1 PHYS 705: Classical Mechanics Hamilton-Jacobi Equation 2 Hamilton-Jacobi Equation There is also a very elegant relation between the Hamiltonian Formulation of Mechanics and Quantum Mechanics. To do that, we need to derive the Hamilton-Jacobi equation. The Hamilton-Jacobi equation also represents a very general method in solving mechanical problems. To start… Let say we are able to find a canonical transformation taking our 2n phase q, p space variables i i directly to 2n constants of motion, i , i i.e., Qi i Pi i 3 Hamilton-Jacobi Equation One sufficient condition to ensure that our new variables are constant in time is that the transformed Hamiltonian K shall be identically zero. If that is the case, the equations of motion will be, K Qi 0 Pi K Pi 0 Qi Recall that we have the new and old Hamiltonian, K and H, relating through the generating function F by: F K H t 4 Hamilton-Jacobi Equation With K identically equals to zero, we can write: F H qpt, , 0 (*) t For convenience in later calculations, we shall take F to be of Type 2 so that its independent variables are the old coordinates q i and the new momenta P i . In the same notations as previously, we can write FqQPt ,,, FqPt2 (,,) QPi i Now, in order to write H in terms of this chosen set of variables qPt ,, , we will replace all the p i in H using the transformation equation: F2 qPt,, pi qi 5 Hamilton-Jacobi Equation F F Then, one can formally rewrite Eq. (*) as: note that 2 since t t FqPt ,, FqPt2 (,,) QPi i F2 F 2 F 2 Hqq1, ,n ; , , ; t 0 q1 qn t Fq ,, t p 2 This PDE is known as the Hamilton-Jacobi Equation. i qi Notes: -Since P i i are constants, the HJ equation constitutes a partial differential equation of (n+1) independent variables: q1,,, qtn - We used the fact that the new P’s and Q’s are constants but we have not specified in how to determine them yet. - It is customary to denote the solution F 2 by S and called it the Hamilton’s Principal Function. 6 Hamilton-Jacobi Equation Suppose we are able to find a solution to this 1st order partial differential equation in (n+1) variables… FSSqq2 1,,,n 1,, n ; t where we have explicitly write out the constant new momenta: Pi i Recall that for a Type 2 generating function, we have the following partial derivatives describing the canonical transformation: Sq ,, t S pi pi ( T 1) qi qi Sq,, t S Q Qi i ( T 2) i Pi i 7 Hamilton-Jacobi Equation After explicitly taking the partial derivative on the RHS of Eq. (T1) Sq ,, t pi qi and evaluating them at the initial time t 0 , we will have n equations that one can invert to solve for the n unknown constants i in terms of the initial conditions qp 0, 0 , t 0 , i.e., i i qp0, 0, t 0 Similarly, by explicitly evaluating the partial derivatives on the RHS of Eq. (T2) at time t 0 , we obtain the other n constants of motion i Sq ,, t Qi i i qq0, tt 0 8 Hamilton-Jacobi Equation With all 2n constants of motion i , i solved, we can now use Eq. (T2) again to solve for q i in terms of the i , i at a later time t. Sq ,, t qqi i ,, t i i p Then, with ii , , and q i known, we can use Eq. (T1) again to evaluate i in terms of i , i at a later time t. Sq ,,,, t t p pp,, t i q i i i The two boxed equations constitute the desired complete solutions of the Hamilton equations of motion. 9 Hamilton Principal Function and Action Now, let consider the total time derivative of the Hamilton Principal function S, dSq ,, t SS qi dt qi t Using Eq. (T1) and the Hamilton-Jacobi equation itself, we can replace the two partial derivatives on the right by p i and H, S S dS pi H 0 t pqi i H L qi dt So, the Hamilton Principal Function is differed at most from the action by a constant, i.e., S Ldt constant 10 Example: Solving Harmonic Oscillator with HJ m Recall, we have: 1 2 0 x H pmqE2 2 2 2 k m kq2 2m f( q ) kq U( q ) 2 The Hamilton-Jacobi equation gives: 2 1 SS SS m2 2 q 2 0 Hq, ; t 0 q t 2mq t Recall that under this scheme, we have: K 0 Q and P 11 Example: Solving Harmonic Oscillator with HJ m 1 H pmqE2 2 2 2 2 k m 2m 0 x kq2 f( q ) kq U( q ) When H does not explicitly depend on time, 2 we can generally try this trial solution for S: SqtWq ,,, t (q and t are separable) With this trial solution, we have S t and the HJ equation becomes: 2 2 1 SS 2 2 2 1 W 2 2 2 recall: LHS H m q 0 m q 2mq t 2m q so H 12 Example: Solving Harmonic Oscillator with HJ m The previous equation can be immediately integrated: 2 2 1 W W 2 2 2 m2 2 q 2 mq 2 m q 2m q 2 W 2 2 2 2m m q q 2 2 2 W 2 m m2 2 q 2 dq and S 2 m m q dq t 13 Example: Solving Harmonic Oscillator with HJ m We don’t need to integrate yet since we only need SS to use , in the transformation eqs: q Sq ,, t S 2 2 2 #1 eq: i Q 2 m m q dq t i 1 dq m dq 2m t t 2 2m m2 2 q 2 2 1 m 2 q 2 2 m Letting x q , we can rewrite the integral as: 2 1 dx 1 1 m t tarcsin x arcsin q 2 1 x 2 14 Example: Solving Harmonic Oscillator with HJ m 1 1 m tarcsin x arcsin q 2 Inverting the arcsin, we have 2 qt sin t ' with ' m 2 Sq ,,,, t t nd p The 2 transformation equation gives, i q i S #2 eq: p2 m m2 2 q 2 dq t 2m m2 2 q 2 q q 15 Example: Solving Harmonic Oscillator with HJ Substituting , we have, m q t m 2 2 pt2 m 1 sin2 t ' 2 2 m pt2 m 1sin 2 t ' with ' pt2 m cos2 t ' pt 2 m cos t ' 16 Example: Solving Harmonic Oscillator with HJ 2 m qt sin t ' m 2 pt 2 m cos t ' Completing the calculations, we still need to link the two constants of motion , to initial conditions: @ t0, q q0 , p p 0 Recall that the constant of motion is H=E of the system so that we can calculate this by squaring q and p at time t 0 : 1 2 2 2 2 p0 m q 0 2m q 1 And, we can calculate at time t 0 by dividing the 2 eqs: 0 tan ' p0 m 17 Another example in using the Hamilton-Jacobi Method (t-dependent H) (G 10.8) Suppose we are given a Hamiltonian to a system as, p2 H mAtq where A is a constant 2m Our task is to solve for the equation of motion by means of the Hamilton’s principle function S with the initial conditions t0, q0 0, pmv 0 0 The Hamilton-Jacobi equation for S is: SS Hq; ; t 0 q t 18 Example in using the Hamilton-Jacobi Method Recall that the Hamilton’s principle function S is a Type 2 generating function with independent variables qPt,, S SqPt ,, with the condition that the canonically transformed variables are constants, i.e., Qand P For a Type 2 generating function, we have the following partial derivatives for S: Sq ,, t Sq ,, t p ( T 1) Q ( T 2) q 19 Example in using the Hamilton-Jacobi Method Writing H out explicitly in the Hamilton-Jacobi equation, we have 2 1 S S mAtq 0 (*) 2mq t S Hq; ; t q Here, we assume a solution to be separatable of the form: Sq ,;, t f tqgt Substituting this into the above Eq (*), we have, 1 2 f, t mAtq f ' , tqgt ' 0 ( ' is time 2m derivative here) 20 Example in using the Hamilton-Jacobi Method 1 2 f, t mAtq f ',t qg' t 0 2m Concentrating on the q dependent terms, they have to be independently add up to zero… So, we have, mAt 2 f', t mAt ft, f 2 0 And, requiring the remaining two terms adding up to zero also, we have 1 gt' ft2 2m Substituting f (t) from above, we have 22 2 2 4 1mAt 1 mAt 2 2 gt' f0 mAtff 0 0 2m 2 2 m 4 21 Example in using the Hamilton-Jacobi Method Integrating wrt time on both sides, we then have, mA2 t 5 Af f 2 g t 0 t3 0 t g 40 6 2m 0 Putting both f (t) and g(t) back into the Hamilton’s principle function, we have, 2 2 5 2 mAt mAt Af0 3 f 0 S q, ; t f0 q t t g 0 2 40 6 2m Since the Hamilton-Jacobi Equation only involves partial derivatives of S, g 0 can be taken to be zero without affect the dynamics and for simplicity, we will take the integration constant f 0 to be simply , i.e., f0 2 2 5 2 mAt mAt A3 Sqt, ; q tt 2 40 6 2m 22 Example in using the Hamilton-Jacobi Method mAt2 mA2t 5 A 2 We can then solve for using Eq.