In quantum mechanics each operator corresponds to some physical observable, but say we have the operators $X,Y,Z,H, \operatorname{CNOT}$. I understand how these gates act on qubits, but what do they actually represent in terms of a physical observable?


2 Answers 2


I agree with the main points that Niel makes: not all operators are observables, and the purpose of the ones you list is typically to transform states, not to be measured. However, the operators you list happen to be hermitian (allowing them also to represent observables) as well as unitary (allowing them to represent transformations), so in this case we can still play devil's advocate, and ask if they have physical interpretations as measurable quantities. So, interpreting your list as observables to be measured...

  • $X$, $Y$, and $Z$ are easy: these represent unitless versions of spin measurement (or the analogous quantity, if your system is not a spin system) along the three spatial coordinate directions.
  • $H$ also represents a unitless version of a spin measurement, but along an axis exactly between the $x$ and $z$ axes.
  • $\operatorname{CNOT}$ is the hardest, and I don't have a clean physical interpretation. It has eigenvalues $\pm1$ (3 eigenvectors with eigenvalue +1, and one with eivenvalue -1), and as an observable it should represent correlative properties of the two qubits. We could state what it literally measures, in terms of the four dimensions that it acts on, but this information will be less informative than just looking at its matrix form.

All this being said, typically the only one of the five that is actually measured is $Z$, and occasionally the other coordinate spin measurements.


In quantum mechanics, not all operators are observables. Many operators are observables, and in the first year or two of treatments in some physics courses you will only see operators which are observables; but not all operators of interest are observables.

The operators you have mentioned all happen to be Hermitian, and could therefore be interpreted as observables. This is particularly useful for the operators $X$, $Y$, and $Z$, which up to a scalar factor of $2\hbar$ you may be more familiar with as orbital angular momentum components. However, in the context of quantum computation, the more important property of the operators $\{X,Y,Z,H,\mathrm{CNOT}\}$ is that they are unitary: that is, their eigenvalues are all roots of unity, and so they represent possible finite-time evolutions of a quantum system.

Finite-time evolutions of the system are of course governed by the wave-equation. Specifically, from the time-independent Schrödinger equation, we have $$ i\hbar \frac{\mathrm d}{\mathrm d t} \lvert \psi(t) \rangle = \hat{\mathrm{H}} \, \lvert \psi(t) \rangle$$ which has solutions of the form $\lvert \psi(t) \rangle \,=\, \hat{\mathrm{U}}(t) \: \lvert \psi(0)\rangle$, for some initial state-vector $\lvert \psi(0)\rangle$ and where $\hat{\mathrm{U}}(t)$ is a unitary operator given by $$ \hat{\mathrm{U}} (t) \,=\, \exp\bigl(-i \:\!\hat{\mathrm{H}} \,t\big/\hbar\bigr) \,=\, \sum_{n \geqslant 0} \frac{\bigl( -i\!\:\hat{\mathrm{H}}\,t \big/ \hbar\bigr)^n}{n!} $$ whose eigenvalues are of the form $\mathrm e^{-iE_k t / \hbar}$ for eigenvalues $E_k$ of $\hat{\mathrm H}$. Such an operator is called a unitary operator, and had the property that $\hat{\mathrm U}(t){}^{-1} = \hat{\mathrm U}(-t) = \hat{\mathrm U}{}^\dagger$. The gates $X$, $Y$, $Z$, $H$, and $\mathrm{CNOT}$ are all operators of this form, where it so happens that the eigenvalues are $\pm 1$.

All unitary operators can be obtained from the exponentiation of a Hermitian operator in this way; though in many cases we prefer to consider them as products if a sequence of different operators $\hat{\mathrm U} = \hat{\mathrm U}_N(t_N)\,\cdots\,\hat{\mathrm U}_2(t_2)\,\hat{\mathrm U}_1(t_1)$, representing evolution according to piecewise-constant Hamiltonians. (Realising such a unitary evolution will be more complicated in practise, of course, just as computing with classical computers is more complicated than just applying 'AND' and 'OR' gates to bits represented by electrical voltages in wires, but this is the picture of quantum computation to first-order.)

  • $\begingroup$ Is there an intuitive way to think of them physically though ? As in what is it that we are actually observing when we measure them ? $\endgroup$
    – bhapi
    Apr 27, 2019 at 15:37
  • $\begingroup$ That's the point --- these are intended to be propogators, not observables. You don't measure unitary operators, you use them to describe the evolution of (closed) systems. $\endgroup$ Apr 27, 2019 at 16:32
  • $\begingroup$ @can'tcauchy it is really a coincidence that the operators you list are measurable at all, so Niel's main point stands. In general, transformations on a quantum state are not physically measurable; however, in the case of those you list, they in principle are, so see my answer below. $\endgroup$
    – Will
    Apr 27, 2019 at 17:44

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