Quantum Mechanics in One Hour

Transcription

1 July 13, 001 Revision Quantum Mechanics in One Hour Orlando Alvarez Department of Physics University of Miami P.O. Box Coral Gables, FL 3314 USA Abstract Introductory lecture given at PCMI 001. Contents 1 Introduction 1 Framework 1 3 Schroedinger and Heisenberg Pictures 5 4 Symmetries 6 5 One Dimensional Quantum Mechanics 7 6 The Simple Harmonic Oscillator 7 alvarez@physics.miami.edu

2 1 Introduction The normal historical exposition of Quantum Mechanics is to begin with classical hamiltonian (symplectic) mechanics and quantize the system and obtain quantum mechanics. I only have a one hour exposition and much of the physical and historical motivation will be left out. Notation: We will be working in a complex Hilbert space H. Theinner product in this Hilbert space will be denoted by (, ). We adopt the physics convention that the inner product is anti-linear in the first slot and linear in the second slot: (aφ, bψ) = āb(φ, ψ). To avoid confusion, I will adhere to the mathematics convention that the adjoint of an operator A is denoted by A. Let me inform you that in the physics literature the adjoint is usually denoted by A. What follows is not the most general form of quantum mechanics but it will suffice for our purposes. Framework There are three major ingredients in any physical system. 1. The states of the system S.. A prescription for dynamical evolution. 3. A set of observables, i.e., quantities you would like to measure experimentally. To construct a quantum mechanical theory you beginwithacomplexhilbert space H. The Hilbert space is infinite dimensional in many important physical examples. The states of the system S correspond to the points in the projectivized Hilbert space PH. Remember, if ψ 1,ψ H\{0} then you define an equivalence relation by ψ 1 ψ if there exists λ C, λ 0suchthatψ 1 = λψ.apointintheprojectivized Hilbert space PH is an equivalence class [ψ]. Quantum mechanics is a probabilistic theory and not deterministic. You can ask the following question that does not makesensein classical physics. Question 1. What is the probability that the state [ψ] is in the state [φ]? The answer to this question is one of the central postulates of quantum mechanics. Let P φ : H Hbe the orthogonal projector onto the 1-dimensional linear subspace spanned by φ. 1

3 Postulate 1(Probability). The probability that the state [ψ] is in the state [φ] is given by P φ ψ ψ = (φ, ψ) φ ψ. (1) Note that this expression is independent of the choice of representatives for [φ] and [ψ] andthatittakes values in the closed interval [0, 1]. In physics you will often encounter an implicit assumption that the representatives ψ and φ have unit norm. In this case the formula above simplifies to (φ, ψ). What is the motivation for Postulate 1? First we note that the probability that the state [ψ] beinthestate[ψ] is1asexpected. Next, let ψ have unit normalization and let ξ 1,ξ,ξ 3,... be a complete orthonormal basis for H. Wecanwrite ψ = c n ξ n where n c n =1. () Note that P ξn ψ = c n ξ n and sothe probability that [ψ] isin[ξ n ]is c n. From the equation above we see that the total probability is 1 when decomposed over a complete orthonormal basis. Question. How does a state evolve in time? Postulate (Evolution). The dynamicalevolution ofstates isgiven by a 1-parameter group of unitary transformations U(t) :H H. Ifψ(0) is a representative for the initial state then the state at time t is determined by ψ(t) =U(t)ψ(0). n The 1-parameter group property tells us that U(0) = I, U(t)U(t )=U(t + t ), and U( t) =U(t) 1. (3) Problem 1. Show that if ψ 1 (0) ψ (0) then ψ 1 (t) ψ (t). Therefore, we have a well defined induced map [U(t)] : S S. Note that if ψ(0) =1then ψ(t) =1and wedonot have loss of probability as the system evolves. Problem. Show that time evolution preserves probabilities. Namely, if the probability that [ψ(0)] is in the state [φ(0)] is p then the probability that [ψ(t)] is in the state [φ(t)] is also p.

4 Problem 3 (The Hamiltonian). Show that there exists a self-adjoint operator 1 H, called the hamiltonian, such that i du dt U(t) 1 = H. (4) Show that H is independent of t. Weoftenwrite U(t) =e ith. Notice that knowing H is equivalent to knowing U. Problem 4 (The Schroedinger Equation). Show that time development of a state [ψ] is given by the solution to i dψ (t) =Hψ(t). (5) dt The hamiltonian H plays a central role. It is the most important observable of the system. The eigenvectors of H are central for for the following reason. Let φ 0,φ 1,... be a complete orthonormal basis of eigenvectors of H with respective eigenvalues E 0,E 1,... given by Hφ n = E n φ n. (6) Assume we prepare a system in a state [ψ(0)] where ψ(0) = φ k.itiseasytoseethat ψ(t) =e iekt φ k,therefore [ψ(t)] = [ψ(0)]. We conclude that the states [φ n ]corresponding to the eigenvectors of H do not evolve! They are called the stationary states of the hamiltonian. The eigenvalues of H are the allowed energies of the system. If you begin in a stationary state, you remain in the same stationary state. We say that the stationary state [φ k ]hasenergy E k. Before the invention of quantum mechanics (c.a. 195), there was the Old Quantum Theory developed by Planck, Einstein, Bohr, Sommerfeld anddebroglie. Bohr postulated the existence of such stationary states heuristically and derived the spectrum of hydrogen using a variety of ad hoc assumptions. He said that only certain classical orbits were allowed determined by the condition that the angular momentum be an integral multiple of Planck s constant. Allknown reasonable systems have the property that the energy is bounded from below. Postulate 3 (Boundedness of the Hamiltonian). The spectrum of the hamiltonian H is bounded from below. If we order the eigenvectors and eigenvalues such that E 0 E 1 E then E 0 is called the ground state energy. If the multiplicity (degeneracy) of E 0 is one then φ 0 is called the ground state or the vacuum. 1 Physicists are generally imprecise about the hamiltonian and just say that it is hermitian. They really mean that it is self-adjoint. 3

5 Problem 5. Assume ψ(0) = (φ 0 + φ 1 ) /. What is the probability that [ψ(0)] is in the state [φ 0 ]? What is ψ(t)? What is the probability that [ψ(t)] is in [φ 0 ]. Notethat [ψ(t)] [ψ(0)] in general. In any physical system therearequantities that you would like to measure experimentally. The set of thesequantities is called the set of observables O of the system. In aquantum mechanical system an observable A Ois a self-adjoint operator. Assume asystemisinastate[ψ] andwetry to measure the observable A. Whattoweget? We first need a few concepts. Given an A Oand a state [ψ], the real number A [ψ] = (ψ,aψ) (ψ,ψ) (7) is called the expectation value of A in the state [ψ]. Sometimes there is a distinguished state such as the vacuum and then you abbreviate and say, The vacuum expectation value of A is A. If α 1,α,... is a complete set of eigenvectors of A with respective eigenvalues a 1,a,... then A αn = a n. When the state of a system is an eigenvector of A O then the expectation value of the observable is the corresponding eigenvalue. If we choose a unit normalized ψ then we have that ψ = c n α n where c n =1. Note that A [ψ] = a n c n. (8) Problem 6 (Uncertainty). Fix [ψ] S and consider the positive semi-definite hermitian operator ( ) Var(A) [ψ] = A A [ψ]. Show that Var(A) [ψ] =0if and only if ψ is an eigenvector of A with eigenvalue [ψ] A [ψ]. You should try to understand the analogy of the above with the concept of the variance in probability theory. The founders of quantum mechanics put all this together and concluded the following. Postulate 4 (Measurement). Assume the system is in state [ψ] and you measure the observable A. Forclarity of statement, we assume the eigenvalues of A have multiplicity one. The measurement of A will always yield an eigenvalue. The probability of obtaining eigenvalue a n is P αn ψ / ψ. 4

6 If this measurement experiment is repeated many times, the expected value of A will be A [ψ],seeequation (8). It is very important that you try to make sense of the following problem. I could have been more precise but I want you to try figure our what are the ins and outs. Problem 7. Assume A and B are observables. Explain the statement, You can measure A and B simultaneously if [A, B] =0. Problem 8 (Generalized Heisenberg Uncertainty Relation). Let A, B O,show that Var(A)[ψ] Var(B)[ψ] 1 [A, B] [ψ] [ψ] 4 [ψ]. (9) What does the generalized Heisenberg uncertainty relation tell you about attempting to make simultaneous measurements of observables? The Heisenberg uncertainty relation is the case where A = Q, theposition operator, and B = P,themomentum operator, and [Q, P] =ii (usually written [Q, P] =i). The uncertainty relation is usually written in the physics literature as ( Q)( P ) 1/. What can you say in this simpler case if you try to construct a state [ψ] where you specify Q witharbitrary precision. Problem 9. Try to understand well the connection between Var and measurements. 3 Schroedinger and Heisenberg Pictures There are two viewpoints to evolution. We havebeendiscussingtheschroedinger viewpoint where you assume the state evolvesintimebut the observables do not. You can study the time evolution of expectation values A [ψ(t)] = (ψ(t),aψ(t)) (ψ(t),ψ(t)). For example if A = Q the position operator the Q [ψ(t)] tells you where you expect to find the particle at time t. Theequation above may equivalently be written as (ψ(0), (U(t) 1 AU(t)) ψ(0)). (ψ(0),ψ(0)) You can equivalently think of this as having a fixed state [ψ(0)] and an evolving selfadjoint operator A H (t) =U(t) 1 AU(t). This is called the Heisenberg picture of quantum mechanics. The operator A H satisfies the Heisenberg evolutionequation da H = i[h, A H (t)]. (10) dt Note that H H = H and is also timeindependent. 5

7 Schroedinger Picture states evolve in time operators do not evolve Heisenberg Picture states do note evolve in time operators evolve in time Most elementary courses in quantum mechanics use the Schroedinger picture. This is not convenient in a relativistic theory because it singles out a choice of time. In relativistic theories one fixes the state and one looks at the evolution of the operator A H. There are additional operators in the theory, the spatial momentum operators, that generate spatial translations, i.e., spatialderivatives, and you can make everything covariant as we will see in the field theory lectures. 4 Symmetries Asymmetry 3 of the system is a unitary transformation V that commutes with the time evolution operators U(t), i.e., VU(t)V 1 = U(t). Differentiating at t =0weseethat V commutes with the hamiltonian, i.e., VHV 1 = H. Assume you have a 1-parameter group of symmetries V (θ) thatcommuteswithtimeevolution. It is standard in the physics literature to express the 1-parameter group in terms of a generator Q that is taken to be a self-adjoint operator and write V (θ) =e iθq.youcaneasily verify that the observable Q commutes with the hamiltonian, i.e., [Q, H] =0. Youshould now go back and reexamine Problem 7. Under the action of V, youhaveψ ψ(θ) = V (θ)ψ. A A(θ) =V (θ)av (θ) 1.Notethat Similarly an operator da dθ = i[q, A(θ)]. (11) Setting θ =0we see that the infinitesimal action 4 of the 1-parameter group on A is given by i[q, A]. This is a very important observation that is used all the time. Since the physical states are S = PH, projective representations will play a central role. There is a very complicated covariant Schroedinger formulation for relativistic field theories developed by Tomonaga and Schwinger. 3 To discuss time reversal invariance you have to also consider anti-unitary operators. Refer to any standard quantum mechanics textbook for a discussion of this issue. 4 As D. Freed emphasized in his lectures, left and right actions are different. This is the reason for the relative minussignbetween (10) and (11). 6

8 5 One Dimensional Quantum Mechanics Here we discuss the quantum mechanics of a particle moving in 1-dimension. The Hilbert space is the square integrablefunctions H = L (R). The hamiltonian is given by (Hψ)(q)= 1 d ψ + V (q)ψ(q) (1) m dq where q is a standard coordinate for R, m is the mass and V is called the potential function 5.Thepositionobservable Q is the operator multiplication by q : (Qψ)(q)= qψ(q). The momentum observable is the operator: P = i d. Note that these are dq unbounded operators. They satisfy the canonical commutation relations [Q, P] = ii. The word canonical here is likeinclassical mechanics where canonical coordinates are the Darboux coordinates (q,p) with canonical Poisson bracket {q,p} = 1. The abstract three dimensional Lie group with Lie algebra consisting of the operators {I,Q,P} satisfying the canonical commutation relations has a unique nontrivial irreducible unitary representation up to isomorphism (Stone - von Neumann Theorem). The Hilbert space is L (R)andtheoperator realizations are the ones we just discussed. Consider the self-adjoint operator χ [a,b] defined by ( χ[a,b] ψ ) ψ(q) if q [a, b], (q) = 0 if q [a, b]. You can think oftheoperator χ [a,b] as multiplication by the characteristic function for the interval [0, 1]. What is the interpretation of χ [a,b]? I claim thatyoushould [ψ] interpret this as the probability of finding the particle in [a, b]. To see this choosea unit normalized representative ψ, i.e., R ψ(q) dq =1,then b χ[a,b] [ψ] = ψ(q) dq, see Figure 1. Thus ψ(q) dq is the probability density. Note that Q [ψ] = R q ψ(q) dq is really the expectation value of Q inthe sense ofstandard probability theory. a 6 The Simple Harmonic Oscillator The simple harmonic oscillator (SHO) is probably the most important physical system because it is the starting point for much of our understanding of quantum mechanics 5 We assume V is bounded from below. 7

9 a b q Figure 1: What χ [a,b] [ψ] computes. and quantum field theory. This is one dimensional motion with hamiltonian d H = 1 m dq + 1 kq, (13) where k is called the spring constant. This corresponds to a classical restoring force F = kq. Classically the general solution to the harmonic oscillator is q(t) = A cos(ωt)+bsin(ωt) whereω = k/m. Tosolvetheproblem quantum mechanically we observe that the hamiltonian may be written as H = 1 P + 1 m kq. Remember that [Q, P] =i. Define new operators by Q =(mk) 1/4 Q and P =(mk) 1/4 P. Note that [ Q, P ]=i and we can write H = ω H where H = 1 ( ) P + Q. (14) Define operators a and a called respectively the lowering (destruction, annihilation) operator and raising (creation) operator by a = Q + i P, a = Q i P. (15) Problem 10. Compute [a, a ].Show that H = a a + 1 I. (16) The operator N = a a is called the number operator. [N, a ]=+a. Show that [N, a] = a and Problem 11. Note that the operator N is positive semi-definite. Assume φ is an eigenvector of N with eigenvalue λ. Show that the state a φ is an eigenvector with eigenvalue λ +1. Show that the state aφ is an eigenvector with eigenvalue λ 1. Do 8

10 you see why these operators are called raising and lowering operators. Explain why there must exists a non-zero vector φ 0 with the property aφ 0 =0. Give φ 0 unit normalization and define states φ n = 1 n! (a ) n φ 0.Show that φ n =1 and that Nφ n = nφ n. The hamiltonian H = ω(n + 1 )haseigenvectors φ n with eigenvalues E n =(n+ 1 )ω. Since Nφ n = nφ n and E n =(n + 1 )ω we can interpret the state φ n as being made up of n non-interacting excitations each with energy ω built on a ground state (vacuum) with energy 1 ω.thisviewpoint is key in quantum field theory as we shall see in future lectures. Problem 1. Using the definition of a write down the first order differential equation aφ 0 =0.What is its solution? Show that the inequality (9) is saturated by [φ 0 ]. 9