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?

  • $\begingroup$ "How do we know the unitarity constraint is the only constraint" -> We know that there exist (small) universal sets of one- and two- qubit operators, and we know by combining them we can get any unitary operator. So once you accept you can make the universal set (which, again, is small an easily verified) you can make any unitary operator. $\endgroup$ Aug 3, 2021 at 19:58

3 Answers 3


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

  • 2
    $\begingroup$ I like the answer! Perhaps one should add that with unitary transformations the norm is preserved (this is there "only" as a mathematical expression, section 2 last part), this is important since e.g. the probability of the physical system is always within [0,1]. $\endgroup$
    – P_Gate
    Aug 3, 2021 at 7:52
  • $\begingroup$ What do you mean with quantum gates being built from unitary operators? $\endgroup$ Aug 3, 2021 at 21:25

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.

  • 3
    $\begingroup$ If you change "add all components" to "add squared absolute values" and explain why it is important for that sum to be 1, then you'll probably make a good answer. $\endgroup$ Aug 3, 2021 at 7:20
  • 1
    $\begingroup$ Just to add that a unitary transform preserve a vector length, i.e. the sum of probabilities remains 1, and an angles among vectors. $\endgroup$ Aug 3, 2021 at 11:46

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.