Path integrals and the classical approximation D. E. Soper University of Oregon 4 November 0 I offer here some background for Sections.5 and.6 of J. J. Sakurai, Modern Quantum Mechanics. Introduction There is more than one way to understand quantum mechanics. In one way, we study a certain partial differential equation, the one particle Schrödinger equation. More generally, we have states ψt) in a vector space with evolution according to id/) ψt) = H ψt). Here H is an operator on the space of states. Possible measurements and symmetry operations are represented by other operators. In the simplest case, this formulation is the same as the one particle Schrödinger equation. This is the Schrödinger picture for quantum mechanics. There is an alternative way of looking at the same physical content in which the states do not depend on time, but the operators do. This is the Heisenberg picture. There is another way of understanding quantum mechanics, invented by Richard Feynman: the path integral formulation. In the simplest case, we can consider Feynman s formulation in the case of a single particle moving in three dimensions. We still have operators and states, but the emphasis is not on the operators and states and we do not use the Schrödinger equation directly. Instead, we express the amplitude for the particle to get from one place to another as an integral over all paths that it might have taken to do that. This formulation is not wonderfully useful from the point of view of calculation for non-relativistic quantum mechanics, but it does provide us with some insights. In relativistic quantum field theory, the path integral formulation is often the most useful way of expressing the theory. For a certain class of problems, it is also directly useful for calculations. Copyright, 0, D. E. Soper soper@uoregon.edu
Propagator for the Schrödinger equation We consider the propagator We take the hamiltonian to be K x F, t F ; x 0, ) = x F e iht F ) x0. ) H = p op + V x op ). ) The subscripts op indicate that p op and x op are operators. I often use hats for this, but hats and vector signs do not live happily together.) The propagator K x F, t F ; x 0, ) gives the amplitude for a particle at position x 0 at an initial time to be found at position x F at a later time t F. This definition is for t F >. For t F <, one usually defines K = 0. If the particle was originally in a general state ψt0 ), with wave function x0 ψ0), then at time t F its wave function is xf ψtf ) = d x 0 K x F, t F ; x 0, ) x 0 ψt0 ). 3) Thus the propagator tells us quite generally how the particle propagates in time. We can find this exactly for a free particle. For convenience we can set = 0 and x 0 = 0. Then if V = 0 a simple calculation gives K x, t; 0, 0) = [πi] 3/ 3 The path integral m t ) 3/ exp i m t x ). 4) We can derive the Feynman path integral in a simple and elegant fashion. Divide the time interval into small increments of size t. There are a very large number N of time increments, with t = t F )/N. We define intermediate times t i for i =,..., N and we denote t N = t F, so t i = t i + t for i =,..., N. We then have K x F, t F ; x 0, ) = x F e ih t e ih t e ih t x0. 5)
with N factors of exp ih t). Between each pair of factors exp ih t), we introduce a complete sum over position eigenstates, K x F, t F ; x 0, ) = d x N d x d x xf e ih t xn x e ih t 6) x x e ih t x0 Physically, x i is going to represent where the particle is at time t i. Now we approximate xi+ e ih t xi xi+ exp i t p op ) exp i t V x op ) ) xi This is not exact because p op does not commute with x op. However, the error is of order t), which is small for t 0. 3 After making this approximation, we can evaluate the matrix element without further approximation. We insert an integral over momentum eigenstates, then complete the square in the exponent. This gives xi+ e ih t xi xi+ exp i t p op ) xi exp i t V xi ) ) = d p x i+ exp i t p op ) p p xi e i t V x i ) = d p x i+ p p xi exp i t p ) e i t V x i) d p = π) expi p x 3 i+ x i )) exp i t p ) 8) e i t V x i) m ) 3/ = πi t [ ) m xi+ x i exp i t V x i )]) t With this result, we have K x F, t F ; x 0, ) = D[x] expis[x]). 9) xt N )= x F x )= x 0 3 We could make the error smaller by putting exp i t V x op )/) on the left and exp i t V x op )/) on the right, but that doesn t change the formalism very much. 3. 7)
Here D[x] = xt N )= x F x )= x 0 m πi t ) N 3N/ i= d x i, 0) where we understand that in the integrand we should set xt N ) and x ) as indicated. Here also S[x] represents the classical action associated with the path x: S[x] = N i=0 tf { ) t m xi+ x i V x i )} t { ) d xt) m V xt))}. ) We use the discrete approximations shown to the integration over paths and to the action, then take the limit t 0. At least that is what we should do in principle. I do not address whether this process converges. Thus with the path integral formulation, the propagator K x F, t F ; x 0, ) is an integral over all paths that get from x 0 to x F in time t F. The integrand is expis[x]), where x here denotes the path and S[x] is the classical action for that path. In classical mechanics, the action S[x] also appears. One considers all possibilities for paths from from x 0 to x F in time t F. The path that nature chooses is the one for which S[x] does not change if one makes a small variation δx away from the classical path x cl. We see that in quantum mechanics, all paths appear. The particle gets from x 0 to x F in time t F every way it can. The amplitude is the sum of the amplitudes for each path x, weighted by expis[x]). There can be constructive interference among paths and there can be destructive interference among paths. 4 Method of stationary phase Let us look at what seems to be a completely different subject, performing integrals by the method of stationary phase. Suppose that we want to approximately evaluate an integral of the form I = dφ expiνsφ)) gφ). ) 4
Here φ is a D-dimensional vector with components φ i and the integral means dφ = dφ dφ dφ D. 3) In the integrand, g is some well behaved function of φ. In the exponent of the integrand S is some well behaved function of φ, which I assume is real for real φ. The function Sφ) is multiplied by a parameter ν that we consider to be large, ν. Suppose that at some point φ = φ cl, the gradient of S vanishes, [ ] Sφ) = 0 4) φ i φ=φ cl This is a point of stationary phase. Suppose, for convenience, that there is only one point of stationary phase. Otherwise, we should apply the method to each such point and sum the results. We recognize that contributions to the integral from regions of φ not near φ cl will decrease as ν increases because of the rapidly oscillating phase. However, this cancellation is inhibited near the point of stationary phase. Therefore, we expand about φ = φ cl. We have Sφ) = Sφ cl ) + φ i φ j D ij +, 5) where i,j φ i = φ i φ cl,i 6) and [ ] Sφ) D ij = 7) φ i φ j φ=φ cl This gives after taking the factors that do not depend on φ out of the integral and changing the integration variables from φ to φ), ) I expiνsφ cl )) gφ cl ) d φ exp iν φ i D ij φ j. 8) How can we perform this integral? We note that D is a real, symmetric matrix. Therefore it has a complete set of real eigenvectors v n, D ij vn j = λ n vn i. 9) j 5 i,j
The eigenvectors can be chosen to be normalized and orthogonal with a real inner product), vmv i n i = δ mn. 0) i We can write φ i = a n vn i. ) n We change integration variables to the expansion coefficients a n. For the exponent, we have iν φ i D ij φ j = iν a n a m v i md ij vn j i,j n,m i,j = iν a n a m λ n v mv i n i n,m i = iν a n a m λ n δ mn n,m = iν λ n a n. For the integration, we have d φ i = det v n i n ) da n. 3) Here we have the determinant of the matrix v with matrix elements vn. i This matrix is an orthogonal matrix, i vi mvn i = δ mn, so its determinant is. Thus our integral is ) I expiνsφ cl )) gφ cl ) da exp iν λ n a n. 4) The integral is now a product of many one dimensional integrals, I expiνsφ cl )) gφ cl ) [ da n exp iν )] λ na n. 5) n We can perform each integral by changing variables to ξ n, i a n = λ n ν ξ n. 6) 6 n
This gives I expiνsφ cl )) gφ cl ) n [ i λ n ν dξ n exp ξ n) ]. 7) That is We recognize that Thus I expiνsφ cl )) gφ cl ) n [ ] πi λ n ν. 8) λ n = det D. 9) n [ ] D πi I expiνsφ cl )) gφ cl ). 30) det D ν This is the leading result. You should verify for yourself that if you expand g and the remaining exponential factor in powers of φ, you will generate corrections that modify the leading result by multiplying it by a power series + c /ν + c /ν +. The essential point is that, inside the integral, δφ is effectively of order / ν. Exercise 4. Consider the one dimensional integral where and I = dφ expiνsφ)) gφ), 3) Sφ) = + φ + φ 4 3) gφ) = + φ 4. 33) Use the method of stationary phase to evaluate I for large ν at the leading order. Taking ν = 5, ν = 0, and ν = 0, compare to the exact integral evaluated by numerical integration. If you use Mathematica for the numerical integration, I suggest MaxRecursion 0 to help it get a good answer with an oscillatory integrand.) 7
5 Classical approximation to the path integral Consider our path integral for the propagator, K x F, t F ; x 0, ) = D[x] expis[x]). 34) xt N )= x F x )= x 0 Let x cl t) be the classical path with x cl ) = x 0 and x cl t F ) = x F. Define Then We have S[x] = S[x cl ] + tf tf { m d x cl t) { m + Oδx) 3 ). δ xt) = xt) x cl t). 35) δ x ) = δ xt F ) = 0. 36) dδ xt) } + V x cl t)) δ xt) ) i j V x cl t))) δxt) i δxt) j } 37) The first term is the classical action. In the second term, I have integrated by parts using δ x ) = δ xt F ) = 0. Then the second term vanishes because x cl t) obeys the classical equation of motion. This is the principle of stationary action. Since the action is the phase, we are now applying the method of stationary phase to the integral. In the third term, we can integrate by parts also, so that S[x] = S[x cl ] + tf { δxt) i mδ ij + Oδx) 3 ). d i j V x cl t))) Define the operator { D = mδ d ij } i j V x cl t))) 8 } δxt) j 38). 39)
Taking over our previous results, we have the classical approximation to our propagator K x F, t F ; x 0, ) N det D expis[x cl ]). 40) Here N is a normalization factor coming from the definition of the integration over paths. We should let t be finite, so that we have a finite dimensional integral. Then D us really a finite dimensional matrix. In principle, we are supposed to calculate with t finite and then take a limit t 0. We can note immediately that for the simple case of a free particle that starts at x 0 = 0 at time = 0, the classical action is S = t 0 dτ m x t = m t x. 4) Thus the exponent in Eq. 40) matches the exponent in our exact solution for K x, t; 0, 0) in Eq. 4). Exercise 5. Another case for which it is pretty easy to evaluate the classical action is the simple harmonic oscillator in one dimension, H = p /) + mω /)x. The classical approximation for the propagator Kx F, t F ; x 0, 0), should be exact: the action is a quadratic function of x so the order δx) 3 contributions to the action vanish. Take = 0 and find S[x cl ] for the path that starts at x 0 and reaches x F at time t F. Use this to calculate the exponential part of Kx F, t F ; x 0, 0). There is also a factor N / detd that is independent of x 0 and x F and that multiplies the exponential. Don t bother to calculate this factor. Compare your answer to the exact result given in Eq..6.8) of Sakurai. 6 The action and the classical approximation Let us try to derive this classical approximation another way. We seek an approximate solution to the Schrödinger equation [ i t + ] V x) K x, t; x 0, ) = 0. 4) 9
The boundary condition is that K x, t; x 0, ) approaches δ x x 0 ) at t. Given the result 40), we may suspect that the classical action for classical paths that go from x 0 at time to x at time t has something to do with this. Let us call this function S x, t). That is [ t ) m d xcl τ) S x, t) = dτ V x clτ))]. 43) dτ Here x cl τ) is the classical path, with x cl ) = x 0, x cl t) = x. 44) We need some properties of this function. First, consider a new classical path x cl τ) + δ x cl τ) that starts at x 0 at time but gets to some new place x + δ x at time t. we have t [ δs x, t) = dτ m d x clτ) dδ x ] clτ) δ x cl τ) dτ dτ V x cl τ)). 45) This is δs x, t) = t dτδ x cl τ) [ ] m d x cl τ) V dτ x cl τ)) + m δ x d x clt). 46) The first term vanishes because x cl τ) obeys the equation of motion. The second term remains because δ x cl τ) does not vanish at τ = t. We conclude that δs x, t) = δ x p cl t), 47) where p cl τ) = m d x clτ) dτ Since this holds for any δ x, we have. 48) S x, t) = p cl t). 49) We also need S x, t)/ t. Consider keeping one path, x cl τ), but extending the time to t + δt, so that the particle gets to a new place, x + δ x. Then δ x = d x clt) δt. 50) 0
The action changes by On the other hand, δs = [ ] p clt) V x cl t)) δt. 5) δs = S x, S x, t) t) δ x + δt [ t = p cl t) d x ] clt) S x, t) + δt t 5) Comparing these, we have S x, t) t = p clt) V x cl t)) p cl t) d x clt) = p clt) V x cl t)). The right hand side is the negative of the energy E on the path. Thus 53) S x, t) t Now we can return to the Schrödinger equation. Let = E. 54) K x, t; x 0, ) = exp[ia x, t)]. 55) Then A t ) A i + A V x) = 0. 56) I propose that under certain circumstances the semiclassical approximation ), we can neglect A. Let s try it and then come back to see under what conditions this is a good approximation. With the semiclassical approximation, we have Let s try A x, t) = S x, t). I claim that A t ) A V x) 0. 57) S t ) S V x) = 0. 58)
To see this, just use our previous results, giving E p clt) V x cl t)) = 0. 59) Since E is just the kinetic energy p /) plus the potential energy V, Eq. 58) is indeed true. In Eq. 59), we consider the path x cl τ) that gets from x 0 at time to x at time t. We then evaluate x cl τ) and p cl τ) at τ = t. The results depend on x and t, so we could write this equation as E x, t) p x, t) V x) = 0. 60) Now, let s look at the semiclassical approximation. Suppose that E, V, and p /) are large. Then our wave function has closely spaced wiggles in space and time. In particular, p is big, or the quantum wavelength λ = / p is small. However, let s suppose that V x) is slowly varying and that x x 0 λ, so that x is far from the singular point where all the paths start. Let s denote the distance characteristic of the variation of V or x x 0 by R. Then we consider the situation in which λ R. The distance characteristic of variation of p x, t) is controlled by classical mechanics, so it should be R: p x, t) p x, t). 6) R That is S) R S, 6) while S λ. 63) Then since R λ we have S) /Rλ) while S) /λ, so S) S), 64) That is our approximation. We see that the solution to the Schrödinger equation in the classical limit is K x, t; x 0, ) = ρ x, t) exp[is x, t)]. 65) The classical action appears in the exponent. There is a normalization factor ρ in front. If ρ is real and if we call K x, t; x0, ) = ψ x, t), then ρ = ψ. To find ρ, we need the next order approximation. We will return to that question after we have learned a little more.
7 More about the classical action It is easy to interpret the classical action as an approximate quantum mechanical phase. We write, using p cl = m d x/, { t ) S x, t) = m d xcl t) V x clt))} { t ) d xcl t) = m ) m d xcl t) V x clt))} 66) t = d x t clt) p cl t) E = x x 0 d x p cl x) t )E. In the first term we integrate along the path in space that the particle takes, with p being a vector pointing along the path whose magnitude is p cl x) = E V x)), 67) where E is the classical energy for the path. We accumulate phase from a factor expi p d x) for each step along the path. In the second term, we have the classical energy E times the total time needed to traverse the path, corresponding to a factor exp ie) for each step of the path. We can write this as where S x, t) = W x, E) t )E. 68) W x, E) = x x 0 d x p cl x). 69) We consider this to be a function of the final position x and the energy E. Let us find how W x, E) varies if we vary x and E: δw x, E) = δs + Eδt + t )δe = [ p δ x Eδt] + Eδt + t )δe = p δ x + t )δe. 70) 3
Thus W x, E) = p cl x), W x, E) E = t ). 7) We use W x, E) if we want to find an energy eigenfunction in the classical limit. For instance, we can think of a wave function that represents waves that start from a source at location x 0. If we move x 0 to somewhere far away near the z axis, then we would have waves that start as plane waves for large negative z and then get bent by a potential V x) that is zero for large x but can bend the waves when they get to the region where the potential acts. This corresponds to unbound quantum states. We can also think of bound states, so that we want to find bound state energy eigenvalues. Then our approximation is part of the WKB approximation described in Sakurai. However, for a bound state in one dimension, there is a point where the particle runs out of kinetic energy and turns around. At the turning point, p cl = 0. At that point, the semiclassical approximation is not working. Thus we need to use real quantum mechanics to match wave functions in the classically allowed region and the classically forbidden region. To find an energy eigenstate, we want to solve [ ] + V x) E ψ x) = 0. 7) We put ψ x) = exp[ia x)]. 73) Then ) A i A + V x) E = 0. 74) We approximate this in the small wavelength limit as ) A + V x) E 0. 75) We try the solution A x) = W x, E). Since W = p cl, we get p cl x) + V x) E = 0. 76) 4
Thus this ansatz solves our equation. We see that the solution to the Schrödinger equation for an energy eigenfunction in the classical limit is ψ x) = ρ x, E) exp[iw x, E)]. 77) The function W from Eq. 69) appears in the exponent. There is a normaliztion factor ρ in front. If ρ is real, then ρ = ψ. To find ρ, we need the next order approximation. 8 The next order approximation We have two semiclassical treatments. One considers quantum particles that start at x 0 at time and reach position x at time t. These particles can have any energy. The other treatment considers particles that have energy E. They travel from x 0 to x, but we do not ask how long it takes. In each case, there is a normalization ρ such that ψ = ρ. Our treatment has not been at the level of approximation that enables us to find ρ, but we can analyze that now. The constant energy case is easiest. Let ψ x) = exp iw x, E) + ) logρ x, E)). 78) The Schrödinger equation 7) is equivalent to i W + ) ρ ρ i W + ) log ρ +V x) E = 0. 79) Counting ρ and W as being slowly varying, we can approximate this by W ) With W = p cl x), this is p cl x) i ρ ρ W i ρ ρ p cl x) i W + V x) E = 0. 80) i p cl x) + V x) E = 0. 8) We have kept the big terms here and the first not-so-big terms. As we have already arranged, the big terms cancel E = p /) + V ) and we are left with the not so big terms, i ρ x) ρ x)) p cl x) i p cl x) = 0. 8) 5
This is more understandable if we write Then p cl x) = m v cl x). 83) v cl x) ρ x) + ρ x) v cl x) = 0. 84) Given v cl x), one could pretty easily solve this for ρ, at least numerically. To see what the equation means, write it as [ρ x) v cl x) ] = 0. 85) Imagine lots of particles. The particles at x have density ρ x) and move with velocity v cl x). This equation says that the number of particles is conserved. For example, if v cl x) always points in the same direction, then ρ is smallest where v cl is biggest. If v cl is constant, but all of the particles are moving radially outward from the origin, then ρ / x. Now let s try the time dependent case. We write ψ x) = exp is x, t) + ) logρ x, t)). 86) The the Schrödinger equation 7) is equivalent to 0 = S t + i ρ + ρ t + i S + log ρ i S + ) ρ ρ ) V x). 87) Counting ρ and W as being slowly varying, we can approximate this by 0 = S t + i ρ ρ t ) S i + S ρ ρ + i S V x). 88) With S/ t = E and S = p cl x), this is 0 = E + i ρ ρ t p cl + i ρ p cl ρ + 6 i p cl V x). 89)
As we have already arranged, the big terms cancel E = p /) + V ) and we are left with the not-so-big terms, 0 = i ρ ρ t + i ρ p cl ρ + This is more understandable if we write i p cl. 90) p cl x, t) = m v cl x, t). 9) Then ρ t + v cl ρ + ρ v cl = 0. 9) To see what the equation means, write it as ρ x, t) t + [ρ x, t) v cl x, t) ] = 0. 93) This equation says that the number of particles is conserved. We need some boundary conditions to actually solve for ρ. For example, for free particles starting from x 0 = 0 at time = 0, we have v = x/t and, using Eq. 4), ρ /t 3. This does satisfy the conservation equation. 9 The path integral with a magnetic field Our derivation of the path integral was for a hamiltonian /)+V. For a particle in a time independent electric and magnetic field, the hamiltonian is H = i qa x)) + qφ x). 94) Here q is the charge of the particle and φ and A are the scalar and vector potentials φ = E, A = B. 95) We can derive the Feynman path integral in this case also. As before, we divide the time interval into a large number N of small increments of size t. We then have K x F, t F ; x 0, ) = x F e ih t e ih t e ih t x0. 96) 7
with N factors of exp ih t). Between each pair of factors exp ih t), we introduce a complete sum over position eigenstates, K x F, t F ; x 0, ) = d x N d x d x xf e ih t xn x e ih t 97) x x e ih t x0 Physically, x i is going to represent where the particle is at time t i. Now we examine xi+ e ih t xi = [ x i+ ]) exp i t p op qa x 98) op )) + qφ xop ) x i. We need to approximate this. We set each instance of x op equal to a number x = x i+ + x i )/. This is not exact, but it is OK to first order in the small quantity t. Then xi+ e ih t xi = [ x i+ exp i t p op qa x) ) ]) xi 99) + qφ x). After making this approximation, we can evaluate the matrix element without further approximation. We insert an integral over momentum eigenstates, giving xi+ e ih t xi = π) 3 = π) 3 d p [ x i+ p exp i t d p e i p x i+ x i ) exp d p exp i t i t p qa x) ) ]) p + qφ x) xi [ p qa x) ) ]) + qφ x) ) ]) v p + qφ x) [ p qa x). 00) Here we have defined v = x i+ x i t. 0) 8
We can complete the square in the exponent: p qa x) ) v p = p m v qa x) ) m v q v A x). 0) Then [ ]) xi+ e ih t xi exp i t m v + q v A x) q φ x) ) d k exp i t Here = π) 3 m πi t ) 3/ exp i t k [ m v + q v A x) q φ x) ]). 03) With this result, we have K x F, t F ; x 0, ) = D[x] expis[x]). 04) xt N )= x F x )= x 0 D[x] = xt N )= x F x )= x 0 m πi t ) N 3N/ i= d x i, 05) where we understand that in the integrand we should set xt N ) and x ) as indicated. Here S[x] represents the classical action associated with the path x: S[x] = tf { m d xt) ) + q d xt) A xt)) q φ xt)) }. 06) This is the action for a smooth path. What we get from our derivation is a discrete approximation to this, in which the integral over t is really as sum. If you study classical mechanics with a charged particle moving in an electric and magnetic field, then this S is the action that you use. We can note something very interesting about this. In the term involving A we can write d xt) = d x. 07) 9
Then S[x] = tf { ) d xt) m q φ xt))} + q d x A x). 08) The first term depends on how the motion of the particle along the path depends on the time t, but the second term is more geometrical: it is the line integral of A along the path that the particle traces out, but the time dependence doesn t matter. This is especially interesting when the particle is moving in a region in which there is no magnetic field. Then in a classical approximation, the classical path is determined by the electric field φ, while the vector potential A does not affect the classical path. Nevertheless, the quantum wave function accumulates a phase q d x A x). This is the basis of the Aharonov-Bohm effect, which is described in Sakurai. 0