Quantum Mechanics I - Session 5 Apil 7, 015 1 Commuting opeatos - an example Remine: You saw in class that Â, ˆB ae commuting opeatos iff they have a complete set of commuting obsevables. In aition you saw that a set of commuting obsevables is complete if togethe thei eigenvalues completely specify the state of the system. Let us consie a physical system, the Hilbet space of which can be efine by the span of thee othogonal states: u 1, u, u 3. In the basis efine by these states (in the same oe) we efine two opeatos: 1 0 0 H = ω 0 1 0 0 0 1 1. Can H, B epesent obsevable quantities? 1 0 0 B = b 0 0 1 (1) 0 1 0. Show that H an B commute. What is the most geneal fom of a matix that is commutable with H? 3. Fin a basis of simultaneous eigenstates fo H an B. 4. Ae {H, B} a complete set of commuting opeatos? What about {H, B}? Solution 1. Both H an B ae symmetical an compose of eal numbes only. Theefoe taking the tanspose an complex conjugate of each of those matices leaves them the same. This means that H = H +, B = B + an theefoe H, B ae Hemitian an can epesent obsevable (physical) quantities.. u 1 is an eigenstate of both H an B: H u 1 = ω u 1, B u 1 = b u 1. () It follows that HB u 1 = BH u 1. It is left only to show that the commutation hols also in the sub-space spanne by u, u 3. Since both H an B leave any state that is a 1
supeposition of u, u 3 within the sub-space efine by these vectos, we may look at the euction of H, B to this sub-space. We efine H as the opeato H limite to this sub-space: ( ) 1 0 H = ω = ωi. (3) 0 1 Since H is popotional to the ientity opeato, it commutes with any opeato in this sub-space. We theefoe conclue that [H, B] = 0. In oe to consie the geneal fom of an opeato M that commutes with H, we stat by noting that H has two eigenvalues: λ 1 = ω coesponing to the eigenvecto u 1 an λ = ω which has a egeneacy level of an coespons to any two linealy inepenent states (that can be chosen to be othogonal) in the sub-space spanne by u, u 3. Theefoe,M must also have u 1 as an eigenstate an two othe eigenstates that ae in the sub-space spanne by u, u 3. We may use this oppotunity to show that in geneal if M, H ae Hemitian an commutable opeatos an u 1, u ae two eigenstates of H with iffeent eigenvalues λ 1, λ then u 1 M u = 0. This is because: HM MH = 0 0 = u 1 HM MH u = (λ 1 λ ) u 1 M u 0 = u 1 M u (4) This means that in geneal M has the fom: M 11 0 0 M = 0 M M 3 (5) 0 M 3 M 33 3. As mentione above, u 1 is a simultaneous eigenstate of B an H. We ae now left with fining the emaining two states in the sub-space spanne by u, u 3. Confine to this sub-space B is given by: ( ) 0 1 B = b (6) 1 0 Nomalize eigenstates of this opeato ae: φ 1 = 1 ( u 1 + u ), φ = 1 ( u 1 u ) (7) coesponing to eigenstates b, b accoingly. These ae natually also eigenstates of H since they belong to the egeneate sub-space of H, in which any state is an eigenstate with eigenvalue ω. Notice that in H: u, u 3 ae egeneate an u 1 has a unique eigenvalue wheeas in B: u 1, φ 1 ae egeneate an φ has a unique value. Theefoe by measuing both B an H fo ou physical system, we may lift the egeneacy that exists in each by itself. 4. Since any two matching eigenvalues of H an B coespon to a single eigenstate, thee is no nee fo anothe opeato to lift the egeneacy. Theefoe {H, B} is a complete set of commuting obsevables. The set {H, B} howeve, is not a complete. Since all the eigenvalues of H ae ω, it cannot help in lifting the egeneacy in B between u 1 an φ 1.
functions of opeatos What is the meaning of f(â) whee  is an opeato? f(â) is itself an opeato that can be efine by its Taylo seies. Assume the Taylo seies of the function f(x) is: f(x) = i a i x i (8) then f(â) is simply: f(â) = a i  i. (9) i Question Assume  an ˆB ae opeatos that commute with thei commutato: [Â, [Â, ˆB]] = [ ˆB, [Â, ˆB]] = 0 an f is an analytic function of its agument. Show that: Solution Let us stat fom f( ˆB) = ˆB : [Â, f( ˆB)] = f ( ˆB)[Â, ˆB]. (10) [Â, ˆB ] = [Â, ˆB] ˆB + ˆB[Â, ˆB] = [Â, ˆB] ˆB = [Â, ˆB]f ( ˆB) (11) Using inuction, let us assume that the claim is coect fo f( ˆB) = ˆB n an show that it is also tue fo f( ˆB) = ˆB n+1 : [Â, ˆB n+1 ] = [Â, ˆB n ] ˆB + ˆB n [Â, ˆB] = [Â, ˆB]n ˆB n 1 ˆB + [ Â, ˆB] ˆB n = [Â, ˆB](n + 1) ˆB n (1) Theefoe the claim is coect fo any n an by ecomposing any function to its Taylo seies, we see that it is coect fo any analytic f( ˆB). 3 Diffeentiation of an opeato Let A(t) be an opeato which epens on an abitay vaiable t. By efinition, the eivative of A(t) with espect to t is given by the limit: A(t) A = lim A(t + t) A(t) t 0. (13) t The matix elements of A(t) in an abitay basis of t-inepenent vectos u i ae functions of t: u i A u j = A ij (t). (14) Let us call ( ) A = u i A u j the matix elements of A ij ae: ( ) A = ij A ij (15) 3
Thus we obtain a vey simple ule: to obtain the matix elements epesenting A, all we must o is take the matix epesenting A an iffeentiate each of its elements (without changing thei places). Example: Diffeentiating e At Let us calculate the eivative of e At. By efinition we have: e At = n (At) n. (16) n! Diffeentiating the seies by tem we obtain: eat = n=0 n tn 1 A n n! = A n=1 (At) n 1 (n 1)! = [ (At) n 1 ]A (17) (n 1)! n=1 We ecognize insie the backets the seies which efines e At. The esult is theefoe: eat = Ae At = e At A. (18) The oe can be change because e At is a function of A. This is not the case if one is inteeste in iffeentiating an opeato such as e At e Bt. In this case we obtain: (eat e Bt ) = Ae At e Bt + e At Be Bt (A + B)e At e Bt. (19) 4 The uncetainty elations an atomic paametes Using uncetainty elations it is possible to unestan the stability of atoms an even to eive simply the oe of magnitue of the imensions an the enegy of the hyogen atom in its goun state. Let us consie, theefoe, an electon in the coulomb fiel of a poton, which we shall assume to be stationay at the oigin of the cooinate system. When the two paticles ae sepaate by a istance, the potential enegy of the electon is: V () = e Assume that the state of the electon is escibe by a spheically symmetic wave function whose spatial extent is chaacteize by o (this means that the pobability of pesence is pactically zeo fo o ). The potential enegy coesponing to this state is then on the oe of: V = e (1) 0 Fo V be as low as possible, it is necessay to take 0 as small as possible. That is, the wave function must be as concentate as possible about the poton. Howeve, it is also necessay to take the kinetic enegy into account. This is whee the uncetainty pinciple comes in: if the electon is confine within a volume of linea imension o, the uncetainty p in its momentum (0) 4
is at least of the oe of / 0. Theefoe even if the aveage momentum is zeo, the kinetic enegy T associate with the state une consieation is not zeo: T T min = ( p) = m e m e 0 () If we take o smalle in oe to ecease the potential enegy, the minimum kinetic enegy () inceases. The lowest total enegy compatible with the uncetainty elation is thus the minimum of the function: E min = T min + V = e m e 0 0 (3) The minimum is obtaine when: an is equal to: E min 0 = m e 3 0 + e 0 = 0 0 = m e e a 0. (4) E 0 = m ee 4 (5) Eq. 4 povies the aius fo the goun state of the hyogen atom in the Boh moel an Eq. 5 povies the coect enegy fo this state. This situation is vey iffeent than what we ae use to in Classical mechanics whee the lowest enegy (minus infinity) is eache at = 0. Thus, we can say that it is the uncetainty elation which enables us to unestan, as it wee, the existence of atoms. 5 Compatibility of the Boh moel fo the atomic levels with the uncetainty pinciple Let us look at the atomic levels beyon the goun state in Boh s moel. Apat fo the quantization of the pemitte obits, the Boh moel escibes the motion of the electon classically. The momentum p = mv of an electon at an obit of aius is elate to the aius by the quantizatuion conition: L = p = n. (6) Fo us to be able to speak in this way of an electon tajectoy in classical tems, the uncetainties in its position an momentum must be negligible compae to an p espectively: which woul mean that: Howeve, the uncetainty pinciple imposes:, p p (7) p p 1. (8) p p p = n = 1 n. (9) Combining Eqs. 8, 9 we see that n 1 is necessay fo the semi-classical pictue of the atom pesente in the Boh moel to be compatible with the uncetainty pinciple. At the lowe levels, the teatment of this moel is too simplistic. 5