Quantum Information Methods for Many-Body Physics · Notes Sec. 2.1
This lesson gives a precise meaning to a “uniformly random” unitary and introduces the object we will learn to compute: the average of \(U^{\otimes k}A\,(U^\dagger)^{\otimes k}\) over all unitaries. Haar invariance reduces that integral to a symmetry problem, and the symmetry turns out to be described by permutations, the subject of the next two lessons.
Lebesgue measure on \(\R\) is translation invariant: shifting a set does not change its length. Because \(\R\) is non-compact, however, its total length is infinite, so there is no translation-invariant probability distribution on the whole real line. Compact groups avoid this obstruction.
The unitary group \(\U(\dd)\) is compact. We abbreviate its normalized Haar measure by \(\rmd U\) and write
Thus “uniform” means invariant under a change of origin in the group, not that one has assigned an equal positive probability to each of infinitely many matrices. We write \(U\sim\Haar(\U(\dd))\) when \(U\) is sampled from this measure.
The action of \(\U(\dd)\) on normalized state vectors is transitive: given \(\ket\psi\) and \(\ket\varphi\), extend each vector to an orthonormal basis and map the first basis to the second. The resulting unitary satisfies \(U\ket\psi=\ket\varphi\). Consequently \(U\ket0\), with Haar-random \(U\), defines the unitarily invariant distribution on pure states. We call its samples Haar-random states.
Let \(f\) be a function of a unitary. Its ensemble average is
\[ \EE_{U\sim\Haar(\U(\dd))}[f(U)]:=\int_{\U(\dd)}f(U)\,\rmd U. \]For example, prepare \(\ket\psi\), apply \(U\), and measure in the basis \(\{\ket i\}_{i=0}^{\dd-1}\). The Born probability of outcome \(i\) is \(p_U(i)=|\sandwich{i}{U}{\psi}|^2\). The first Haar twirl, derived later in the course, gives
\[ \EE_U[p_U(i)]=\tr\!\left[\out i\,\EE_U(U\out\psi U^\dagger)\right]=\tr\!\left(\out i\,\frac{\one}{\dd}\right)=\frac1\dd. \]This statement fixes only the mean. Fluctuations require the second moment:
\[ \begin{gathered} \Var_U[p_U(i)]=\EE_U[p_U(i)^2]-\frac1{\dd^2},\\ p_U(i)^2=\tr\!\left[\out i^{\otimes2}\,U^{\otimes2}\out\psi^{\otimes2}(U^\dagger)^{\otimes2}\right]. \end{gathered} \]We will find \(\Var_U[p_U(i)]=(\dd-1)/(\dd^2(\dd+1))\), and return to its physical interpretation when we discuss anticoncentration.
Let \(\H=\H_A\otimes\H_B\), \(\rho=U\out\psi U^\dagger\), and \(\rho_A=\tr_B\rho\). On two copies, let \(\Swap_A\) exchange the two \(A\) factors and act trivially on the two \(B\) factors. The swap trick states that
To see it without diagrammatic notation, choose product bases and expand
\[ \tr(\rho_A^2)=\sum_{a,a',b,b'}\rho_{ab,a'b}\,\rho_{a'b',ab'}, \]which is exactly the index contraction on the right-hand side. Hence purity, although nonlinear in one copy of \(\rho\), is a linear observable on two copies.
Both examples are balanced polynomials: every term contains the same number \(\kk\) of matrix entries of \(U\) and of complex-conjugated entries of \(U\). Every balanced polynomial of bidegree \((\kk,\kk)\) is a linear combination of matrix elements of
\[ p_{A,B}(U):=\tr\!\left[B\,U^{\otimes\kk}A\,(U^\dagger)^{\otimes\kk}\right],\qquad A,B\in\L(\H^{\otimes\kk}). \]The task of this part of the course is to evaluate \(\MO_\kk(A)\) without introducing coordinates on \(\U(\dd)\). To see why, it helps to do it once with coordinates.
Consider
\[ \int_{\U(2)} U A U^\dagger\,\rmd U \qquad \text{for} \quad A=\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}. \]The global phase cancels in \(UAU^\dagger\), so it is enough to parametrize \(\SU(2)\) as
\[ U(\theta,\alpha,\beta)=\begin{pmatrix} e^{\i\alpha}\cos\theta & e^{\i\beta}\sin\theta \\ -e^{-\i\beta}\sin\theta & e^{-\i\alpha}\cos\theta \end{pmatrix},\qquad 0\le\theta\le\pi/2,\quad 0\le\alpha,\beta<2\pi, \]with normalized measure \(\rmd U=\frac{\sin(2\theta)}{4\pi^2}\,\rmd\theta\,\rmd\alpha\,\rmd\beta\). For the upper-left entry,
\[ (UAU^\dagger)_{11}=\cos^2\!\theta\,a_{11}+\sin^2\!\theta\,a_{22}+\sin\theta\cos\theta\left(e^{\i(\alpha-\beta)}a_{12}+e^{-\i(\alpha-\beta)}a_{21}\right). \]The phase integrals kill the last two terms, and the two remaining polar integrals are equal:
\[ \int_0^{\pi/2}\sin(2\theta)\cos^2\!\theta\,\rmd\theta=\int_0^{\pi/2}\sin(2\theta)\sin^2\!\theta\,\rmd\theta=\frac12. \]The off-diagonal entries vanish by the same phase average. Therefore
\[ \int_{\U(2)} U A U^\dagger\,\rmd U=\frac{\tr(A)}{2}\,\one. \]The calculation is elementary for \(\dd=2\) and \(\kk=1\), but coordinates become unwieldy as either parameter grows. The symmetry argument below scales much better.
The key observation in evaluating \(\MO_\kk(A)\) is that it commutes with all unitaries \(U^{\otimes\kk}\), \(U\in\U(\dd)\).
Proof. If \(A\) commutes with all \(U^{\otimes\kk}\), then \(\MO_\kk(A)=A\), so every element of \(\Comm_\kk\) is of the form \(\MO_\kk(A)\). Conversely, if \(B=\MO_\kk(A)\), then by the invariance of the Haar measure
\[ \begin{aligned} U^{\otimes\kk}B\,(U^\dagger)^{\otimes\kk}&=\int_{\U(\dd)}(UV)^{\otimes\kk}A\,\big((UV)^\dagger\big)^{\otimes\kk}\rmd V\\ &=\int_{\U(\dd)}V^{\otimes\kk}A\,(V^\dagger)^{\otimes\kk}\rmd V=B\qquad\forall\,U\in\U(\dd). \end{aligned} \]Thus \(B\in\Comm_\kk\). □
A central step in understanding the commutant is to realize that every \(U^{\otimes\kk}\) commutes with any operator \(\PR_\pi\) that permutes the \(\kk\) copies of the Hilbert space \(\H\) according to the permutation \(\pi\). For instance, \(U^{\otimes2}\) clearly commutes with a swap of the two copies. Hence the permutation operators \(\PR_\pi\) are contained in the \(\kk\)-fold unitary commutant. These operators form a representation of the symmetric group \(\SG_\kk\). A powerful result from representation theory says that \(\U(\dd)\) and \(\SG_\kk\) are dual to each other; among other things, every element of the commutant is a linear combination of permutation operators:
We will not seek a deeper understanding of Schur–Weyl duality in this course, but take it as the motivation for a technique: it lets us express \(\MO_\kk(A)\) as a linear combination over permutations, with coefficients determined by their combinatorics. This is the framework of Weingarten calculus, which evaluates many relevant integrals over the unitary group exactly. Permutations will therefore play a key role throughout the course, not for abstract mathematical reasons, but because they offer a concrete and computable handle on random quantum processes.
The next lesson collects what we need about permutations themselves: notation, composition, cycles and their counting. The one after it lets permutations act on \(\kk\) copies of a quantum system, where traces of \(\PR_\sigma\) turn into counting cycles.
ExercisesPractice the permutation toolkit on QuBite, the course's exercise platform.
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