Can somebody please explain me what is a Haar random state? I am not able to find any friendly resource to read about it.


2 Answers 2


Typically this is a slight abuse of notation. One can have a unitary operator $U$ chosen from some Haar measure, such as the circular unitary ensemble. Then, taking some fiducial state $|\psi_0\rangle$, a "Haar-random state" would be $|\psi_U\rangle=U|\psi_0\rangle$ for $U$ randomly chosen according to the Haar measure. If you want to find the average value of some observable $O$ using such a Haar-random state, you'd say $$\bar{ O}=\int dU \langle \psi_U| O|\psi_U\rangle=\int dU \langle \psi_0|U^\dagger OU|\psi_0\rangle,$$ so you can also look at this in a Heisenberg picture in which the operators $O$ get distributed according to the Haar measure through $U^\dagger OU$ and the fiducial state $|\psi_0\rangle$ remains unchanged. In the above I've assumed the measure to be normalized, with $\int dU=1$, which can always be facilitated.

The reason I see the unitaries as more fundamental than the states is that it is the unitaries that form a group (the composition of two unitaries $U_1 U_2$ is a unitary, while the composition of two states $\rho_1 \rho_2$ is not a state). Further, it is the set of unitaries that define the space of states. Take, for example, a single harmonic oscillator, spanned by the Fock states $|n\rangle$. What is a random distribution for these states? Should each coefficient $\psi_m$ in the state $$|\psi\rangle=\sum_{n=0}^\infty \psi_n|n\rangle$$ be distributed in the same way? That would require a Haar measure on the infinte-dimensional unitary group $\lim_{N\to\infty}U(N)$, and I'm not sure whether that exists. Should we take a state with equal probability of being everywhere in phase space; i.e., should it have a uniform quasiprobability distribution? That would make sense in some physical scenarios. These questions can only be answered by specifying a group of unitaries and fiducial state.

For example, one can choose the group of unitaries corresponding to displacements $D(\alpha)=\exp(\alpha a^\dagger-\alpha^* a)$, which comprise the Heisenberg group, and the fiducial state $|0\rangle$. Then the uniform set of states is the set of coherent states $D(\alpha)|0\rangle$. For another fiducial state such as $|1\rangle$, the uniform distribution gives a different set of states $D(\alpha)|1\rangle$, which are also (over)complete in the Fock space, but do not contain any of the states $D(\alpha)|0\rangle$! So specifying the Hilbert space or set of states is not enough to determine a random set of states; it is the unitary group and the fiducial state that matter.

We could even generalize this to Gaussian operations, which comprise the symplectic group, acting on the very same Fock space. Then we get a different Haar measure over this new group, which creates all of the Gaussian states from the fiducial state $|0\rangle$, but creates a whole set of non-Gaussian states from the fiducial state $|1\rangle$! Just telling us the fiducial state is not enough to specify the group, just telling us the fiducial state and its Hilbert space (here the Fock space spanned by $|n\rangle$) is not enough; one must equip the space with a method for travelling around in it, which is provided by the Haar measure for the group of transformations in which one is interested. Why do we have different sets of random states but only one measure? Because the measure comes from the unitary group, while the set of random states comes from a fiducial state and the Haar measure for the chosen unitary group.

As a separate set of examples, we can consider random distributions of $N$ qubits. These can be obtained using the unitary group $U(2^N)$, as the Hilbert space has dimension $2^N$. Alternatively, one can consider only the permutation-symmetric group of unitaries, which act on the $(N+1)$-dimensional subspace of the unsymmetrized Hilbert space. Then a random distribution could invoke the unitary group $U(N+1)$, essentially randomizing the coefficients of the state in some basis of this Hilbert space. But then we can also remember that a symmetric set of $N$ qubits furnishes us with a representation of $SU(2)$, so we can distribute these states randomly using the Haar measure from $SU(2)$. These could all begin with the same fiducial state and yet would give radically different distributions. This is why specifying the group matters.

Incidentally, the $N$ qubit example uses unitary groups throughout, so each set of states can be parametrized by some coordinates on a hypersphere. The harmonic oscillator example is different, as the hypersphere would need to be infinite-dimensional, or else it must be a plane (as in the set of displacements, with a planar phase space) or parametrized by the symplectic group. These should hopefully show how fundamental the unitary groups are to specifying the set of random states one wishes to distribute using a Haar measure.

  • $\begingroup$ I'd say it is a generalization rather than an abuse of terminology. $\endgroup$ Commented Jan 24, 2022 at 14:54
  • $\begingroup$ @NorbertSchuch do you prefer "induced measure?" $\endgroup$ Commented Jan 24, 2022 at 15:12
  • $\begingroup$ why is this abuse of notation or necessitates a notion of induced measure? Pure states (projectively speaking) live on the surface of some hypersphere, e.g. for a single-qubit this is the Bloch sphere, so all we have to do is sample from the latter manifold uniformly i.e. from the Haar measure of the hypersphere. This is a well-defined notion without needing to invoke a uniformly-drawn unitary (i.e. drawn from the Haar measure of this different, larger space) acting on a reference state to "project" it down. $\endgroup$
    – nervxxx
    Commented Jan 24, 2022 at 20:09
  • $\begingroup$ @QuantumMechanic I'd say "Haar random state" is completely fine, if you consider "Haar random" to mean "a measure which is invariant under the action of any unitary" - this definition applies equally to "Haar random unitaries" and "Haar random states". $\endgroup$ Commented Jan 24, 2022 at 21:14
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    $\begingroup$ @nervxxx the manifold of pure states is the complex projective space $\mathbb CP^{d-1}$, not the complex sphere. You have to identify points on the sphere which differ by a complex phase. $\mathbb CP^{d-1}$ has an induced metric from $\mathbb C^d$ which is the Fubini-Study metric. Using that metric, you can define a measure on $\mathbb CP^{d-1}$ which would be the "uniform measure" on pure states. The reason why it's uniform is that the isometry group of FS is the projective unitary group $PU(d)$. Thus, in the end it's the same as acting with Haar-random unitaries up to a phase. $\endgroup$ Commented Jan 25, 2022 at 12:49

Please read my paper. We have discussed Haar uniformity. It will help. https://iopscience.iop.org/article/10.1088/1367-2630/ac37c8/meta

Section 2.1 "Haar uniform random states and probability density functions."

  • 3
    $\begingroup$ hello and welcome to the site. Would you mind adding some more details about the relevance of the linked paper to the question at hand? As it stands, this is essentially a link-only answer and might be better suited as a comment rather than as an answer. The relevance of this paper is also not immediately obvious reading the title, so some context might be warranted $\endgroup$
    – glS
    Commented Jul 3, 2022 at 14:41

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