## Hamilton-Jacobi and Schrodinger equations

Quantum mechanics is a theory significantly larger than its standard textbook formulation. Quantum mechanics is categorical in nature and we need to go beyond parochial points of view and understand the links between different formulations and the categorical approach. Last time I presented a concrete problem in the phase space formulation, and today I want to arrive at the usual Hilbert space formalism by deriving Schrodinger's equation. There is no better way to do that than the follow the original paper by Schrodinger which I was pleasantly surprised to discover it is very clearly written and understandable.

The prerequisite of Schrodinger's approach is the Hamilton-Jacobi formalism and I will start with a short introduction of it.

The Hamiltonian equations of motion are:

$$\dot p = -\nabla_{x} H$$
$$\dot x = \nabla_{p} H$$

We want to explore changes of variables: $$(p,x) \rightarrow (P(p,x), X(p,x))$$ such that the Hamiltonian equations of motion remain valid in the new coordinates (those are called canonical transformations). This means that that differential form $$\omega = dp \wedge dx$$ is preserved:

$$dp \wedge dx = d(pdx) = d(PdX) = dP \wedge dX$$

which demands: $$d(pdx - P dX) = 0$$ or $$pdx - PdX = dR$$ So far so good, but we want to have a change of coordinates to simplify things. In particular we want the new Hamiltonian to be zero which will make the new $$X$$ and $$P$$ constants. So we have to see what will happen when we have time-dependent canonical transformations. By notation we say $$f_t(P,X) = f(P,X,t)$$. Then:

$$dS_t = pdx - P_t dX_t$$

Because
$$dR = \partial_t R dt + dR_t$$
$$dX = \partial_t X dt + dX_t$$

we get:

$$pdx - Hdt = PdX - (H + P\partial_t X + \partial_t R)dt + dR$$
which means that $$H+ P\partial_t X + \partial_t R$$ is the transformed Hamiltonian for the new coordinates: $$K(P(t), X(t))$$ The link between the new and old Hamiltonian is given by the use of a generating function $$S$$:

$$K = H + \partial_t S$$

where $$\nabla_x S = p$$ and  $$S = S(P,x,t)$$

If we demand $$K = 0$$ then $$P$$ and $$X$$ are constants. $$K = 0$$ is known as the Hamilton-Jacobi equation:

$$\partial_t S(x,t) + H(\nabla_x S(x,t), x) = 0$$

Because this is derived from the phase space Hamiltonian formalism, this equation is another formulation of classical mechanics in configuration space. So how does Mr. Schrodinger arrive at his equation which describes quantum mechanics?

 Erwin SchrĂ¶dinger

The original paper can be found here and as I said it is remarkably accessible. The physical idea is that of the optical-mechanical analogy and the Eikonal equation in geometric optics because according to de Broglie the particles are waves (of an unspecified kind). In the short wavelength limit the waves becomes rays (and the particles are tracing a well defined path according to classical mechanics).

If we start from the Hamilton-Jacobi equation:

$$\frac{\partial S}{\partial t} + \frac{1}{2m}{(\frac{\partial S}{\partial x})}^2 + V = 0$$
$$p = \frac{\partial S}{\partial x}$$

we have

$$\frac{\partial S}{\partial t} = -H = -(T+V)$$
$$dS = \frac{\partial S}{\partial x}dx +\frac{\partial S}{\partial t } dt = p dx - H dt = L dt$$ - the Lagrangian and
$$S = \int_{t_0}^{t} L dt$$ - the action

With a usual decomposition: $$S = -Et + Q(x)$$:
$$\frac{\partial S}{\partial t} = -E$$

and from the H-J equation and our decomposition

$$|\nabla S| = \sqrt{2m(E-V)}$$ - which is the Eikonal equation.

Then Schrodinger considered surfaces of equal values for $$S$$ to see how this would propagate. The normal on those surfaces is:

$$dn = \frac{dS}{\sqrt{2m(E-V)}}$$

and the velocity:

$$u = \frac{E}{\sqrt{2m(E-V)}}$$

Those surfaces almost respect a wave equation but to take de Broglie's idea seriously Schrodinger forced the solution to be a wave:

$$\psi = A e^{\frac{i S}{\hbar}} = A e^{\frac{-iEt}{\hbar} + \frac{iQ(x)}{\hbar}}$$

respecting a wave equation according to the geometric optics insight (thus modifying the Hamilton-Jacobi equation in the process):

$$\Delta \psi -\frac{\ddot \psi}{u^2} = 0$$

Because $$u^2 = \frac{E^2}{2m(E-V)}$$, Schrodinger's equation is:

$$(-\frac{\hbar^2}{2m} \Delta + V )\psi = E \psi$$

It is informative to see what we get if we directly plug the ansatz $$\psi = A e^{\frac{i S}{\hbar}}$$ into the Hamilton-Jacobi equation. Simple calculations yields:

$$(-\frac{\hbar^2}{2m \psi} \Delta + V )\psi = E \psi$$

Conversely, if we plug the ansatz into the Schrodinger equation we recover the Hamilton-Jacobi equation and an additional term proportional with $$\hbar$$.

Schrodinger's equation and the Hamilton-Jacobi equations are cousins and they both are equations in the configuration space.

There are typical mistakes people make about them. For example working with a single particle, one gets the temptation to identify the particle with its wavefunction, This mistake is cured when considering the wavefunction of N particles where such identification is impossible because we are working in configuration space. Other mistake is "deriving" Schrodinger equation directly from Hamilton-Jacobi, like in this paper: http://arxiv.org/pdf/1204.0653v1.pdf. If that were true, quantum mechanics would be nothing but classical mechanics. In this paper the mistake occurs in Eq. 4.5 because total and partial derivatives cannot be interchanged. Here is a simple counterexample: $$x(t) = e^t$$

$$\frac{\partial \frac{d x}{d t}}{\partial x} = \frac{\partial x}{\partial x} = 1$$ but
$$\frac{d \frac{\partial x}{\partial x}}{dt} = \frac{d 1}{dt} = 0$$

The original triumph of Schrodinger's equation (derived from free particle considerations) was its application to the bound states of the atom. There the matrix mechanics of Heisenberg was developed to eliminate unobservable classical concepts like the classical path and later on Schrodinger proved the equivalency of the two seemingly different approaches.

But what is the link between Schrodinger's equation and the categorical formulation of quantum mechanics using the abstract products $$\alpha$$ and $$\sigma$$? First we need to pass from the Schrodinger picture where the wavefunction evolves in time and the operators are stationary (recall that in Hamilton-Jacobi $$\dot P = \dot Q = 0$$) to the Heisenberg picture where the states are stationary and the operators change in time. Here the time evolution of the operators is given by the abstract product $$\alpha$$:

$$\dot A = H\alpha A$$

The categorical formulation of quantum mechanics is a formulation in a Hilbert (and phase space when possible) in the Heisenberg picture.