hbar}$, which is a unitary operator. Furthermore, if we ignore the global phase of the quantum state, we can suppose all the quantum gates to be special unitary, i.e., the determinant of them is 1.
