The Hamilton-Jacobi equation

Jordan Bell
April 16, 2014

1 Example of free particle in one dimension

Define L(q,v)=m2v2, where m is a nonzero constant. Fixing two times t0<t1, we define the action for a path γ in by

S(γ) = t0t1L(γ(t),γ˙(t))𝑑t
= m2t0t1(γ˙(t))2𝑑t.

Suppose that γ is satisfies the Euler-Lagrange equation for the Lagrangian L. That is,


and here Lv=mv and Lq=0, so




so γ¨(t)=0, and hence γ(t)=at+b for some constants a,b. If we are given the conditions γ(t0)=q0 and γ(t1)=q1, then a solution of the Euler-Lagrange equation that satisfies these conditions must be


The action of this path is


The dimensions of the right hand side are


which are indeed the dimensions that action ought to have. If instead of talking about action that is a function of paths we talk about action that is a function of the end point of a motion and the time at which the motion ends, taking the time and location at which the motion starts as fixed, then


for which




These satisfy


If we write




2 Motivation for the Hamilton-Jacobi equation

Suppose that γ is a path that satisfies the Euler-Lagrange equation for some Lagrangian L. If we perturb the path to start at the same position at time t0 but to end at q+δq instead of at q, then the perturbed path is sγ(s)+(δγ)(s), where (δγ)(0)=0 and (δγ)(t)=δq. Then, first doing a Taylor approximation in which we drop all powers of δγ or δγ˙ higher than the first and then using the Euler-Lagrange equation, and using Einstein summation notation,

S(γ+δγ)-S(γ) = t0tL(γ+δγ,γ˙+δγ˙)-L(γ,γ˙)ds
= t0tLqi(γ,γ˙)δγi+Lvi(γ,γ˙)δγi˙ds
= t0t(ddtLvi(γ,γ˙))δγi+Lvi(γ,γ˙)δγi˙ds
= t0tddt(Lvi(γ,γ˙)δγi)𝑑s
= Lvi(γ(t),γ˙(t))(δγi)(t)-Lvi(γ(t0),γ˙(t0))(δγi)(t0)
= Lvi(γ(t),γ˙(t))δq.

We have not been precise about what we mean by perturbing a path, but what we have obtained suggests that if we think of S as a function of the endpoint of a path and the time at which the path ends rather than as a function of a path itself, we have


If on the other hand we fix the point q at which a path ends and change the time at which it arrives at this point from t to t+δt, then, doing a Taylor expansion and dropping all powers of δt higher than the first,


But γ(t+δt)=q, so




Defining H=Lvivi-L, what we have done suggests that


Then, using Sqi=pi, with which H(q,p)=H(q,Sq), and using St=-H(q,p), we have


We call this equation the Hamilton-Jacobi equation.

To precisely sort out where the Hamilton-Jacobi equation comes from and what it means, the only place I can imagine that does an adequate job is Abraham and Marsden.11 1 Abraham and Marsden, Foundations of Mechanics, second ed. Certainly there are other sources that present this more precisely than I have presented it, but it is almost universal to thoughtlessly confound the variables on which H or S depends with paths; that is, to write things like Hq and also to think of q not as a point but rather as a path which for each time goes through a particular point, in which case one has no certain way of knowing whether dqdt=0, as is the case for the derivative of any fixed point, or to say that q is a path and that dqdt is a tangent vector at the point q(t) on the path. If one plainly states that what one has said is only suggestive of how symbols work together then one does not need to apologize for the absence of precision, but there is a foul area between suggestive symbol manipulation and actual precision in which one tricks oneself into believing that one has given a precise presentation, and this is the path followed by some presentations of the Hamilton-Jacobi equation.

3 Harmonic oscillator



The Hamilton-Jacobi equation for this Hamiltonian is


We set S(q,t)=S0(q)-Et; for this to make sense presumes that there is in fact a constant E and a function S0 so that S(q,t)-S0(q) depends just on t. With this, Sq=S0q and St=-E, and so the Hamilton-Jacobi equation becomes




Supposing that S0 is nonnegative we get


a primitive of which is

S0 = 2mE-m2ω2q2𝑑q
= Eω(arcsinmωq2mE+mωq2mE1-(mωq2mE)2)

for which


But S0E=t (using the expression involving S and Et), so ωt=arcsinmωq2mE, hence


and therefore


As well,


and using the above expression for q this becomes


We have thus written q and p as functions of E and t. Since for a particular trajectory the energy is fixed, on a particular trajectory the position and momentum have thus been expressed as functions of t.

4 Schrödinger equation



called the Schrödinger equation. If H(q,p)=p22m+V(q), and p=-i, then p2=-2Δ and the Schrödinger equation is


Supposing that there is a solution ψ of the form ψ=eiS, we get

ieiSiSt = -22mq(eiSiSq)+V(q)eiS
= -22m(eiS(iSq)2+eiSi2Sq2)+V(q)eiS.

Diving both sides by eiS gives


Taking 0 yields the equation


which is the Hamilton-Jacobi equation.

The above derivation of the Hamilton-Jacobi equation from the Schrödinger equation is suggestive symbol manipulation. Rather than stating that we assume ψ=eiS, I could have written that we “make an Ansatz”; “ein Ansatz” means “an approach”, and is used to mean a guess which may work out. Of course, if there is some problem for which one knows there is a unique solution and we find an explicit solution starting from some unjustified assumption, then we don’t need to have justified the assumption because we can explicitly check that what we have found is a solution. But this is the only situation where there is precision to “making an Ansatz”. Otherwise, to talk about an Ansatz is a sophisticated sounding way of saying “we make an assumption”, and after making this assumption we have no guarantee that anything we end up with need make any sense. This does not mean that it is useless to make unjustified assumptions; but it is deceitful to smuggle Ansätze into the realm of proved things, and confuses those who later would rely on one’s work.