Continuous Quantum Systems

Transcription

1 Chpter 8 Continuous Quntum Systems 8.1 The wvefunction So fr, we hve been tlking bout finite dimensionl Hilbert spces: if our system hs k qubits, then our Hilbert spce hs n dimensions, nd is equivlent to C n. This follows becuse set of qubits hs finite number of sttes: it is only possible to mesure ech qubit in the stte 1 or 0. However, qubits with their finite sttes re not the only thing tht quntum mechnics cn del with. Certinly we could try to mesure the position of quntum prticle, nd the possible outcomes lie on continuum. In wht follows we describe how to del with continuous quntum sttes. We must now expnd our notion of Hilbert spce, since the dimension (ie. number of bsis sttes) runs to infinity. A continuous observble, such s position x, must be represented by n infinitedimensionl mtrix x x 0 ˆx = x where x j denotes ll possible positions on line, in the limit where j becomes continuous vrible. If the prticle is sitting t known position, x p, then its stte, ψ, cn be represented in the position-bsis by the infinite-dimesionl vector ψ = x p =(0, 0,...,0, 1, 0,...,0, 0), where only the p th position is nonzero. Of course, the prticle s stte might lterntively be composed of n rbitrry superposition of position sttes: where = 1. ψ = 0 x x 1 + The mtrix/vector nottion becomes extremely wkwrd s we ttempt to cope with n infinite number of infinitesimlly-spced bsis sttes. To del with this, suppose the prticle s stte, ψ is some rbitrry superposition of infinitesimlly-spced position eigensttes. If we now sk, Wht 73

2 74 CHAPTER 8. CONTINUOUS QUANTUM SYSTEMS is the probbility-density tht the prticle will lie t n rbitrry position, x, represented by the position eigenstte, x? Just s in the finite cse, the nswer is the inner product x ψ. Since x is continuous vrible, this inner product is continuous function of x. This leds us to define ψ(x) = x ψ, nd it is clled the wvefunction of the prticle with respect to position. Now, rther thn struggle to tediously write down infinite superpositions of infinitesimlly-spced bsis sttes, we need only specify the continuous function ψ(x). This contins ll of the complex informtion of the infinite-dimensionl superposition of sttes. Since ψ is unit vector in n infinite-dimensionl Hilbert spce, then ψ(x) must stisfy the condition ψ ψ = lim x j 0 x j = ψ x j x j ψ x j = ψ x x ψ dx = ψ(x) dx =1 The opertor tht represents position, X, now opertes on the inner product, ψ(x), to yield the eigenvlue eqution Xψ(x) =xψ(x), where x is sclr. 8. The Schrödinger Eqution The fundmentl eqution of quntum mechnics is the Schrödinger Eqution, stumbled upon by physicist Erwin Schrödinger in 195. The Schrödinger eqution tells us how quntum prticle in continuous system should behve. The eqution is very difficult to solve, in fct in most rel situtions it is impossible to solve. For prticle free to move in one dimension, the Schrödinger eqution reds HΨ(x, t) =i d Ψ(x, t) dt where H is the Hmiltonin, or energy opertor, of prticle tht cn move in one dimension. Clssiclly the energy of prticle is simply the sum of its kinetic nd potentil energy. The totl energy of prticle with mss m is well defined in terms of the momentum p nd position x of the prticle, nd is given by E(p, x) = p m + V (x) Here, p /m is the kinetic energy nd V (x) is the clssicl potentil energy of the prticle t position x. The form of V (x) depends upon wht interctions the prticle is subjected to (e.g. n electron in mgnetic field, or free photon). Exctly how to get from the clssicl energy function, E(p, x) to the quntum mechnicl energy opertor, H, is not totlly obvious. We will rely on n xiom of quntum mechnics tht we will try to justify (but not derive) lter on. Axiom: If the clssicl energy opertor for system is E(p, x), then the quntum mechnicl Hmiltonin cn be written s H = E(ˆp, ˆx), where ˆp nd ˆx re the quntum mechnicl momentum nd position opertors, respectively. In the position bsis, the ˆx opertor is simply the function x, wheres the ˆp opertor is ˆp = i x. To relly understnd this, we need few exmples.

3 8.. THE SCHRÖDINGER EQUATION 75 The Free Prticle Our first exmple will be tht of free prticle in 1 dimension. Here free mens tht the prticle is subject to no potentil interctions with other prticles. Schrödinger s eqution sys HΨ(x, t) =i Ψ(x, t) Becuse the prticle is free, its clssicl energy is just its kinetic energy p m ; the potentil energy V (x) = 0. Thus, nd H = p m = m x m x Ψ(x, t) =i Ψ(x, t) At first glnce, this eqution looks dunting nd difficult to solve, it is in fct second order prtil differentil eqution. However, mthemticl trick clled seprtion of vribles mkes the eqution firly esy to solve. The mthemticl formlism of seprtion of vribles is not necessry for this course, so if the following discussion is not helpful to you, feel free to skip it. The importnt result we derive is tht the Schrödinger eqution cn be seprted into into two prts: ψ(x) =Eψ(x) m x i φ(t) =Eφ(t) where Ψ(x, t) =ψ(x)φ(t). Here s how it works. Becuse the derivtives in the Schödinger eqution re with respect to different vribles, there is n esy wy to solve the eqution. Becuse the right hnd side nd the left hnd side of the eqution depend on different vribles entirely, we cn sy tht the time dependence of Ψ is independent of x dependence: Ψ(x, t) =Ψ(x, 0)Ψ(0,t). For if this were not true, then when we chnge x without chnging t, the right hnd side of the Shrödinger eqution chnges differently from the left hnd side. In eqution form, this trnsltes to: Ψ(x, 0) [Ψ(0,t)] = [Ψ(x, 0)] i m x Ψ(0,t) If we let ψ(x) =Ψ(x, 0) nd φ(t) =Ψ(0,t), we see tht chnging x does not chnge either φ(t) or φ(t): these terms re constnt with respect to ψ(x). Thus we cn seprte the bove eqution into two prts, one tht describes the time dependence, nd one tht describes the position dependence. ψ(x) =Eψ(x) m x i φ(t) =Eφ(t)

4 76 CHAPTER 8. CONTINUOUS QUANTUM SYSTEMS The first eqution is clled the time-independent Schrödinger eqution, nd it is firly esy to solve. The constnt E is used becuse the Hmiltonin on the right hnd side is the energy of the prticle: Hψ = Eψ. The solution to the time independent Schrödinger eqution is ψ k (x) =e ikx, ψ k (x) =e ikx where k = me/. Ifweletk run negtive, then we only need to think bout the first solution. The solution to the time portion of the Schrödinger eqution is φ k (t) =e iωt where ω = E/ = k /m. Becuse the time dependent eqution is not dependent upon the potentil energy, this is lwys the time dependence of system. To turn solution of the the time independent Schrödinger euqtion into time dependent solution, just tck on n e iek/ to ech k th energy eigenstte. Now, becuse Ψ(x, t) = ψ(x)φ(t), the finl Ψ k (x, t) =e i(kx ωt) We see tht for ech k (or E) there is solution, which gives us continuous set of solutions to the Schrödinger eqution. We cn think of the set Ψ k s bsis for the possible sttes of our prticle, since liner combintions of solutions to liner differentil eqution re lso solutions to the sme eqution. Prticle in Box Another clssic exmple where Schrödinger s eqution is ctully solvble is the prticle in box, lso known s the infinite squre well. In this problem, the prticle is free move however it likes within single line segment (this is its box), but is not llowed to leve. While this sitution is unrelistic nd does not occur in nture, it is hlf decent wy to pproximte n tom. You cn think of n electron in hydrogen tom, for exmple, s being trpped in box. The potentil ner the centrl proton is much lower thn fr from it. It might be stretch to sy tht the this is the sme s our prticle in box, but they certinly re similr. The wy to describe the prticle in box is to sy tht the potentil inside the box is 0, while the potentil outside is infinite: it would tke infinite energy for the prticle to exist outside of its little line segment, thus it cnnot exist outside of the box. V (x) = 0 if 0 <x< otherwise Inside the box, the Hmiltonin is H = Ψ(x, t) = 0. m. x Outside the box we simply mndte tht

5 8.. THE SCHRÖDINGER EQUATION 77 Furthermore, to mke sure tht the Schrödinger eqution mkes sense, we will require tht Ψ(x, t) is continuous. Becuse ψ(x) = 0 outside of the box, we require tht: ψ(0) = ψ(l) =0 We first solve the time-independent Schrödinger eqution without worrying bout the bove restriction: d ψ(x) =Eψ(x) dx We lredy solved this, nd our solution ws ψ k (x) =Ae ikx + Be ikx. But before we cn cll this done we need to impose our boundry condition, i.e. the restriction tht ψ(0) = ψ() = 0. To get rid of the complex exponentils nd mke life little esier, we recll tht for some C nd D, 1 Ae ikx + Be ikx = C sin(kx)+d cos(kx) So ψ k (x) =C sin(kx) +D cos(kx) for some C nd D. To find the conditions on C nd D, we impose our boundry conditions: ψ k (0) = C sin(0) + D cos(0) = D But ψ k (0) = 0 so D = 0, nd we cn forget bout the cosine solution. condition tells us: The second boundry ψ k () =C sin(k) =0 This cn only be stisfied when k = nπ, wheren is n integer. And becuse k = me/, the only llowed energies re E n = n π. Thus, our (lmost finl) set of solutions is, for ech integer n, nπx ψ n (x) =C sin with energye n = n π Notice how the quntiztion of energy levels, the fct tht there is discreet set of energy eigenvlues, nd quntiztion of bsis sttes simply flls out of the mth. This kind of phenomenon is wht gives quntum mechnics its nme. There is one lst step before we cn cll it dy. We need to mke sure tht psi n ψ n = 1 for ll n. This is clled normlizing the wvefunctions. ψ n ψ n = l 0 ψ n (x) dx =1 l 0 C sin nπx dx = C =1 1 If you hve never seen this before, it is not too bd of n exercise to find C nd D with the identity e ikx = cos(kx)+i sin(kx).

6 78 CHAPTER 8. CONTINUOUS QUANTUM SYSTEMS so tht C = /. With everything normlized nd ll boundry conditions ccounted for, we finlly hve our proper set of energy eigenfunctions nd eigenvlues: Qubits ψ n (x) =C sin nπx with energye n = n π While it is importnt to hve this bckground in quntum mechnics to better understnd quntum informtion science, there is direct connection between the solution to the prticle in box problem nd qubits. We hve discussed severl exmples of qubits in previous lectures, one of which used the ground stte nd first excited stte of n tom s the two sttes of qubit. Becuse the prticle in box is simple model for hydrogen tom, we will discuss hydrogen tom qubits in the context of the squre well. To obtin qubit from prticle in box system, we cn construct our stndrd bsis 0 nd 1 by restricting our stte spce to the bottom two eigensttes: 0 = 1 = πx sin, E 1 = π sin πx, E = 4 π Physiclly this would men demnding tht the energy in the box be less thn or equl to E, mening tht the prticle could never hve ny overlp with ψ n for n>. This cn be done in the lb by cooling the tom to very low tempertures (perhps lser cooling). In fct, given the right length of the box, the lowest two sttes of the hydrogen tom re pproximted very well by the infinite squre well. Becuse of this, we cn do some clcultions to tke look t wht hydrogen tom qubit looks like. An rbitrry qubit superposition of the electron stte cn be written s ψ = α 0 + β 1 = α πx sin + β sin πx The time evolution of this stte t some lter time t cn be tcked on by multiply ech eigenstte by e ien/, s noted in the free prticle problem: This cn be rerrnged to become: ψ(t) = α 0 e ie 1t/ + β 1 e ie t/

7 8.. THE SCHRÖDINGER EQUATION 79 or ψ(t) = e ie 1t/ α 0 + β 1 e i(e E 1 )t/ ψ(t) = e ie 1t/ α 0 + β 1 e i( E)t/ The importnt point to notice here is tht s time psses, the phse difference between the two qubit sttes differs by rte tht is proportionl to E, the energy difference between them. For tomic systems this is pretty fst rte, since E = 10 ev corresponds to frequency of ν = E h = Hz. This is very close to the frequency of opticl light, nd thus tomic qubits re controlled opticlly vi interction with light pulses.