Given that the global phases of states cannot be physically discerned, why is it that quantum circuits are phrased in terms of unitaries and not special unitaries? One answer I got was that it is just for convenience but I'm still unsure.

A related question is this: are there any differences in the physical implementation of a unitary $U$ (mathematical matrix) and $ V: =e^{i\alpha}U$, say in terms of some elementary gates? Suppose there isn't (which is my understanding). Then the physical implementation of $c\text{-}U$ and $c\text{-}V$ should be the same (just add controls to the elementary gates). But then I get into the contradiction that $c\text{-}U$ and $c\text{-}V$ of these two unitaries may not be equivalent up to phase (as mathematical matrices), so it seems plausible they correspond to different physical implementations.

What have I done wrong in my reasoning here, because it suggests now that $U$ and $V$ must be implemented differently even though they are equivalent up to phase?

Another related question (in fact the origin of my confusion, I'd be extra grateful for an answer to this one): it seems that one can use a quantum circuit to estimate both the modulus and phase of the complex overlap $\langle\psi|U|\psi\rangle$ (see https://arxiv.org/abs/quant-ph/0203016). But doesn't this imply again that $U$ and $e^{i\alpha}U$ are measurably different?

  • 2
    $\begingroup$ It is more philosophically accurate to say projective unitary group $PU$ instead. That is because the operation is to take an arbitrary unitary matrix and lose the phase vs the subset for which that phase is $1$. The maps go $SU \to U \to PU$ so they are on opposite sides of the arrows. $\endgroup$
    – AHusain
    Commented May 12, 2018 at 22:04
  • $\begingroup$ @AHusain Which are "The maps"? In terms of quotienting out, it will go $U\to SU\to PU$. $\endgroup$ Commented May 13, 2018 at 10:48
  • 1
    $\begingroup$ No. SU is the subset with determinant 1, so it includes with a map into U. PU is the quotienting out. You can take a projective unitary and give a representative in SU with determinant 1, but that is not automatic. $\endgroup$
    – AHusain
    Commented May 14, 2018 at 15:31

4 Answers 4


Even if you only limit yourself to special-unitary operations, states will still accumulate global phase. For example, $Z = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$ is special-unitary but $Z \cdot |0\rangle = i |0\rangle \neq |0\rangle$.

If states are going to accumulate unobservable global phase anyways, what benefit do we get out of limiting ourselves to special unitary operations?

are there any differences in the physical implementation of a unitary $U$ (mathematical matrix) and $V :=e^{i\alpha}U$, say in terms of some elementary gates?

As long you're not doing anything that could make the global phases relevant, they can have the same implementation. But if you're going to do something like, uh-

add controls to the elementary gates

Yeah, like that. If you do stuff like that, then you can't ignore global phases. Controls turn global phases into relative phases. If you want to completely ignore global phase, you can't have a black box "add a control" operation modifier.

  • $\begingroup$ Thanks, but doesn't an "add a control" modifier exist for gates in a universal gate set and you could first decompose $U$ and $V$ into these gates in order to add control, e.g. c-$X$ is the CNOT gate. $\endgroup$
    – wdc
    Commented May 18, 2018 at 20:59
  • 1
    $\begingroup$ @Daochen Yes you can do that, but it's not an example of adding a control while ignoring the sub-operation's global phase. You will have to explicitly decide on the global phase of the sub-operation when deciding what exactly the overall controlled operation should do and how to decompose it. $\endgroup$ Commented May 18, 2018 at 21:58

The fact that quantum gates are unitary, is rooted in the fact that the evolution of (closed) quantum systems is by the Schrödiner equation. For a time interval in which we are trying to realise a particular unitary transformation at a constant rate, we use the time-independent Schrödinger equation:

$$ \tfrac{\mathrm d}{\mathrm dt} \lvert \psi(t) \rangle = \tfrac {1}{i\hbar}H \lvert \psi(t) \rangle, $$

where $H$ is the Hamiltonian of the system: a Hermitian matrix, whose eigenvalues describe energy eigenvalues. In particular, the eigenvalues of $H $ are real. The solution to this equation is

$$ \lvert \psi(t) \rangle = \exp\bigl(-i H t/\hbar\bigr) \lvert \psi(0) \rangle $$ where $U = \exp(-iHt/\hbar)$ is the matrix which you obtain by taking the eigenvectors of $H$, and replacing their eigenvalues $E$ with $\mathrm{e}^{iEt/\hbar}$. Thus, from a matrix with real eigenvalues, we get a matrix whose eigenvalues are complex numbers with unit norm.

What would it take for this evolution to specifically be a special unitary matrix? A special unitary matrix is one whose determinant is precisely $1$; that is, whose eigenvalues all multiply to $1$. This corresponds to the restriction that the eigenvalues of $H$ all sum to zero. Furthermore, because the eigenvalues of $H$ are energy levels, whether the sum of its eigenvalues is equal to zero depends on how you have decided to fix what your zero energy point is — which is in effect a subjective choice of reference frame. (In particular, if you decide to adopt the convention that all of your energy levels are non-negative, this implies that no interesting system will ever have the property of the energy eigenvalues summing to zero.)

In short, gates are unitary rather than special unitary, because the determinant of a gate does not correspond to physically meaningful properties — in the explicit sense that the gate arises from the physics, and the conditions which correspond to the determinant of the gate being 1 is a condition of one's own reference frame and not the physical dynamics.


When writing gates for, for example, a quantum circuit diagram, you could always write them using the convention of having determinant one (from the special unitary group), but it's just a convention. It makes no physical difference to the circuit that you implement. As said elsewhere, whether what you naturally produce corresponds directly to the special unitary is really a choice of convention, and where you define your 0 energy to be.

As for the issue when you start implementing controlled-$U$, there is an interesting comparison to be made. Let's say we define $V=e^{i\alpha}$. How can we implement controlled-$V$ in terms of controlled-$U$? You apply controlled-$U$ and then, on the control qubit, you apply the phase gate $\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{i\alpha} \end{array}\right)$. There are two things to observe here. First, the difference is on the control qubit rather than the target qubit. The target qubit, where you're implementing the $U$, doesn't really care about the difference in phase. It's the control-qubit that's hit by the phase gate. The second is that I didn't write the phase gate as a special unitary. Of course, I could have written it as $\left(\begin{array}{cc} e^{-i\alpha/2} & 0 \\ 0 & e^{i\alpha/2}\end{array}\right)$ but I didn't because the way that I chose to write it was notationally more convenient - less writing for me, and hopefully more immediately obvious to you why it works.


Plain and simple answer: In the absence of decoherence, state vectors evolve according to $|\psi(t)\rangle = e^{-iHt}|\psi(0)\rangle$ for a Hamiltonian $H$. This is what a "gate" is doing. Hamiltonians have to be Hermitian, so this transformation is unitary. Hamiltonians do not have to have eigenvalues that sum to 0, so the transformation does not have to be special unitary.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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