Dirac-Feynman-action principle and pseudo-differential operators

1. Feb 9, 2009

jostpuur

I have encountered some mathematical difficulties when examining a one dimensional system defined by a Lagrange's function

$$L(x,\dot{x}) = M|\dot{x}|^{\alpha} - V(x),$$

where $\alpha > 1$ is some constant. The value $\alpha=2$ is the most common, but I am now interested in a more general case. The velocity and canonical momentum are related by equations

$$p = \alpha M |\dot{x}|^{\alpha - 2} \dot{x},$$

$$\dot{x} = (\alpha M)^{\frac{1}{1 - \alpha}} |p|^{\frac{2-\alpha}{\alpha - 1}} p.$$

and the Hamilton's function for the system is

$$H(x,p) = (\alpha - 1)\alpha^{\frac{\alpha}{1-\alpha}} M^{\frac{1}{1-\alpha}} |p|^{\frac{\alpha}{\alpha - 1}} + V(x).$$

When one attempts to write down the Schrödinger's equation

$$i\hbar\partial_t\Psi(t,x) = H(x,-i\hbar\partial_x)\Psi(t,x),$$

one encounters the difficulty of interpreting the operator $|\partial_x|^{\frac{\alpha}{\alpha - 1}}$. To my understanding, the only reasonable answer is to define this as a pseudo-differential operator, meaning as a multiplication operator in the Fourier space. So the Hamilton's operator is

$$H(x,-i\hbar\partial_x)\psi(x) = (\alpha - 1)\alpha^{\frac{\alpha}{1-\alpha}} M^{\frac{1}{1-\alpha}} \int \frac{dp\;dx'}{2\pi\hbar} |p|^{\frac{\alpha}{\alpha - 1}} e^{\frac{i}{\hbar}(x-x')p}\psi(x') \; + \; V(x)\psi(x).$$

If $\alpha=2$, or if $\frac{\alpha}{\alpha - 1}$ is an even number in some other way, then one can check through integration by parts, that this definition gives the ordinary differential operators. But in general, one cannot reduce this operator, defined through Fourier transforms, into the ordinary differentiation operations.

If $\alpha=2$, an alternative way to arrive at the Schrödinger's equation is available, through the action principle. When the particle propagates from point $x'$ to the point $x$, in some small time $\tau$, the corresponding action is given by

$$S(x',x,\tau) = \int\limits_0^{\tau} du\;L\big(x' + (x-x')\frac{u}{\tau}, \frac{x-x'}{\tau}\big)$$
$$= M\tau^{1-\alpha}|x'-x|^{\alpha} \;-\; \tau\big(V(x) + O(x'-x)\big)$$

and the time evolution of the wave function is given by

$$\Psi(t+\tau, x) = C_{\tau} \int dx'\; \exp\Big(\frac{i}{\hbar}S(x',x,\tau)\Big)\Psi(t,x') \;+\; O(\tau^2).$$

With value $\alpha=2$ one can compute

$$\Psi(t+\tau, x) = C_{\tau} \int dx'\; \exp\Big(\frac{iM}{\hbar\tau} (x'-x)^2\Big) \Big(1 \;-\; \frac{i\tau}{\hbar}\big(V(x) \;+\; O(x-x')\big) \;+\; O(\tau^2)\Big)$$
$$\Big(\Psi(t,x) \;+\; (x'-x)\partial_x\Psi(t,x) \;+\; \frac{1}{2}(x'-x)^2\partial_x^2\Psi(t,x) \;+\; O((x'-x)^3)\Big) \;+\; O(\tau^2)$$
$$=C_{\tau}\sqrt{\frac{\hbar\pi\tau}{2M}}(1+i)\Big(\big(1 \;-\; \frac{i\tau}{\hbar}V(x)\big)\Psi(t,x) \;+\; \frac{i\hbar\tau}{4M}\partial_x^2\Psi(t,x) \;+\; O(\tau^2)\Big) \;+\; O(\tau^2).$$

After choosing

$$C_{\tau} = \Big(\sqrt{\frac{\hbar\pi\tau}{2M}}(1+i)\Big)^{-1}$$

we get

$$\Psi(t+\tau,x) = \Psi(t,x) \;-\; \frac{i\tau}{\hbar}\Big(-\frac{\hbar^2}{4M}\partial_x^2 \;+\; V(x)\Big)\Psi(t,x) \;+\; O(\tau^2)$$
$$\implies\quad i\hbar\partial_t\Psi(t,x) = \Big(-\frac{\hbar^2}{4M}\partial_x^2 \;+\; V(x)\Big)\Psi(t,x).$$

My question is, that is it possible to derive the Schrödinger's equation, containing pseudo-differential operator, when $\alpha\neq 2$, somehow from the action principle. The standard computation with $\alpha=2$ doesn't generalize immediately, because the Taylor series of $\Psi$ around the point $x$ will never produce the pseudo-differential operators.

2. Feb 13, 2009

Avodyne

Given a hamilonian H(p,x), the path-integral formula for the transition amplitude is

$$\langle x_2|e^{-iH(t_2-t_1)/\hbar}|x_1\rangle=\int{\cal D}x\,{\cal D}p\,\exp\!\left[i\int_{t_1}^{t_2}dt\,[p\dot x - H(p,x)]/\hbar\right]$$

where the integral is over all paths in phase space (position and momentum) that begin at position $x_1$ at time $t_1$ and end at position $x_2$ at time $t_2$ (with arbitrary momenta at these times). If the hamiltonian is quadratic in $p$, then the path integral over the momentum is gaussian, and can be performed exactly. This then yields the usual classical action in the exponent. But if the hamiltonian is not quadratic in $p$, this step cannot be carried out explicitly.

3. Feb 13, 2009

RedX

I just wanted to add that the part of the Hamiltonian that is quadratic in p can't depend on x for you to take passage into the action. Otherwise the path integral would be the action multiplied by the Sqrt[1/x] or whatever the Gaussian pre-factor is, I think.

This trick only applies to quadratic Hamiltonians though. To me it just seems to be lucky that it works. The integral of a quadratic Gaussian happens to be the value of the integrand at the maximum point: this is exactly the definition of the Legendre transform:

http://en.wikipedia.org/wiki/Legendre_transform

4. Feb 16, 2009

jostpuur

Avodyne, I don't believe that you understood what my original problem was, because it was not that I would have encountered some well defined integral, whose computation would have difficult or impossible. The problem is that from what principle is the pseudo-differential operator supposed to emerge, because it is not emerging from Taylor expansion of $\psi$ at least. Anyway, your post made me think about the action in Hamiltonian picture, and it could be that it is the solution to my original problem, so thanks anyway . It could be that the key is to write the time evolution using the action in terms of the Hamiltonian, integrate first over spatial coordinate, and then over the momenta. That is a kind of computation that could make pseudo-differential operators emerge. I haven't yet tried to carry out the computation, though.