Quantum Mechanics for Mathematicians: Energy, Momentum, and the Quantum Free Particle

Transcription

1 Quantum Mechanics for Mathematicians: Energy, Momentum, and the Quantum Free Particle Peter Woit Department of Mathematics, Columbia University November 28, 2012 We ll now turn to the problem that conventional quantum mechanics courses generally begin with: that of the quantum system describing a free particle moving in physical space R 3. The state space H will be a space of complex-valued functions on R 3, called wave-functions. There is one crucial sort of observable to understand: the momentum operator. This operator will have the same relationship to spatial translations as the Hamiltonian does to time translations. In both cases, the operators are given by the Lie algebra representation corresponding to the unitary representation on the quantum state space of groups of translations (translation in the three space and one time directions respectively). One way to motivate the quantum theory of a particle is that, whatever it is, it should have the same sort of behavior under the translational and rotational symmetries of space-time. Invoking the classical relationship between energy and momentum used in non-relativistic mechanics relates the Hamiltonian and momentum operators, giving the conventional Schrödinger differential equation for the wave-function of a free particle. We will examine the solutions to this equation, beginning with the case of periodic boundary conditions, where spatial translations in each direction are given by the compact group U(1) (whose representations we have studied in detail). 1 Energy, momentum and space-time translations We have seen that it is a basic axiom of quantum mechanics that the observable operator responsible for infinitesimal time translations is the Hamiltonian operator H, a fact that is expressed as i d ψ = H ψ dt When H is time-independent, one can understand this equation as reflecting the existence of a unitary representation (U(t), H) of the group R of time transla- 1

2 tions on the state space H. U(t) is a homomorphism from the group R to the group of unitary transformations of H. It is of the form U(t) = e i th Knowing the time-dependence of a state is equivalent to knowing the action of the group R of time-translations on the state, which means knowing what U(t) is. We get a Lie algebra representation of R by taking the time derivative of U(t), which gives us d dt U(t) t=0 = i H As the Lie algebra representation coming from taking the derivative of a unitary representation, i H will be skew-adjoint, so H will be self-adjoint. The minus sign is a convention, and the factor of depends on the choice of units of time and energy. Once one has chosen a unit of either time or energy, it would be natural to choose the other unit so as to make = 1. Since ψ(t) = U(t) ψ(0) taking this derivative also gives us the equation d dt ψ(t) t=0 = i H ψ(0) and for time-independent H the Schrödinger equation is just the statement that this holds for all t. If we have a group R that is explicitly represented as a group of translations on functions in a variable u, then the Lie algebra is also R, represented as elements proportional to d du. The way exponentiation relates the Lie algebra and Lie group is reflected here by Taylor s formula: one can think of translation by a as e a d du since Taylor s formula says that f(a) = e a d df a2 d 2 f du f(0) = f(0) + a (0) + du 2! du 2 (0) + There clearly are issues of convergence here which we are not entering into. Since we now want to describe quantum systems that depend not just on time, but on space variables q = (q 1, q 2, q 3 ), we will have an action by unitary transformations of not just the group R of time translations, but also the group R 3 of spatial translations. We will define the corresponding Lie algebra representations using self-adjoint operators P 1, P 2, P 3 that play the same role for spatial translations that the Hamiltonian plays for time translations. They will satisfy analogs of the Schrödinger equation as follows: Definition. Momentum operators For a quantum system with a unitary representation of the group R 3 of spatial translations in variables (q 1, q 2, q 3 ) on the state space H, the momentum operators P 1, P 2, P 3 are defined by P 1 = i q 1, P 2 = i q 2, P 3 = i q 3 2

3 These are given the name momentum operators since we will see that their eigenvalues have an intepretation as the components of the momentum vector for the system, just as the eigenvalues of the Hamiltonian have an interpretation as the energy. Note that the same choice of relationship between units (a factor of ) is made here as in the case of the Hamiltonian. The same factor of i appears, turning the skew-adjoint differentiation operators into self-adjoint momentum operators, but the convention for the sign choice is the opposite here from the case of the Hamiltonian. The reason for this comes from considerations of special relativity. We will review special relativity next semester, but for now we just need the relationship it gives between energy and momentum. Space and time are put together in Minkowski space, which is R 4 with indefinite inner product < (u 0, u 1, u 2, u 3 ), (v 0, v 1, v 2, v 3 ) >= u 0 v 0 u 1 v 1 u 2 v 2 u 3 v 3 Energy and momentum are the components of a Minkowski space vector (p 0 = E, p 1, p 2, p 3 ) with norm-squared given by the mass-squared: < (E, p 1, p 2, p 3 ), (E, p 1, p 2, p 3 ) >= E 2 p 2 = m 2 This is the formula for a choice of space and time units such that the speed of light is 1. Putting in factors of the speed of light c to get the units right one has E 2 p 2 c 2 = m 2 c 4 Two special cases of this are: For photons, m = 0, and one has the energy momentum relation E = p c For velocities v small compared to c (and thus momenta p small compared to mc), one has E = p 2 c 2 + m 2 c 4 = c p 2 + m 2 c 2 c p 2 2mc = p 2 2m This is the non-relativistic limit, and we use this energy-momentum relation to describe particles with velocities small compared to c. Next semester we will discuss quantum systems that describe photons, as well as other possible ways of constructing quantum systems for relativistic particles. For now though, we will stick to the non-relativistic case. To describe a quantum non-relativistic particle, we write the Hamiltonian operator in terms of the components of the momentum operator, in such a way that the eigenvalues of the operators (the energy and momentum) will satisfy the classical equation E = p 2 2m : H = 1 2m (P P2 2 + P3 2 ) = 1 2m P 2 = 2 2m ( 2 q q q3 2 ) 3

4 The Schrödinger equation then becomes: i 2 ψ(q, t) = t 2m ( 2 q q q3 2 )ψ(x, t) = 2 2m 2 ψ(q, t) This is an easily solved simple constant coefficient second-order partial differential equation. One method of solution is to separate out the time-dependence, by first finding solutions ψ E to the time-independent equation Hψ E (q) = 2 2m 2 ψ E (q) = Eψ E (q) with eigenvalue E for the Hamiltonian operator and then using the fact that ψ(q, t) = ψ E (q)e i te will give solutions to the full-time dependent equation i ψ(q, t) = Hψ(q, t) t The solutions ψ E (q) to the time-independent equation are just complex exponentials proportional to e i(k1q1+k2q2+k3q3) = e ik q satisfying 2 2m ( i)2 k 2 = 2 k 2 2m = E We see that the solutions to the Schrödinger equation are all linear combinations of states k labeled by a vector k which are eigenstates of the momentum and Hamiltonian operators with P a k = k a k, H k = 2 2m k 2 k These are states with well-defined momentum and energy p a = k a, E = p 2 2m so satisfy exactly the same energy momentum relations as those for a classical non-relativistic particle. While the quantum mechanical state space H contains states with the classical energy-momentum relation, it also contains much, much more. A general state will be a linear combination of such states. At t = 0 one has ψ = k c k e ik q 4

5 where c k are complex numbers, and the general time-dependent state will be ψ(t) = k c k e ik q k 2 it e 2m or, equivalently in terms of momenta p = k ψ(t) = p c p e i p q e i p 2 2m t 2 Periodic boundary conditions and the group U(1) We have not yet discussed the inner product on our space of states when they are given as wave-functions on R 3, and there is a significant problem with doing this. To get unitary representations of translations, we need to use a translation invariant, Hermitian inner product on wave-functions, and this will have to be of the form < ψ 1, ψ 2 >= C ψ 1 (q)ψ 2 (q)dq R 3 for some constant C. But if we try and compute the norm-squared of one of our basis states k we find k k = C (e ik q )(e ik q )dq = C R 3 1 dq = R 3 As a result, while our basis states k are orthogonal for different values of k, there is no value of C which will make them of norm-squared one. In the finite dimensional case, a linear algebra theorem assure us that given a self-adjoint operator, we can find an orthonormal basis of its eigenvectors. In this infinite dimensional case this is no longer true, and a much more sophisticated formalism (the spectral theorem for self-adjoint operators ) is needed to replace the linear algebra theorem. This is a standard topic in treatments of quantum mechanics aimed at mathematicians emphasizing analysis, but we will not try and enter into this here. One place to find such a discussion is section 2.1 of [2]. One way to deal with the normalization problem is to replace the noncompact space by one of finite volume. We ll show how this works for the simplified case of a single spatial dimension, since once one sees how this works for one dimension, treating the others the same way is straight-forward. In this one dimensional case, one replaces R by the circle S 1. This is equivalent to the physicist s method of imposing periodic boundary conditions, meaning to define the theory on an interval, and then identify the ends of the interval. One can then think of the position variable q as an angle φ and define the inner product as < ψ 1, ψ 2 >= 1 ψ 1 (φ)ψ 2 (φ)dφ 0 5

6 and take as state space H = L 2 (S 1 ) the space of complex-valued square-integrable functions on the circle. Instead of the translation group R, we have the standard action of the group SO(2) on the circle. Elements g(θ) of the group are rotations of the circle counterclockwise by an angle θ, or if we parametrize the circle by an angle φ, just shifts φ φ + θ Recall that in general we can construct a representation on functions from a group action on a space by π(g)f(x) = f(g 1 x) so we see that this rotation action on the circle gives a representation on H π(g(θ))ψ(φ) = ψ(φ θ) If X is a basis of the Lie algebra so(2) (for instance ( taking ) the circle as the unit 0 1 circle in R 2, rotations 2 by 2 matrices, X =, g(θ) = e 1 0 θx ) then the Lie algebra representation is given by taking the derivative so we have π (X)f(φ) = d dθ f(φ θ) θ=0 = f (φ) π (X) = d dφ This operator is defined on a dense subspace of H = L 2 (S 1 ) and is skew-adjoint, since (using integration by parts) < ψ 1, ψ 2 >= 1 = = < ψ 1, ψ 2 > ψ 1 d dφ ψ 2dφ ( d dφ (ψ 1ψ 2 ) ( d dφ ψ 1)ψ 2 )dφ The eigenfunctions of π (X) are just the e inφ, for n Z, which we will also write as state vectors n. These are also a basis for the space L 2 (S 1 ), a basis that corresponds to the decomposition into irreducibles of L 2 (S 1 ) as a representation of SO(2) described above. One has (π, L 2 (S 1 )) = n Z (π n, C) 6

7 where π n are the irreducible one-dimensional representations given by π n (g(θ)) = e inθ The theory of Fourier series for functions on S 1 says that one can expand any function ψ L 2 (S 1 ) in terms of this basis, i.e. ψ = ψ(φ) = + n= c n e inφ = + n= c n n where c n C. The condition that ψ L 2 (S 1 ) corresponds to the condition + n= c n 2 < on the coefficients c n. Using orthonormality of the n we find c n = n ψ = 1 0 e inφ ψ(φ)dφ The Lie algebra of the group S 1 is the same as that of the group (R, +), and the π (X) we have found for the S 1 action on functions is related to the momentum operator in the same way as in the R case (up to a sign, which is a convention anyway). So, we can use the same momentum operator which satisfies P = i d dφ P n = n n By changing space to the compact S 1 we now have momenta that instead of taking on any real value, can only be integral numbers times. Solving the Schrödinger equation as before, we find i 2 P ψ(φ, t) = t 2m Eψ E (φ) = 2 2m ψ(φ, t) = 2 2m d 2 dφ 2 ψ E(φ) an eigenvector equation, which has solutions n, with E = 2 2m n2 The general solution to the Schrödinger equaton is ψ(φ, t) = + n= 7 c n e inφ n2 i e 2m t 2 ψ(φ, t) φ2

8 with the c n determined from the initial condition of knowing the wave-function at time t = 0, according to the Fourier coefficient formula c n = 1 0 e inφ ψ(φ, 0)dφ To get something more realistic, we need to take our circle to have an arbitrary circumference R, and we can study our original problem by considering the limit R. To do this, we just need to change variables from φ to φ R, where φ R = Rφ The momentum operator will now be P = i d dφ R and its eigenvalues will be quantized in units of R. The energy eigenvalues will be E = 2 n 2 2mR 3 The group R and the Fourier transform In the previous section, we used the trick of replacing the group R of translations by a compact group S 1, and then used the fact that unitary representations of this group are labeled by integers. This made the analysis rather easy, with H = L 2 (S 1 ) and the self-adjoint operator P = i φ behaving much the same as in the finite-dimensional case: the eigenvectors of P give a countable orthonormal basis of H. If one wants to, one can think of P as an infinite-dimensional matrix. Unfortunately, in order to understand many aspects of quantum mechanics, we can t get away with this trick, but need to work with R itself. One reason for this is that the unitary representations of R are labeled by the same group, R, and we would like to exploit this symmetry. What plays the role here of n = e inφ, n Z will be the k = e ikq, k R. These are functions on R that are irreducible representations under the translation action (π(a) acts by translation by a) π(a)e ikq = e ik(q+a) = e ika e ikq We can try and mimic the Fourier series decomposition, with the coefficients c n that depend on the labels of the irreducibles replaced by a function f(k) depending on the label k of the irreducible representation of R. Definition. Fourier transform The Fourier transform of a function ψ is given by Fψ = ψ(k) = 1 e ikq ψ(q)dq 8

9 The definition makes sense for ψ L 1 (R), Lebesgue integrable functions on R. For the following, it is convenient to instead restrict to the Schwartz space S(R) of functions ψ such that the function and its derivatives fall off faster than any power at infinity (which is a dense subspace of L 2 (R)). For more details about the analysis and proofs of the theorems quoted here, one can refer to a standard textbook such as [1]. Given the Fourier transform of ψ, one can recover ψ itself: Theorem. Fourier Inversion For ψ S(R) the Fourier transform of a function ψ S(R), one has ψ(q) = F ψ = 1 e ikq ψ(k)dk Note that F is the same linear operator as F, with a change in sign of the argument of the function it is applied to. Note also that we are choosing one of various popular ways of normalizing the definition of the Fourier transform. In others, the factor of may appear instead in the exponent of the complex exponential, or just in one of F or F and not the other. The operators F and F are thus inverses of each other on S(R). One has Theorem. Plancherel F and F extend to unitary isomorphisms of L 2 (R) with itself. words ψ(q) 2 dq = ψ(k) 2 dk Note that we will be using the same inner product on functions on R < ψ 1, ψ 2 >= ψ 1 (q)ψ 2 (q)dq both for functions of q and their Fourier transforms, functions of k. A crucial example is the case of Gaussian functions where Fe α q = e ikq e α q2 2 dq = 1 e α 2 ((q+i 2k α )2 ( ik α )2) dq = 1 e k2 2α = 1 α e k2 2α e α 2 q 2 dq In other A crucial property of the unitary operator F on H is that it diagonalizes the differentiation operator and thus the momentum operator P. Under Fourier transform, differential operators become just multiplication by a polynomial, 9

10 giving a powerful technique for solving differential equations. Computing the Fourier transform of the differentiation operator using integration by parts, we find dψ dq = 1 ikq dψ e dq dq = 1 ( d dq (e ikq ψ) ( d dx e ikq )ψ)dq =ik 1 e ikq ψdq =ik ψ(k) So under Fourier transform, differentiation by q becomes multiplication by ik. This is the infinitesimal version of the fact that translation becomes multiplication by a phase under Fourier transform. If ψ a (q) = ψ(q + a), one has ψ a (k) = 1 e ikq ψ(q + a)dq = 1 e ik(q a) ψ(q )dq =e ika ψ(k) Since p = k, we can easily change variables and work with p instead of k, and often will do this from now on. As with the factors of, there s a choice of where to put the factors of in the normalization of the Fourier transform. We ll make the following choices, to preserve symmetry between the formulas for Fourier transform and inverse Fourier transform: ψ(p) = 1 ψ(q) = 1 pq i e dq e i pq dp Note that in this case we have lost an important property that we had for finite dimensional H and had managed to preserve by using S 1 rather than R as our space. If we take H = L 2 (R), the eigenvectors for the operator P (the functions e ikq ) are not square-integrable, so not in H. The operator P is an unbounded operator and we no longer have a theorem saying that its eigenvectors give an orthornormal basis of H. As mentioned earlier, one way to deal with this uses a general spectral theorem for self-adjoint operators on a Hilbert space, for more details see Chapter 2 of [2]. 3.1 Delta functions One would like to think of the eigenvectors of the operator P as in some sense continuing to provide an orthonormal basis for H. A possible approach to this 10

11 is to work with generalized functions, known as distributions. We won t try and develop here this kind formalism in a rigorous form, but will just explain the non-rigorous form in which it is often used in physics. Given any function g(q) on R, one can try and define an element of the dual space of the space of functions on R by integration, i.e by the linear operator f g(q)f(q)dq (we won t try and specify which condition on functions f or g is chosen to make sense of this). There are however some other very obvious linear functionals on such a function space, for instance the one given by evaluating the function at q = q : f f(q ) Such linear functionals correspond to generalized functions, objects which when fed into the formula for integration over R give the desired linear functional. The most well-known of these is the one that gives evaluation at q = c, it is known as the delta function and written as δ(q c). It is the object which, if it were a function, would satisfy δ(q c)f(q)dq = f(c) To make sense of such an object, one can take it to be a limit of actual functions. For the δ-function, consider the limit as ɛ 0 of g ɛ = 1 e (q c)2 2ɛ ɛ which satisfy g ɛ (q)dq = 1 for all ɛ > 0 (one way to see this is to use the formula given earlier for the Fourier transform of a Gaussian). Heuristically (ignoring obvious problems of interchange of integrals that don t make sense), one can write the Fourier inversion formula as follows ψ(x) = 1 e ikq ψ(k)dk = 1 e ikq 1 + ( e ikq ψ(q )dq )dk = 1 = ( δ(q q)ψ(q )dq e ik(q q ) ψ(q )dk)dq 11

12 Taking the delta function to be an even function (so δ(x x) = δ(x x )), one can interpret the above calculation as justifying the formula δ(q q ) = 1 One then goes on to consider the eigenvectors k = 1 e ikq e ik(q q ) dk of the momentum operator as satisfying a replacement for the finite-dimensional orthonormality relation, with the δ-function replacing the δ ij : k k = 1 ( e ik q 1 )( e ikq )dq = 1 e i(k k )q dq = δ(k k ) As mentioned before, we will usually work with the variable p = k, in which case we have p = 1 e i pq and δ(p p ) = 1 e i (p p )q dq For a more mathematically legitimate version of this calculation, one place to look is Lecture 6 in the notes on physics by Dolgachev[3]. 4 For further reading Pretty much every book about quantum mechanics covers this example of the free quantum particle somewhere very early on, in detail. Our discussion here is unusual just in emphasizing the role of the spatial translation groups and its unitary representations. Discussions of quantum mechanics for mathematicians (such as [2]) typically emphasize the development of the functional analysis needed for a proper description of the Hilbert space H and of the properties of general self-adjoint operators on this state space. In this class we re restricting attention to a quite limited set of such operators coming from Lie algebra representations, so will avoid the general theory. References [1] Strichartz, R., A Guide to Distribution Theory and Fourier Transforms, World Scientific, [2] Takhtajan, L., Quantum Mechanics for Mathematicians, AMS, [3] Dolgachev, I., Introduction to Physics, Lecture Notes 12