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PHY 305: WKB and Path Integral Formulation of QM in These Notes, We Will Study the Relationship Between Quantum Mechanics and Cl

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PHY 305: WKB and Path Integral Formulation of QM in These Notes, We Will Study the Relationship Between Quantum Mechanics and Cl PHY 305: WKB and Path Integral Formulation of QM In these notes, we will study the relationship between quantum mechanics and classical mechanics. Any new physical theory that replaces an existing one, has to show that it can reproduce all the old results in the limit where the old theory is accurate. In our case, this limit is ~ → 0. Since ~ is a dimensionful quantity 2 −1 [ ~ ] = kg · m · s . (1) which is the units of angular momentum = momentum × distance, or action = energy × time. This means that the classical description should for systems for which the characteristic mass, size, and velocity are such that its ‘action’ is large compared to ~. This important requirement is known as the correspondence principle, and quantum mechanics satisfies it beautifully. 1. Classical Action For simplicity we consider a non-relativistic particle of mass m moving in one dimension, under influence of a potential V (x). We can describe the classical motion of the particle either via a Hamiltonian or a Lagrangian p2 1 H(x, p) = + V (x); L(x, x˙) = mx˙ 2 − V (x) . (2) 2m 2 The equations of motion are respectively the Hamilton equations or the Euler-Lagrange equations ∂H ∂H d ∂L ∂L x˙ = , p˙ = − ; − = 0 . (3) ∂p ∂x dt ∂x˙ ∂x In our specific example, these equations reduce to the familiar form p ∂V x˙ = , p˙ = − (4) m ∂x ∂V which is just Newton’s equation with a force F (x) = − ∂x . 1 The usual formulation of quantum mechanics, via the Schr¨odinger equation and Heisen- berg equations of motion, is most closely related to the Hamilton formulation of classical mechanics. In these notes, we will summarize an alternative formulation of quantum mechan- ics, the Feynman path integral formulation, that is more directly related to the Lagrangian formalism. The Hamiltonian and Lagrangian are related via a Legendre transformation: ∂L H(x, p) = p x˙ − L(x, x˙) , p = , (5) ∂x˙ ∂H L(x, x˙) = p x˙ − H(x, p) , x˙ = . (6) ∂p On the right-hand side, one is instructed to eliminate the velocityx ˙ in favor of the momentum p in the first equation, and eliminate p in favorx ˙ in the second equation. The Euler-Lagrange equation can be derived via the principle of least action. Consider a path x(t) that starts at position x0 at time t0 and ends at position x at time t: (x0 , t 0 ) (x , t ) x(t) The action functional S is defined as the integral over time of the Lagrangian L(x, x˙), evaluated over the whole path Z S = dt L(x, x˙) (7) Now consider a small deviation x(t) → x(t) + δx(t) of this path. The variation of the action functional can be written as Z ∂L ∂L δS = dt0 δx(t0) + δx˙(t0) ∂x ∂x˙ Z ∂L d ∂L Z d ∂L = dt0 δx(t0) − + dt0 δx(t0) ∂x dt0 ∂x˙ dt0 ∂x˙ The first term on the right-hand side is zero for a classical path, that solves the Euler- Lagrange equations. The second term is a total time-derivative. So δS = 0, provided that the variation δx(t) vanishes at the end-points of the path. In case the variation δx(t) does not vanish at the end-points, we have t 0 ∂L δS = δx(t ) = δx(t) p(t) − δx0 p(t0) (8) ∂x˙ t0 2 2. WKB Wavefunction The equations of classical mechanics, as summarized above, are reproduced from the equations of quantum mechanics, by taking the classical limit ~ → 0. This is done as follows. As explained in Chapter 8 of Griffiths, the Schr¨odinger equation in the limit of small ~ is solved via the WKB approximation. To write the WKB wavefunction, we note that for a classical trajectory with energy E, the momentum p can be solved as a function of position x via H(x, p) = E → p = p(x) , p(x) ≡ p2m(E − V (x)) . (9) The WKB wavefunction reads x Z C i 0 0 ΨWKB(x, t) = p exp p(x )dx − E (t − t0) (10) |p(x)| ~ x0 1 where x0 is the position of the particle at some arbitrary initial time t0. In Griffiths it is shown that the above wavefunction solves the time-independent Schr¨odingerequation H Ψ = E Ψ. We now explain how it relates to classical mechanics. Evaluating the action functional (7) for the classical trajectory x(t) of energy E, going from x0 at t0 to position x at time t, yields x Z dx Z S(x, t) = (p − H) dt = p(x0) dx0 − E (t − t ) (11) dt 0 classical path x0 (x0,t0)→(x,t) where we used eqn (6). So we can write the WKB wavefunction as C i ΨWKB(x, t) = p exp S(x, t) (12) |p(x)| ~ So we conclude that: the phase factor of the WKB wavefunction ΨWKB(x, t) is 1/~ times the action functional S evaluated along the classical path that ends at position x at time t. Note that via the WKB wavefunction (12), we can relate the quantum mechanical definition ~ ∂ ∂S p = i ∂x to the classical identity p(x) = ∂x , which can be derived from eqns (8) or (11). 1 Different choices for t0 just change the wavefunction by an overall constant independent of x and t. 3 3. Feynman Path Integral The Feynman path integral is a reformulation of quantum mechanics, that makes use of the Lagrangian formulation of classical mechanics. One of its advantages is that it can be applied more easily to relativistic theories, and to quantum field theories. Here we summarize how it works for a non-relativistic particle in one dimension. Consider the following question: At time t0 the particle is located at x0, what is the probability that at time t it is located at x? In the usual formulation of quantum mechanics, we would answer this question as follows. At time t0 the particle, the particle is in the position eigenstate |x0i. The wavefunction Ψ(x, t) that solves the Schr¨odinger equation, with this initial condition, is i − (t−t0) H Ψ(x, t) = hx| e ~ |x0i (13) The probability to be at the location x at time t is equal to the absolute value squared of this wavefunction. Or more accurately, the probability to be between x and x+dx is |Ψ(x, t)|2dx. As a short warm-up exercise, let us compute Ψ(x, t) in the case of a free particle with Hamiltonian H = p2/2m. Since the Hamiltonian is just a function of the momentum, it is most convenient to solve the Schr¨odinger equation in the momentum representation. To do this, we insert into eqn (13) twice the closure relation Z 1 = dp |pihp| (14) where 1 is the identity operator. So we write Z Z i p2 − (t − t0) Ψ(x, t) = dp dp0 hx|pih p | e ~ 2m | p0ihp0| x0i Using i px e ~ h p | p0i = δ(p − p0) , hp|xi = √ (15) 2π~ we can easily perform the momentum integral and find Z i p2 dp (x − x0)p − (t − t0) Ψ(x, t) = e ~ 2m 2π~ r im i m(x − x )2 = exp 0 (16) 2π~(t − t0) ~ 2(t − t0) The phase in the exponent equals 1/~ times the action S(x, t) of the path from x0 to x. 4 Now let us turn to the derivation of the Feynman path integral formula. Schematically, it reads as follows i i 0 − (t−t0) H X S(x(t )) hx| e ~ |x0i = e ~ (17) paths x(t0) (x0,t0)→(x,t) 0 The sum on the right-hand side is over all trajectories x(t ) that start at x0 at time t0 and end at x at time t, and S(x(t0)) is the action functional (7) evaluated for the trajectory x(t0). To derive this formula, we start by dividing the time interval between the initial time t0 and final time t into N small time intervals of size ∆t such that t − t0 = N∆t (18) Correspondingly, we can write i i i i i − (t−t0) H − N∆t H − ∆t H − ∆t H − ∆t H e ~ = e ~ = e ~ e ~ . e ~ (19) | {z } N times Define tn = t0 +n∆t, and denote the position x(tn) at this instant by xn. So (xN , tN ) = (x, t). x 1 x x = x . n x n+1 N x .. 0 t t t t t = t 0 1 n n+1 N As indicated by eqn (18) and the above figure, the idea is to divide the time evolution from t0 to t into N small time intervals, and first do the compute the amplitude for each separate small step − i ∆tH hxn+1|e ~ |xni (20) Then afterwards, we will glue all small steps together by repeated use of the formula Z − i 2∆tH − i ∆tH − i ∆tH hxn+1|e ~ |xn−1i = dxn hxn+1|e ~ |xnihxn|e ~ |xn−1i (21) This gluing identity uses the closure formula 1 = R dx |xihx|. Eventually, we will take the limit N → ∞, ∆t → 0, while keeping the product (18) fixed. 5 Let us evaluate the matrix element (20). Z − i ∆t H − i ∆t H(x,p) hxn+1|e ~ |xni = dpn hxn+1|pni hpn|e ~ |xni i i Z pnxn+1 − pnxn + H(xn, pn)∆t e ~ e ~ = dpn √ √ 2π~ 2π~ Z i (xn+1−xn) dpn ∆t pn − H(xn, pn) = e ~ ∆t (22) 2π~ In the first step we used eqn (14), the second eqn (15) and that hpn |H(x, p)|xni = H(xn, pn).
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