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After assuming a wavefunction of a form:

$$ \psi \approx A \exp{\left(i \frac{S(x)}{\hbar}\right)}$$

and letting

$$S = \hbar^0 S_0 + \hbar^1 S_1 + \hbar^2 S_2 +...$$

The odd-indexed terms of the action for a one-dimensional potential in the time-independent Schrodinger equation do not require integration if $p$ is known and differentiable. However the even-terms require non-trivial integration(which is extensively more computationally taxing). p is defined by:

$$ p = \sqrt{2 m ( E - V(x))} $$

The terms:

$$ S_0^\prime = p$$ $$ \boxed{S_0 = \int dx\, p =\pm \int dx \sqrt{2m(E-V(x))} } $$ $$ S_1^\prime = \frac{i}{2}\frac{1}{p}\frac{d p}{d x} $$ $$ \boxed{S_1 = \frac{i}{2} ln(p) } $$ $$ S_2^\prime = \frac{1}{8 p^3} \left(\frac{d p}{d x} \right)^2 - \frac{1}{4} \left(\frac{1}{p^2} \frac{d^2 p}{d x^2} - \frac{1}{p^3} \left(\frac{d p}{d x}\right)^2\right)$$ $$ \boxed{S_2 = \int dx \left(\frac{1}{4 p^2} \frac{d^2 p}{d x^2} + \frac{3}{8 p^3} \left( \frac{d p}{d x}\right)^2 \right)}\mathrm{\,requires\,\,integration} $$ Now for $S_3$: $$ S_3^\prime = -\frac{i}{8 p^3} \frac{d^3 p}{d x^3} + \frac{3}{4} \left( \frac{i}{p^4} \frac{d p}{d x} \frac{d^2 p}{d x^2} - \frac{1}{p^5} \left( \frac{d p}{d x}\right)^3 \right)$$ After making an educated guess that: $$ \frac{d}{dx} \left( -\frac{i}{8 p^3} \frac{d^2 p}{d x^2} + \frac{3 i}{16 p^4} \left(\frac{d p}{d x}\right)^2\right) = -\frac{i}{8 p^3} \frac{d^3 p}{d x^3} + \frac{3}{4} \left( \frac{i}{p^4} \frac{d p}{d x} \frac{d^2 p}{d x^2} - \frac{1}{p^5} \left( \frac{d p}{d x}\right)^3 \right)$$ Then $$\int dx S_3^\prime = \int dx \frac{d}{dx} \left( -\frac{i}{8 p^3} \frac{d^2 p}{d x^2} + \frac{3 i}{16 p^4} \left(\frac{d p}{d x}\right)^2\right)$$ Clearly because of my guess $$\boxed{S_3 = -\frac{i}{8 p^3} \frac{d^2 p}{d x^2} + \frac{3 i}{16 p^4} \left(\frac{d p}{d x}\right)^2} $$ How do we know that $S_2$ cannot be retrieved the same way? Surely I cannot try an ansatz for every possible function that could be a candidate for $S_2$. This problem has come up in general for different types of problems outside the scope of quantum mechanics. Is it only a consequence of this equation(1-D time-independent Schrodinger equation) that we have odd-indexed and even-indexed terms this way? Would it be different for 3-D?

John M
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  • Note that the even-indexed ones are real while the odd-indexed-ones are imaginary. So the even ones decide on the phase while the even ones decide on the normalization. The normalization is sort of dictated by algebraic constraints but the phase may be arbitrary and stores the main dynamical information - it's the phase whose derivatives are dictated by the Schrodinger equation etc. This is just a heuristic observation, not a full proof, however. – Luboš Motl Mar 29 '13 at 07:15
  • The governing equation is what is used to find $S_2, S_3,...$ after expanding the action in terms of powers of $\hbar$ – John M Mar 30 '13 at 04:09

1 Answers1

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The governing Schrödinger equation

$$(e^{\frac{i}{\hbar}S})^{\prime\prime}~=~-k(x)^2 e^{\frac{i}{\hbar}S} ~\Leftrightarrow~ S^{\prime2} ~=~p(x)^2+i\hbar S^{\prime\prime} $$

can be turned into a fixed point equation

$$S^{\prime} ~=~ \sqrt{p^2+i\hbar S^{\prime\prime}} ~=~ \sqrt{p^2+i\hbar \frac{d}{dx}\sqrt{p^2+i\hbar S^{\prime\prime}}} ~=~ \sqrt{p^2+i\hbar \frac{d}{dx}\sqrt{p^2+i\hbar \frac{d}{dx}\sqrt{p^2+i\hbar S^{\prime\prime}}}} ~=~ \ldots $$

Expanding

$$S^{\prime}~=~\sum_{n=0}^{\infty} (i\hbar)^{n}S^{\prime}_n,$$

one gets

$$S^{\prime}_0 ~=~ p,\qquad S^{\prime}_1 ~=~ \frac{p^{\prime}}{2p},\qquad S^{\prime}_2 ~=~ \frac{p^{\prime\prime}}{4p^2}-\frac{3(p^{\prime})^2}{8p^3}, $$

$$ S^{\prime}_3 ~=~ \frac{p^{\prime\prime\prime}}{8p^3}-\frac{3p^{\prime}p^{\prime\prime}}{4p^4}+\frac{3(p^{\prime})^3}{4p^5},\qquad, \ldots $$

OP then asks if there is a method to check if $f=S^{\prime}_n$ is a total derivative? Yes, one can check if the Euler-Lagrange operator

$$E(f)~=~\sum_{n=0}^{\infty} \left( - \frac{d}{dx}\right)^n \frac{\partial f}{\partial p^{(n)}}$$

vanishes on $f=S^{\prime}_n$. E.g.

$$E(S^{\prime}_0)~=~1~\neq~ 0, \qquad E(S^{\prime}_1)~=~ 0,\qquad E(S^{\prime}_2)~\neq~ 0,\qquad E(S^{\prime}_3)~=~ 0,\qquad\ldots. $$

Here we are using the fact that a local functional can be written as a boundary term if and only if the Euler-Lagrange equation vanishes identically. See also this Phys.SE answer.

At least in this way it is possible to check operationally order by order in $n$ whether $S^{\prime}_n$ is a total derivative or not. We have not checked the calculations beyond OP's claims, nor investigated OP's conjectures.

Community
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Qmechanic
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  • That good, that answers one of my questions. The next goes a little deeper. How come the odd terms can by found by taking a derivative or consequent derivatives of p and the even terms have to be integrated? What does this mean physically? – John M Mar 31 '13 at 14:23
  • Comments: (i) Note that a factor of $i$ has been absorbed into the def. of $S^{\prime}_n$ as compared to OP's def. (ii) It might be interesting to note that there exists an action principle $I=\int ! dx\left(\overline{\psi}^{\prime}\psi^{\prime} -k(x)^2\overline{\psi}\psi\right)$ for Schroedinger eq., where $\psi\equiv e^{\frac{i}{\hbar}S}$. – Qmechanic Mar 31 '13 at 19:35