**An** **Elementary** **Pro****of** **of** **Stone**’s **Theorem** **for** **Unitary** **Matrices****Stone** **Theorem**. Let U(·) be a measurable, matrix-valued function mapping from R into the n×n unitarymatrices, such that, **for** all real numbers s and t, U(s + t) = U(s)U(t). There is a self-adjoint matrix H suchthat U(t) = e iHt . In particular, there is a unitary matrix V and there are real numbers λ 1 , . . . , λ n suchthatU(t) = V E(t)V ∗ ,where E(t) is the diagonal matrix with main-diagonal entries e iλ1t , . . . , e iλnt . The matrix H has the **for**mH = V NV ∗ ,where N is the diagonal matrix with main-diagonal entries λ 1 , . . . , λ n .Warm-Up **Theorem**. Let f : R ↦→ C be a measurable function that is bounded on bounded intervals.Suppose that f satisfies f(0) = 1 and, **for** all real s and t, f(s + t) = f(s)f(t). Then f(t) ≡ e αt **for** somecomplex number α.**Pro****of** **of** WUT. For any t ∈ R,There**for**e, **for** any t and t ′ ,f(t) =|f(t) − f(t ′ )| ≤ M∫ 1/2−1/2∫ 1/2−1/2f(t − s) f(s) ds.|f(t − s) − f(t ′ − s)| ds,where M is the supremum **of** |f| on [−1/2, 1/2]. This goes to zero as t ′ → t. There**for**e f is continuous.Let δ > 0 be so small that |t| < δ implies that |1 − f(t)| < 1/10, and fix some t 0 ∈ (0, δ). We can write**for** some unique θ 0 ∈ (−π/4, π/4). Definef(t 0 ) = exp(log |f(t 0 )| + iθ 0 )β ≡ log |f(t 0 )| + iθ 0 .Now, what is f(t 0 /2)? It must satisfy (f(t 0 /2)) 2 = f(t 0 ), and the two numbers that do this are exp(β/2) and− exp(β/2). But f(t 0 /2) cannot be further than a distance **of** 1/10 from 1, so the second root is out. Thusf(t 0 /2) = exp(β/2). Continuing this way, we see that f(t 0 /2 k ) = exp(β/2 k ) **for** all non-negative integersk. The law ‘f(s + t) = f(s)f(t)’ trivially implies that f(mt 0 ) = exp(mβ) **for** all integers m. It’s now aneasy matter to show that f(rt 0 ) = exp(rβ) **for** any dyadic rational r; i.e., any number **of** the **for**m r = n/2 j .Since the dyadic rationals are dense in R, and f and the exponential function are continuous, we get thatf(rt 0 ) = exp(rβ) **for** all real numbers r. If t ∈ R is arbitrary, we can write f(t) = f(rt 0 ) = exp(rβ) **for**r = t/t 0 , implying f(t) = exp(αt) **for** α = β/t 0 . The WUT is proved.Lemma. Let U be an n × n normal matrix with eigenvalues (counted according to multiplicity) λ 1 , . . . ,λ n . If I is the n × n identity matrix, then the operator norm **of** U − I equals max i |1 − λ i |.**Pro****of** **of** Lemma. The matrix U has an orthonormal basis **of** eigenvectors u 1 , . . . , u n , where weassume they are ordered in the same way as the λ i ’s. If x ∈ C n thenUx =n∑λ i 〈x, u i 〉u i11

andThere**for**e‖(U − I)x‖ 2 =Ix = x =n∑〈x, u i 〉u i .1n∑|λ i − 1| 2 |〈x, u i 〉| 21() 2 ∑ n≤ max |1 − λ i | |〈x, u i 〉| 2i1(2= max |1 − λ i |)‖x‖ 2 ,iimplying ‖U − I‖ ≤ max i |1 − λ i |. We get equality by applying U − I to the appropriate eigenvector u i .Lemma. Suppose that U and V are two normal n × n matrices such that U 2 = V and all **of** U’s eigenvalueshave positive real parts. Then U and V have the same eigenvectors: any v ∈ C n is an eigenvector **for** U ifand only if it’s an eigenvector **for** V .Remark. Some restriction on U’s eigenvalues is required. Consider:[ ]1 0U ≡0 −1[ ]1 0V ≡ .0 1Both are normal (they’re even unitary), and U 2 = V , but any non-zero vector is an eigenvector **for** V , whichobviously isn’t true **for** U.**Pro****of** **of** Lemma. **An**y eigenvector **for** U is trivially one **for** V . We show the other direction. Let Uhave an orthonormal basis **of** eigenvectors u 1 , . . . , u n , with corresponding eigenvalues λ 1 , . . . , λ n . Let xbe an eigenvector **for** V with eigenvalue γ. We can writewhich impliesU 2 x =x =n∑〈x, u i 〉u i ,n∑λ 2 i 〈x, u i 〉u i = V x = γx =11n∑γ〈x, u i 〉u i .This can only happen if 〈x, u i 〉 = 0 **for** those i’s such that λ 2 i ≠ γ. The function z ↦→ z2 is one-to-one on theset **of** z’s with positive real parts (where all the λ i ’s lie). There**for**e those λ i ’s **for** which 〈x, u i 〉 ≠ 0 (**for**cingλ 2 i = γ) are equal to the same number—call it λ. So we get:x = ∑〈x, u i 〉u ii:λ i=λ1andand x is an eigenvector **for** U.Ux = ∑i:λ i=λλ〈x, u i 〉u i = λx,**Pro****of** **of** **Stone** **Theorem**. The pro**of** that U(t) is continuous is just like the pro**of** **of** f’s continuityin the WUT. The product law U(s + t) = U(s)U(t) implies that U(0) = I. Let δ > 0 be such that |t| < δimplies ‖U(t) − I‖ < 1/10, and fix a t 0 ∈ (0, δ). We can writeU(t 0 ) = V L 0 V ∗ ,2

where V is unitary and L 0 is a diagonal matrix with entries e iθ1 , . . . , e iθn , and each θ j lies in (−π/4, π/4)(because |1 − e iθj | < 1/10). Important: all **of** U(t 0 )’s eigenvalues have positive real parts! Now, U(t 0 ) =U(t 0 /2) 2 , and ‖U(t 0 /2) − I‖ < 1/10 as well, which means that U(t 0 /2)’s eigenvalues all have positive realparts too. There**for**e V ’s columns—which are U(t 0 )’s orthonormal eigenvectors—are also eigenvectors **for**U(t 0 /2), and we can writeU(t 0 /2) = V L 1 V ∗**for** some diagonal matrix L 1 . This L 1 must satisfy (L 1 ) 2 = L 0 , and its diagonal entries γ j are U(t 0 /2)’seigenvalues: γj 2 = . The γ eiθj j ’s have to be no further than 1/10 from 1, and there**for**e they have to satisfyγ j = e iθj/2 . Continuing as in the WUT, U(t 0 /2 k ) will equal V L k V ∗ , where L k is a diagonal matrix withmain-diagonal entries e iθj/2k . Following the pattern **of** the WUT, we see that U(t) will have the **for**mV L(t)V ∗ , where L(t) is diagonal, with main-diagonal entries e i(θj/t0)t . The theorem is proved.3