- #1

jostpuur

- 2,116

- 19

[tex]

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

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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},

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[tex]

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

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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).

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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),

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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).

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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)

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[tex]

= M\tau^{1-\alpha}|x'-x|^{\alpha} \;-\; \tau\big(V(x) + O(x'-x)\big)

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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).

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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)

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[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)

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[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).

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After choosing

[tex]

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

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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).

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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.