Dirac-Feynman-action principle and pseudo-differential operators

In summary, the conversation discusses mathematical difficulties encountered when examining a one dimensional system defined by a Lagrange's function, where the value of a constant \alpha > 1 is being studied. The velocity and canonical momentum are related by equations, and the Hamilton's function for the system is also discussed. The Schrödinger's equation is then introduced and it is noted that when \alpha \neq 2, the pseudo-differential operator cannot be derived from the action principle. The path-integral formula for the transition amplitude is also mentioned, and it is noted that for quadratic Hamiltonians, the path integral over momentum can be performed exactly. However, if the Hamiltonian is not quadratic in p, this step cannot be carried out
  • #1
jostpuur
2,116
19
I have encountered some mathematical difficulties when examining a one dimensional system defined by a Lagrange's function

[tex]
L(x,\dot{x}) = M|\dot{x}|^{\alpha} - V(x),
[/tex]

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

[tex]
p = \alpha M |\dot{x}|^{\alpha - 2} \dot{x},
[/tex]

[tex]
\dot{x} = (\alpha M)^{\frac{1}{1 - \alpha}} |p|^{\frac{2-\alpha}{\alpha - 1}} p.
[/tex]

and the Hamilton's function for the system is

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

When one attempts to write down the Schrödinger's equation

[tex]
i\hbar\partial_t\Psi(t,x) = H(x,-i\hbar\partial_x)\Psi(t,x),
[/tex]

one encounters the difficulty of interpreting the operator [itex]|\partial_x|^{\frac{\alpha}{\alpha - 1}}[/itex]. 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

[tex]
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).
[/tex]

If [itex]\alpha=2[/itex], or if [itex]\frac{\alpha}{\alpha - 1}[/itex] 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 [itex]\alpha=2[/itex], an alternative way to arrive at the Schrödinger's equation is available, through the action principle. When the particle propagates from point [itex]x'[/itex] to the point [itex]x[/itex], in some small time [itex]\tau[/itex], the corresponding action is given by

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

and the time evolution of the wave function is given by

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

With value [itex]\alpha=2[/itex] one can compute

[tex]
\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)
[/tex]
[tex]
\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)
[/tex]
[tex]
=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).
[/tex]

After choosing

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

we get

[tex]
\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)
[/tex]
[tex]
\implies\quad i\hbar\partial_t\Psi(t,x) = \Big(-\frac{\hbar^2}{4M}\partial_x^2 \;+\; V(x)\Big)\Psi(t,x).
[/tex]

My question is, that is it possible to derive the Schrödinger's equation, containing pseudo-differential operator, when [itex]\alpha\neq 2[/itex], somehow from the action principle. The standard computation with [itex]\alpha=2[/itex] doesn't generalize immediately, because the Taylor series of [itex]\Psi[/itex] around the point [itex]x[/itex] will never produce the pseudo-differential operators.
 
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  • #2
Given a hamilonian H(p,x), the path-integral formula for the transition amplitude is

[tex]\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][/tex]

where the integral is over all paths in phase space (position and momentum) that begin at position [itex]x_1[/itex] at time [itex]t_1[/itex] and end at position [itex]x_2[/itex] at time [itex]t_2[/itex] (with arbitrary momenta at these times). If the hamiltonian is quadratic in [itex]p[/itex], 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 [itex]p[/itex], this step cannot be carried out explicitly.
 
  • #3
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
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 [itex]\psi[/itex] 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 :wink:. 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.
 

Related to Dirac-Feynman-action principle and pseudo-differential operators

1. What is the Dirac-Feynman action principle?

The Dirac-Feynman action principle is a mathematical tool used in quantum field theory to determine the most probable path of a particle from one point to another. It states that the path a particle takes is the one that minimizes the action, which is the integral of the Lagrangian (a function that describes the dynamics of the system) over time.

2. How is the Dirac-Feynman action principle related to quantum mechanics?

The Dirac-Feynman action principle is based on the principles of quantum mechanics, specifically the wave-particle duality of matter. It considers the particle to have a wave-like nature and calculates the probability of its path using the wave function of the particle.

3. What are pseudo-differential operators?

Pseudo-differential operators are mathematical operators that generalize the concept of differentiation to functions that are not smooth or infinitely differentiable. They are used in quantum field theory to describe the dynamics of fields, which are fundamental quantities in quantum mechanics.

4. How are pseudo-differential operators used in the Dirac-Feynman action principle?

In the Dirac-Feynman action principle, pseudo-differential operators are used to represent the wave function of a particle. They are also used in the Lagrangian to describe the dynamics of the system and calculate the action.

5. What is the significance of the Dirac-Feynman action principle and pseudo-differential operators in physics?

The Dirac-Feynman action principle and pseudo-differential operators are essential tools in quantum field theory, which is a fundamental framework for understanding the behavior of matter at the smallest scales. They allow scientists to make predictions about the behavior of particles and fields in a consistent and mathematically rigorous way.

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