A proof for the informational completeness of the rotated quadrature observables
aa r X i v : . [ qu a n t - ph ] D ec A PROOF FOR THE INFORMATIONAL COMPLETENESS OF THEROTATED QUADRATURE OBSERVABLESJ. KIUKAS, P. LAHTI, AND J.-P. PELLONPÄÄAbstra t. We give a new mathemati ally rigorous proof for the fa t that, when S is a densesubset of [0 , π ) , the rotated quadrature operators Q θ , θ ∈ S , of a single mode ele tromagneti (cid:28)eld onstitute an informationally omplete set of observables.1. Introdu tionSin e the pioneering works of Vogel and Risken [15℄ and Smithey et al [13℄ the measurementof the rotated quadratures Q θ , θ ∈ [0 , π ) , has formed one of the major tools in the quantumstate tomography, see, e.g., the ompilation [11℄. The basi idea behind the state re onstru tionis well known: the inverse Radon tranform of the quadrature measurement statisti s allows oneto re onstru t the Wigner fun tion of the state in question and the Wigner fun tion separatesstates, so that the statisti s uniquely determines the state. We do not question the validityof this argument. However, the existing literature, whi h we are aware of, does not give afull justi(cid:28) ation that this pro edure a tually applies to all possible states of a quantum system.Therefore, in this paper, we wish to give a dire t proof of the fa t that the set of quadrature ob-servables Q θ , θ ∈ S , S ⊆ [0 , π ) is dense, is informationally omplete, that is, the measurementout ome statisti s p Q θ T of these observables uniquely determine the state T .There is a beautiful group theoreti al proof of the informational ompleteness of the observ-ables Q θ , θ ∈ [0 , π ) [5℄. This proof builds on a general method of onstru ting informationally omplete sets of observables using the theory of square-integrable representations of unimodularLie groups. The result in question is then obtained as a spe ial appli ation of this theory to theWeyl-Heisenberg group. Due to the pra ti al importan e of the result, we give an alternativedire t proof of it. The proof of this fa t in Se tion 3 forms the main body of this paper.In the (cid:28)nal Se tion 4 we omment on the measurability of the quadrature observables andwe ompare the tomography based on their measurements on the one obtained from the phasespa e measurements. 2. Basi notations and definitionsLet H be a omplex separable Hilbert spa e, and L ( H ) the set of bounded operators on H ,and T ( H ) the set of tra e lass operators. We let k · k denote the tra e norm of T ( H ) . (Theoperator norm of L ( H ) will be denoted by k · k .) When H is asso iated with a quantum system(su h as the single mode ele tromagneti (cid:28)eld), the states of the system are being representedby positive operators T ∈ T ( H ) with unit tra e, density operators, and the observables areasso iated with the normalized positive operator measures de(cid:28)ned on the Borel σ -algebra B ( R )
1f the real line.1 Among them are the onventional von Neumann type of observables, that is,proje tion valued measures P : B ( R ) → L ( H ) , or, equivalently, selfadjoint operators in H .The measurement out ome statisti s of an observable E : B ( R ) → L ( H ) in a state T is givenby the probability measure X tr[ T E ( X )] =: p ET ( X ) .De(cid:28)nition 1. A set M of observables E : B ( R ) → L ( H ) is informationally omplete, if anytwo states S and T are equal whenever tr[ T E ( X )] = tr[ SE ( X )] for all E ∈ M and X ∈ B ( R ) .Thus, the informational ompleteness of a set M of observables means that the totality ofthe measurement out ome distributions p ET , E ∈ M , determines the state T of the system.Clearly, a set M of observables is informationally omplete if and only if T = 0 whenever T isa selfadjoint tra e lass operator with tr[ T E ( X )] = 0 for all E ∈ M and X ∈ B ( R ) . We willuse this hara terization in our proof.Fix {| n i | n ∈ N } to be an orthonormal basis of H . (This is identi(cid:28)ed with the photonnumber basis in the ase where H is asso iated with the single mode ele tromagneti (cid:28)eld.)Here N = { , , , . . . } . We will, without expli it indi ation, use the oordinate representation,in whi h H is represented as L ( R ) via the unitary map H ∋ | n i 7→ h n ∈ L ( R ) , where h n isthe n th Hermite fun tion, h n ( x ) = 1 p n n ! √ π H n ( x ) e − x , and H n is the n th Hermite polynomial.2 Let a and a ∗ denote the usual raising and loweringoperators asso iated with the above basis of H ; they are onsider as being de(cid:28)ned on theirmaximal domain D ( a ) = D ( a ∗ ) = n ϕ ∈ H (cid:12)(cid:12)(cid:12) ∞ X n =0 n |h ϕ | n i| < ∞ o . Then de(cid:28)ne the operators Q = √ ( a ∗ + a ) and P = i √ ( a ∗ − a ) , whi h, in the oordinate rep-resentation are the usual multipli ation and di(cid:27)erentiation operators, respe tively: ( Qψ )( x ) = xψ ( x ) , ( P ψ )( x ) = − i dψdx ( x ) . (Here the bar stands for the losure of an operator, so that e.g. Q is the unique selfadjoint extension of the essentially selfadjoint symmetri operator √ ( a ∗ + a ) ;see [12, Chapter IV℄ or [3, Chapter 12℄ for details on erning the domains of these extensivelystudied operators.) In the ase of the ele tromagneti (cid:28)eld, Q and P are alled the quadratureamplitude operators of the (cid:28)eld. In addition, a = √ ( Q + iP ) and a ∗ = √ ( Q − iP ) (seee.g. [12, p. 73℄). The S hwartz spa e S ( R ) of rapidly de reasing C ∞ -fun tions is in luded in D ( Q ) ∩ D ( P ) = D ( a ) .For a fun tion g : R → R , we let g ( k ) denote the k th derivative of g (with g (0) = g ), providedit exists. We need the following elementary ommutation relations, whi h hold whenever ϕ ∈S ( R ) , and g : R → R is ontinuously di(cid:27)erentiable and bounded: ( g ( Q ) a ∗ − a ∗ g ( Q )) ϕ = √ g (1) ( Q ) ϕ, (1) ( g ( Q ) a − ag ( Q )) ϕ = − √ g (1) ( Q ) ϕ. (2)1Normalized positive operator measure is a map E : B ( R ) → L ( H ) whi h is σ -additive in the weak operatortopology, and has the property E ( R ) = I , that is, for whi h X tr[ T E ( X )] is a probability measure for ea hpositive tra e one operator T .2Hermite polynomials are, of ourse, given by the following re ursion relation: H ( x ) = 1 , H ( x ) = 2 x and H n +1 ( x ) = 2 xH n ( x ) − nH n − ( x ) . 2et N denote the operator a ∗ a . It is selfadjoint on its natural domain D ( N ) = n ϕ ∈ H (cid:12)(cid:12)(cid:12) ∞ X n =0 n |h ϕ | n i| < ∞ o . Now the phase shifting unitary operators are e iθN , and we an de(cid:28)ne the rotated quadratureobservables Q θ by Q θ = e iθN Qe − iθN , θ ∈ [0 , π ) . The spe tral measure of Q θ will be denoted by P Q θ : B ( R ) → L ( H ) .3. The proofWe need the so alled Dawson's integral daw( x ) = e − x Z x e t dt (see e.g. [1, pp. 298-299℄ or [14, Chapter 42℄). The following lemma lists those properties ofDawson's integral that we are going to use. Sin e the Dawson's integral has been studiedextensively, they are probably well known. However, as we were unable to (cid:28)nd these resultsdire tly stated in the literature, we give a proof in the Appendix; a reader familiar with theresults may skip that proof.Lemma 1. (a) daw : R → R is a C ∞ -fun tion, and lim x →±∞ daw ( k ) ( x ) = 0 for all k ∈ N . (b) daw (1) ( x ) = 12 ∞ X n =0 ( − n n !2 n (2 n )! H n ( x ) for all x ∈ R , where H n is the n th Hermite polynomial.Lemma 2. There exists a C ∞ -fun tion f : R → R , su h that(i) ea h derivative f ( n ) of f , n = 0 , , . . . is a bounded fun tion, and(ii) h n | f ( Q ) | n i = δ n , for all n ∈ N , where δ is the Krone ker delta.Proof. We put f = 2 daw (1) . A ording to Lemma 1, f is a C ∞ -fun tion and (i) holds. Lemma1 (b) gives the expansion(3) f ( x ) = ∞ X n =0 ( − n n !2 n (2 n )! H n ( x ) , x ∈ R . Sin e the series in (3) onverges pointwise, we get f ( x ) h ( x ) = ∞ X n =0 ( − n n !2 n (2 n )! p n (2 n )! h n = ∞ X n =0 ( − n n ! p (2 n )! h n ( x ) x ∈ R . In addition, P ∞ n =0 ( n !) / (2 n )! < ∞ , so the series onverges also in L -norm.Noting also that H n h ∈ L ( R ) , we an justify the following omputation. h h n | f ( Q ) h n i = 12 n n ! h H n h | f h i = 12 n n ! ∞ X k =0 ( − k k !2 k (2 k )! h H n h | p k (2 k )! h k i = 12 n n ! n X k =0 ( − k k !2 k (2 k )! h H n h | H k h i = 12 n n ! n X k =0 ( − k k !2 k (2 k )! √ π Z R H k ( x )( H n ( x )) e − x dx = 12 n n ! n X k =0 ( − k k !2 k (2 k )! √ π k + n √ π (2 k )!( n !) ( n − k )!( k !) = n X k =0 ( − k n !( n − k )! k != n X k =0 ( − k (cid:18) nk (cid:19) = lim y → (1 − y ) n = δ n . Here the (cid:28)nite sum after the third equality is obtained by noting that h k is orthogonal to H n h whenever k > n , whi h is due to the fa t that the latter fun tion is a linear ombinationof Hermite fun tions h , . . . , h n . The (cid:28)fth equality follows from the formula 7.375(2) of [6, p.838℄. (cid:3) Now we hoose a fun tion f satisfying the onditions of Lemma 2. The fun tion f will remain(cid:28)xed throughout the rest of the paper.Lemma 3. h n + k | f ( k ) ( Q ) | n i = ( − k √ k k ! δ n for all k, n ∈ N .Proof. We pro eed by indu tion with respe t to k ; the initial step is provided by ondition (ii)of Lemma 2. The indu tion assumption is that for some k ∈ N , the equality h n + k | f ( k ) ( Q ) | n i = ( − k √ k k ! δ n holds for all n ∈ N . We must show that h n + k + 1 | f ( k +1) ( Q ) | n i = ( − k +1 p k +1 ( k + 1)! δ n , n ∈ N . But by using (1), we get h n + k + 1 | f ( k +1) ( Q ) | n i = √ h n + k + 1 | f ( k ) ( Q ) a ∗ | n i − √ h n + k + 1 | a ∗ f ( k ) ( Q ) | n i = p n + 1) h ( n + 1) + k | f ( k ) ( Q ) | n + 1 i − p n + k + 1) h n + k | f ( k ) ( Q ) | n i = − p n + k + 1)( − k √ k k ! δ n = ( − k +1 p k +1 ( k + 1)! δ n for all n ∈ N by the indu tion assumption. (Note, in parti ular, that the (cid:28)rst term in theexpression following the se ond equality is indeed zero by the indu tion assumption, be ause n + 1 > .) (cid:3) Lemma 4. h n + k | f ( k +2 l ) ( Q ) | n i = 2 l ( − k p k l !( l + k )! δ ln , k, l, n ∈ N , n ≥ l. Proof. Now we use indu tion with respe t to l , so the initial step l = 0 is given by Lemma 3.The indu tion assumption is that h n + k | f ( k +2 l ) ( Q ) | n i = 2 l ( − k p k l !( l + k )! δ ln l ∈ N , all k ∈ N , and all n ≥ l . We have to show that this holds also when l isrepla ed by l + 1 . A ordingly, let k ∈ N , and n ∈ N with n ≥ l + 1 . Using the ommutationrelation (2) and the indu tion assumption, we get h n + k | f ( k +2( l +1)) ( Q ) | n i = −√ h n + k | f ( k +2 l +1) ( Q ) a | n i + √ h n + k | af ( k +2 l +1) ( Q ) | n i = −√ n h ( n −
1) + ( k + 1) | f (( k +1)+2 l ) ( Q ) | n − i + p n + k + 1) h n + ( k + 1) | f (( k +1)+2 l ) ( Q ) | n i = −√ n l ( − k +1 p k +1 l !( l + k + 1)! δ l,n − = ( − k p l + 1)2 l √ p k l !( l + 1 + k )! δ l +1 ,n = 2 l +1 ( − k p k ( l + 1)!( l + 1 + k )! δ l +1 ,n . Here the indu tion assumption was applied to the (cid:28)rst term following the se ond equality with n and k repla ed by n − and k + 1 , and to the se ond term with n and k repla ed by n and k + 1 . Here it is essential to note that n ≥ l + 1 > l , whi h makes the se ond term zero. (cid:3) In order to onstru t the proof for the informational ompleteness of the quadratures, weneed some additional tools. First de(cid:28)ne, for ea h (cid:28)xed k ∈ N and X ∈ B ( R ) , V k ( X ) := Z π e − ikθ P Q θ ( X ) dθ ∈ L ( H ) , where the integral is to be understood in the σ -weak operator topology. Indeed, for ea htra e lass operator T , the map θ tr[ T P Q θ ( X )] = tr[ T e iθN P Q ( X ) e − iθN ] is ontinuous,3and | tr[ T E Q θ ( X )] | ≤ k T k k P Q θ ( X ) k ≤ k T k , so the integral exists in the σ -weak sense, andrepresents a bounded operator, with k V k ( X ) k ≤ π .Next, noti e that for ea h k ∈ N , the map X V k ( X ) is an operator measure, that is, σ -additive in the weak operator topology. In fa t, if ( X n ) n ∈ N is a sequen e of mutually disjointsets in B ( R ) , then for any l ∈ N , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) l X n =0 e − ikθ h ϕ | P Q θ ( X n ) ϕ i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ l X n =0 h ϕ | P Q θ ( X n ) ϕ i ≤ h ϕ | P Q θ ( ∪ ∞ n =0 X n ) ϕ i ≤ k ϕ k , so the dominated onvergen e theorem an be applied to get ∞ X n =0 h ϕ | V k ( X n ) ϕ i = ∞ X n =0 Z π e − ikθ h ϕ | P Q θ ( X n ) ϕ i dθ = Z π e − ikθ h ϕ | P Q θ ( ∪ ∞ n =0 X n ) ϕ i dθ = h ϕ | V k ( ∪ ∞ n =0 X n ) ϕ i . Let g : R → R be any bounded Borel fun tion. Then the operator integral V k [ g ] = R g dV k an be de(cid:28)ned in the σ -weak sense, as a bounded operator. This follows from the fa t that g is bounded and | V kT | ( R ) ≤ X ∈B ( R ) | tr[ T V k ( X )] | ≤ π k T k for any tra e lass operator T , with | V kT | denoting the total variation of the omplex measure V kT = tr[ T V k ( · )] . (The map X tr[ T V k ( X )] is a omplex measure, be ause the weak and σ -weak operator topologies oin ide in a norm-bounded set.)3 This an easily be seen by e.g. onsidering a positive tra e lass operator T , using its spe tral resolutionand applying the strong ontinuity of the map θ e iθN .5emma 5. For any bounded fun tion g : R → R , and a tra e lass operator T , we have tr[ T V k [ g ]] = 2 π ∞ X n =0 h n | T | n + k i h n + k | g ( Q ) | n i . Proof. First note that h m | V k [ g ] | n i = tr[ | n ih m | V k [ g ]] = R g dV k | n ih m | by de(cid:28)nition. On the otherhand, for ea h X ∈ B ( R ) , we get V k | n ih m | ( X ) = Z π e − ikθ h m | P Q θ ( X ) | n i dθ = h m | P Q ( X ) | n i Z π e − ikθ e iθ ( m − n ) dθ. This equals π h n + k | P Q ( X ) | n i if m = n + k and zero otherwise. Hen e, h n + k | V k [ g ] | n i = 2 π Z g dP Q | n ih n + k | = 2 π h n + k | g ( Q ) | n i , and h m | V k [ g ] | n i = 0 whenever m = n + k . Thus, tr[ T V k [ g ]] = ∞ X n =0 h n | T V k [ g ] | n i = ∞ X n =0 ∞ X m =0 h n | T | m ih m | V k [ g ] | n i = 2 π ∞ X n =0 h n | T | n + k ih n + k | g ( Q ) | n i . (cid:3) Now we are ready to prove the a tual result of the paper.Theorem 1. Let S be a dense subset of [0 , π ) . The set { P Q θ | θ ∈ S } of observables isinformationally omplete.Proof. Let T ∈ L ( H ) be a selfadjoint tra e lass operator, su h that tr[ T P Q θ ( X )] = 0 for all X ∈ B ( R ) and θ ∈ S . Sin e θ tr[ T P Q θ ( X )] is ontinuous and S is dense it follows that tr[ T P Q θ ( X )] = 0 for all X ∈ B ( R ) and θ ∈ [0 , π ) .Let k ∈ N be (cid:28)xed. By the de(cid:28)nition of the operator measure V k , the assumption implies that V kT ( X ) = tr[ T V k ( X )] = R π e − ikθ tr[ T P Q θ ( X )] dθ = 0 for all X ∈ B ( R ) . From the de(cid:28)nition ofthe σ -weak integral V k [ g ] = R g dV k , it follows that tr[ T V k [ g ]] = R g dV kT = 0 for any boundedBorel fun tion g : R → R . Hen e, by using Lemma 5, we get(4) ∞ X n =0 h n | T | n + k i h n + k | g ( Q ) | n i = 0 for any bounded Borel fun tion g : R → R . We show by indu tion with respe t to n that h n | T | n + k i = 0 for all n ∈ N . First, put g = f ( k ) in (4) (re all that f was a fun tion satisfyingthe assumptions of Lemma 2). By Lemma 4, this gives h | T | k i = 0 , whi h proves the initialstep. The indu tion assumption is that for some m ∈ N , h n | T | n + k i = 0 for all n ∈ N , n ≤ m .We have to show that this implies h m + 1 | T | ( m + 1) + k i = 0 . By the indu tion assumptionand (4), we have ∞ X n = m +1 h n | T | n + k ih n + k | f ( k +2( m +1)) ( Q ) | n i = 0 , where we have simply put g = f ( k +2( m +1)) , whi h is again a bounded Borel fun tion. But,a ording to Lemma 4, h n + k | f ( k +2( m +1)) ( Q ) | n i = 0 , n > m + 1 , h ( m + 1) + k | f ( k +2( m +1)) ( Q ) | m + 1 i 6 = 0 , so that ne essarily h m + 1 | T | ( m + 1) + k i = 0 . This ompletes the indu tion proof.We have thus established that h l | T | l + k i = 0 for all l, k ∈ N . Sin e T is selfadjoint, thisimplies that T = 0 , and the proof is omplete. (cid:3) Note that the set S in the previous theorem an be hosen to be ountable (e.g. S =[0 , π ) ∩ Q ). Hen e, in prin iple, it su(cid:30) es to measure a sequen e of quadrature observables inorder to determine the state of the system.Though obvious, it may be worth to note that the quadrature observables Q θ are not in-formationally omplete in the sense of statisti al expe tation, that is, the numbers tr[ T Q θ ] , θ ∈ [0 , π ) , do not, in general, determine the state T ; for instan e, the number states | n i areindistinguishable by the expe tations, h n | Q θ | n i = 0 for all n and for all θ .4. Wigner fun tion vs. phase spa e distributionsA ording to the result in the pre eding se tion, the quadrature observables Q θ , θ ∈ S ( S ⊆ [0 , π ) is dense) onstitute an informationally omplete set of observables, i.e. the measurementstatisti s p Q θ T , θ ∈ S , determine uniquely the state T of the quantum system. The question ofexperimental implementation of these observables is thus of utmost importan e.The balan ed homodyne dete tion with a strong auxiliary (cid:28)eld is a well developed method ofexperimental quantum physi s, and this method is known to yield the measurement statisti sof the quadrature observable Q θ , depending on the phase e iθ of the (one-mode) auxiliary (cid:28)eld.The heuristi physi al argument behind this method is equally well known, see, e.g. [10, 7℄, thedetailed mathemati al justi(cid:28) ation being, however, more involved.If | z i , z = re iθ , is the oherent state of the (one-mode) auxiliary (cid:28)eld, then the a tuallymeasured observable in the balan ed homodyne dete tion is given by a semispe tral measure E z , whose (cid:28)rst moment operator E z [1] is an extension of the restri tion of the quadratureoperator Q θ on the domain D ( a ) of the signal mode operator a , and whose noise operator E z [2] − E z [1] equals with the operator r − N (where N = a ∗ a ). Clearly, these results suggestthat the high amplitude limit of E z is the spe tral measure P Q θ of Q θ , notably sin e the spe tralmeasures are known to be exa tly those semispe tral measures whose noise operators equal zero.There is, indeed, a rigorous quantum me hani al proof of the fa t that in the high amplitudelimit the observable E z tends to the spe tral measure of Q θ , though the a tual meaning of thislimit requires more aution [8℄.The balan ed homodyne dete tion s heme thus allows one to olle t the statisti s of the quad-rature observables. Assuming that the inverse Radon transform an be applied to the olle tionof distributions p Q θ T , θ ∈ [0 , π ) , one obtains the Wigner fun tion f WT of the state T , whi h isa unique representation of T . Clearly, there is no need to use the inverse Radon transformnor the Wigner fun tion, sin e the set of quadrature observables is, in itself, informationally omplete. In any ase, this method of state re onstru tion is unne essarily ompli ated, sin eit requires that one measures in(cid:28)nite number of di(cid:27)erent observables Q θ , θ ∈ S .It is well-known that a beam splitter ombined with the Q and Q π/ -sensitive dete tors onstitutes a measurement of a phase spa e observable E D of the signal mode, with the gen-erating density operator D depending on the state of the (one-mode) auxiliary (cid:28)eld, see, e.g.[4, VII.3.7℄. Using the balan ed homodyne dete tion realization for the Q and Q π/ -sensitivedete tors one indeed obtains, in a rigorous quantum me hani al sense, an eight-port homodyne7ete tor realization of an arbitrary phase spa e observable E D [9℄. A phase spa e observable E D is known to be informationally omplete whenever the generating density operator D issu h that tr[ W qp D ] = 0 for almost all phase spa e points ( q, p ) ∈ R ; here W qp is the Weyloperator, see, e.g. [2℄. In parti ular, if the auxiliary (cid:28)eld is idle, then the measurement statisti obtained from the eight-port homodyne dete tor is simply the Husimi fun tion f HT of the signalstate T . Like the Wigner fun tion, the Husimi fun tion separates states, that is, the phasespa e observable E D , D = | ih | , is informationally omplete. It is obvious that the statere onstru tion via the eight-port homodyne dete tor is a huge simpli(cid:28) ation when omparedwith the re onstru tion using only the balan ed homodyne dete tion te hnique.Appendix: The proof of Lemma 1To begin the proof of the lemma, we (cid:28)rst note that the Dawson's integral is learly a C ∞ -fun tion. We prove (a) by using the expansion daw( x ) = 12 x n − X j =0 (2 j − x ) j + (2 n − n R n ( x ) , where R n ( x ) = e − x x n − ∞ X j =0 x j j !(2 j − n + 1) and n = 1 , , , . . . (see [14, equation 42:6:5, p. 407℄). Sin e ea h derivative is either even orodd, it su(cid:30) es to onsider the limit x → ∞ . Clearly, any derivative of the (cid:28)rst part tends tozero at this limit, for any hoi e of n . As for R n , it is easy to see that for any given k , we get lim x →∞ R ( k ) n ( x ) = 0 for su(cid:30)ently large n . Indeed, let S n ( y ) = ∞ X j =0 y j j !(2 j − n + 1) , so that R n ( x ) = e − x x n − S n ( x ) . Sin e the power series S n ( y ) is learly onvergent for all y ∈ R ,it an be di(cid:27)erentiated k times (for any k ∈ N ) to get S ( k ) n ( y ) = ∞ X j = k y j − k ( j − k )!(2 j − n + 1) . From this we see that | S ( k ) n ( y ) | ≤ e y for any y > and n ∈ N . Now for a (cid:28)xed k ∈ N , put e.g. n = k + 1 . Sin e R ( k ) n ( x ) is learly a (cid:28)nite sum of terms of the form A l,l ′ e − x x n − − l S ( l ′ ) n ( x ) , with − k ≤ l ≤ k , ≤ l ′ ≤ k , and A l,l ′ a onstant, it follows that lim x →∞ R ( k ) n ( x ) = 0 . The proof of(a) is omplete.To prove (b), onsider the series(5) ∞ X n =0 ( − n n !2 n (2 n )! H n ( x ) . We (cid:28)rst note that the well-known relation ddx H l ( x ) = 2 lH l − ( x ) , l = 1 , , . . . , implies(6) d m dx m H n ( x ) = 2 n (2 n − · · · (2 n − m + 1)2 m H n − m ( x ) , m ≤ n. | H n ( x ) | ≤ e x K n √ n ! , where K > is a onstant ([1, 22.14.17, p. 787℄),we get (cid:12)(cid:12)(cid:12)(cid:12) ( − n n !2 n (2 n )! d m dx m H n ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ m K n ! 2 n (2 n − · · · (2 n − m + 1) p (2 n − m )!(2 n )! e x ≤ m K (2 n ) m n ! p (2 n )! e x for all n, m ∈ N , m ≤ n . Now P ∞ n =0 (2 n ) m n ! √ (2 n )! < ∞ for any m ∈ N by the ratio test, whi h showsthat the series obtained by di(cid:27)erentiating (5) m times term by term onverges uniformly inbounded intervals. Consequently, that series represents the m th derivative of the original series(5), the latter onverging to some C ∞ -fun tion F : R → R uniformly in bounded intervals.By using again the formula ddx H l ( x ) = 2 lH l − ( x ) , we get F (2 m ) ( x ) = 2 m ∞ X n = m n !( − n H n − m ) ( x )2 n (2( n − m ))! ; (7) F (2 m +1) ( x ) = 2 m +1 ∞ X n = m +1 n !( − n H n − m ) − ( x )2 n (2( n − m ) − . (8)In order to al ulate the Ma Laurin series of F , we need the expansion(9) (1 − x ) − ( m +1) = ∞ X n =0 (cid:18) m + nn (cid:19) x n , − < x < , whi h an be obtained from the binomial series (see 3.6.9. in [1, p. 15℄).Now, using (7), the identity H n − m ) (0) = ( − n − m (2( n − m ))! / ( n − m )! , n ≥ m (22.4.8 in[1, p. 777℄), and (9) with x = , we get F (2 m ) (0) = 2 m ( − m ∞ X n = m n !2 n ( n − m )! = 2 m ( − m ∞ X n =0 ( n + m )!2 n n != 2 m ( − m m ! ∞ X n =0 (cid:18) n + mn (cid:19) n = 2 m +1 ( − m m ! . Sin e F (2 m +1) (0) = 0 by (8) and the fa t that H l (0) = 0 for odd l , we get the followingMa Laurin series for F :(10) ∞ X m =0 ( − m m !(2 m )! (2 x ) m , x ∈ R . This is exa tly the Ma Laurin series for the (cid:28)rst derivative of the Dawson's integral, sin e daw( x ) = ∞ X m =0 ( − m m !2 m (2 m + 1)! x m +1 (see e.g. [14, p. 406℄).In order to prove that the series (10) a tually onverges to F (pointwise for all x ∈ R ), we haveto show that for ea h x ∈ R , the orresponding remainder F ( k ) ( ξ k ) x k /k ! , where ξ k ∈ [ −| x | , | x | ] ,9oes to zero as k → ∞ . By applying the estimate | H l ( y ) | ≤ Ke y l/ √ l ! to (7) and (8), weget | F (2 m ) ( y ) | ≤ e y m K ∞ X n =0 ( n + m )! p (2 n )! = Ke y m m ! ∞ X n =0 (cid:18) n + mn (cid:19) n ! p (2 n )! ;2 | F (2 m +1) ( y ) | ≤ e y √ m K ∞ X n =0 ( n + m + 1)! p (2 n + 1)! = √ Ke y m ( m + 1)! ∞ X n =0 (cid:18) n + m + 1 n (cid:19) n ! p (2 n + 1)! for all y ∈ R . It is easy to see by indu tion that n ! / p (2 n )! ≤ n/ n − for all n ≥ ; using this,as well as the relation ( m + 1)2 m +1 = P ∞ n =0 (cid:0) m + nn (cid:1) n n whi h is obtained by di(cid:27)erentiating (9)and putting x = , we on lude that | F (2 m ) ( y ) | ≤ Ke y m ( m + 1)! , and | F (2 m +1) ( y ) | ≤ √ Ke y m +1 ( m + 2)! for all y ∈ R . Consequently, | F (2 m ) ( ξ ) x m | (2 m )! ≤ Ke x (2 | x | ) m ( m + 1)!(2 m )! , | F (2 m +1) ( ξ ) x m +1 | (2 m + 1)! ≤ √ Ke x (2 | x | ) m +1 ( m + 2)!(2 m + 1)! , whenever x ∈ R , ξ ∈ [ −| x | , | x | ] and m ∈ N . It is easy to see that for ea h (cid:28)xed x ∈ R , the righthand sides of the inequalities tend to zero as m → ∞ . Hen e, we have shown that F ( x ) = ∞ X n =0 ( − m m !(2 m )! (2 x ) m = daw (1) ( x ) , x ∈ R ..