This Quantum World/Feynman route/Schroedinger at last

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Schrödinger at last
The Schrödinger equation is non-relativistic. We obtain the non-relativistic version of the electromagnetic action differential,



dS=-mc^2\,dt\sqrt{1-v^2/c^2}-qV(t,\mathbf{r})\,dt+(q/c) \mathbf{A}(t,\mathbf{r})\cdot d\mathbf{r},$$

by expanding the root and ignoring all but the first two terms:



\sqrt{1-v^2/c^2}=1-{1\over2}{v^2\over c^2}-{1\over8}{v^4\over c^4}-\cdots\approx 1-{1\over2}{v^2\over c^2}.$$

This is obviously justified if $$v\ll c,$$ which defines the non-relativistic regime.

Writing the potential part of $$dS$$ as $$q\,[-V+\mathbf{A}(t,\mathbf{r})\cdot (\mathbf{v}/c)]\,dt$$ makes it clear that in most non-relativistic situations the effects represented by the vector potential $$\mathbf{A}$$ are small compared to those represented by the scalar potential $$V.$$ If we ignore them (or assume that $$\mathbf{A}$$ vanishes), and if we include the charge $$q$$ in the definition of $$V$$ (or assume that $$q=1$$), we obtain



S[\mathcal{C}]=-mc^2(t_B-t_A)+\int_\mathcal{C} dt\left[{\textstyle{m\over2}}v^2-V(t,\mathbf{r})\right]$$

for the action associated with a spacetime path $$\mathcal{C}.$$

Because the first term is the same for all paths from $$A$$ to $$B,$$ it has no effect on the differences between the phases of the amplitudes associated with different paths. By dropping it we change neither the classical phenomena (inasmuch as the extremal path remains the same) nor the quantum phenomena (inasmuch as interference effects only depend on those differences). Thus



\langle B|A\rangle=\int\mathcal{DC} e^{(i/\hbar)\int_\mathcal{C} dt[(m/2)v^2-V]}.$$

We now introduce the so-called wave function $$\psi(t,\mathbf{r})$$ as the amplitude of finding our particle at $$\mathbf{r}$$ if the appropriate measurement is made at time $$t.$$ $$\langle t,\mathbf{r}|t',\mathbf{r}'\rangle\,\psi(t',\mathbf{r}'),$$ accordingly, is the amplitude of finding the particle first at $$\mathbf{r}'$$ (at time $$t'$$) and then at $$\mathbf{r}$$ (at time $$t$$). Integrating over $$\mathbf{r},$$ we obtain the amplitude of finding the particle at $$\mathbf{r}$$ (at time $$t$$), provided that Rule B applies. The wave function thus satisfies the equation



\psi(t,\mathbf{r})=\int\!d^3r'\,\langle t,\mathbf{r}|t',\mathbf{r}'\rangle\,\psi(t',\mathbf{r}').$$

We again simplify our task by pretending that space is one-dimensional. We further assume that $$t$$ and $$t'$$ differ by an infinitesimal interval $$\epsilon.$$ Since $$\epsilon$$ is infinitesimal, there is only one path leading from $$x'$$ to $$x.$$ We can therefore forget about the path integral except for a normalization factor $$\mathcal{A}$$ implicit in the integration measure $$\mathcal{DC},$$ and make the following substitutions:



dt=\epsilon,\quad v=\frac{x-x'}{\epsilon},\quad V=V\left(t{+}\frac{\epsilon}{2},\frac{x{+}x'}{2}\right).$$

This gives us



\psi(t{+}\epsilon,x)=\mathcal{A}\int\!dx'\,e^{im(x{-}x')^2/2\hbar\epsilon}\, e^{-(i\epsilon/\hbar)V(t{+}\epsilon/2,(x{+}x')/2)}\,\psi(t,x').$$

We obtain a further simplification if we introduce $$\eta=x'-x$$ and integrate over $$\eta$$ instead of $$x'.$$ (The integration "boundaries" $$-\infty$$ and $$+\infty$$ are the same for both $$x'$$ and $$\eta.$$) We now have that



\psi(t+\epsilon,x)=\mathcal{A}\int\!d\eta\,e^{im\eta^2/2\hbar\epsilon}\, e^{-(i\epsilon/\hbar)V(t{+}\epsilon/2,x{+}\eta/2)}\,\psi(t,x{+}\eta).$$

Since we are interested in the limit $$\epsilon\rightarrow0,$$ we expand all terms to first order in $$\epsilon.$$ To which power in $$\eta$$ should we expand? As $$\eta$$ increases, the phase $$m\eta^2/2\hbar\epsilon$$ increases at an infinite rate (in the limit $$\epsilon\rightarrow0$$) unless $$\eta^2$$ is of the same order as $$\epsilon.$$ In this limit, higher-order contributions to the integral cancel out. Thus the left-hand side expands to



\psi(t+\epsilon,x)\approx\psi(t,x)+{\partial \psi\over\partial t}\epsilon,$$

while $$e^{-(i\epsilon/\hbar)V(t{+}\epsilon/2,x{+}\eta/2)}\,\psi(t,x{+}\eta)$$ expands to



\left[1-{i\epsilon\over\hbar}V(t,x)\right]\left[\psi(t,x)+{\partial \psi\over\partial x}\eta+\frac12{\partial^2\psi\over\partial x^2}\eta^2\right]= \left[1-{i\epsilon\over\hbar} V(t,x)\right]\!\psi(t,x)+{\partial \psi\over\partial x}\eta+ {\partial^2\psi\over\partial x^2}{\eta^2\over2}.$$

The following integrals need to be evaluated:



I_1=\int\!d\eta\, e^{im\eta^2/2\hbar\epsilon},\quad I_2=\int\!d\eta\, e^{im\eta^2/2\hbar\epsilon}\eta,\quad I_3=\int\!d\eta\, e^{im\eta^2/2\hbar\epsilon}\eta^2.$$

The results are



I_1=\sqrt{2\pi i\hbar\epsilon/m},\quad I_2=0,\quad I_3=\sqrt{2\pi\hbar^3\epsilon^3/im^3}.$$

Putting Humpty Dumpty back together again yields



\psi(t,x)+{\partial \psi\over\partial t}\epsilon=\mathcal{A}\sqrt{2\pi i\hbar\epsilon\over m} \left(1-{i\epsilon\over\hbar}V(t,x)\right)\psi(t,x) +{\mathcal{A}\over2}\sqrt{2\pi\hbar^3\epsilon^3\over im^3}{\partial^2\psi\over\partial x^2}.$$

The factor of $$\psi(t,x)$$ must be the same on both sides, so $$\mathcal{A}=\sqrt{m/2\pi i\hbar\epsilon},$$ which reduces Humpty Dumpty to



{\partial \psi\over\partial t}\epsilon=-{i\epsilon\over\hbar}V\psi+ {i\hbar\epsilon\over2m}{\partial^2\psi\over\partial x^2}.$$

Multiplying by $$i\hbar/\epsilon$$ and taking the limit $$\epsilon\rightarrow0$$ (which is trivial since $$\epsilon$$ has dropped out), we arrive at the Schrödinger equation for a particle with one degree of freedom subject to a potential $$V(t,x)$$:



i\hbar{\partial \psi\over\partial t}=-{\hbar^2\over2m}{\partial^2\psi\over\partial x^2}+V\psi.$$

Trumpets please! The transition to three dimensions is straightforward:

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