# Derivation of Schrödinger equation in Feynman-Hibbs

I am going through the derivation in chapter 4-1 of "Quantum Mechanics and Path Integrals. Emended Edition" by Feynman and Hibbs. The chapter starts with a proof of the equivalence of the propagator formalism to Schrödinger equation formulation of quantum mechanics.

Let us consider a wavefunction at a time $$t+\varepsilon$$: \begin{align} \psi(x,t+\varepsilon) = \frac{1}{A}\int_{-\infty}^{\infty} \mathcal{K}(x_f,t+\varepsilon;x_i,t) \psi(x,t) dx, \end{align} where $$\varepsilon$$ is an infinitesimally small time interval and $$\frac{1}{A}$$ is the normalisation factor. We approximate the propagator: \begin{align} \psi(x,t+\varepsilon) =&\frac{1}{A} \int_{-\infty}^{\infty} \exp\left[\frac{i}{\hbar} \varepsilon L\left(\frac{x-y}{\varepsilon}, \frac{x+y}{2} \right)\right]\cr &\qquad \;\psi(y,t) dy .\tag{4.3} \end{align} Now let us apply the above to the case of a particle moving in one dimension in a potential $$V(x,t)$$: \begin{align} \psi(x, t+\varepsilon) =& \frac{1}{A} \int_{-\infty}^{\infty} \exp\left[\frac{i}{\hbar} \frac{m}{2} \left(\frac{x-y}{\varepsilon}\right)^2 \varepsilon\right]\cr & \exp\left[- \frac{i}{\hbar} \varepsilon V\left(\frac{x+y}{2},t\right)\right] \psi\!\left(y,t\right) dy .\tag{4.4} \end{align} The first exponential contains the factor of $$\frac{x-y}{\varepsilon}$$ which will cause the propagator to oscillate rapidly when $$y$$ differs from $$x$$ considerably. Let us then make a substitution $$y = x + \eta$$. \begin{align} \psi(x,t+\varepsilon) =& \frac{1}{A} \int_{-\infty}^{\infty} \exp\left[\frac{i}{\hbar} \frac{m}{2} \left(\frac{\eta}{\varepsilon}\right)^2 \varepsilon\right] \cr &\exp\left[- \frac{i}{\hbar} \varepsilon V\left(x+ \frac{\eta}{2},t\right)\right] \psi\!\left(x+\eta,t\right) d\eta.\tag{4.5} \end{align} We expect that the most significant contribution will be yielded by small $$\eta$$ only.

We now expand $$\psi$$ in a Taylor series, keeping terms to first order in $$\varepsilon$$ and to second order in $$\eta$$: \begin{align} \psi (x,t) + \varepsilon \frac{\partial \psi}{\partial t} =& \frac{1}{A} \int_{-\infty}^{\infty} \exp\left[\frac{i}{\hbar} \frac{m}{2} \frac{\eta^2}{\varepsilon}\right] \left[1 - \frac{i}{\hbar} \varepsilon V(x,t)\right]\cr & \left[\psi(x,t) + \eta \frac{\partial \psi}{\partial x} + \frac{\eta^2}{2} \frac{\partial^2 \psi}{\partial x^2}\right] d\eta.\tag{4.6} \end{align} After solving the Gaussian integrals in $$(4.6)$$ and comparing terms linear in $$\epsilon$$ on both sides we arrive at the Schrödinger equation.

I understand the whole proof but am quite perplexed by the statement:

"We need only keep terms of order $$\varepsilon$$. This implies keeping second-order terms in $$\eta$$."

I cannot figure out how it is implied other than wishful thinking that we need a first-order derivative wrt $$t$$ and a second-order derivative wrt $$x$$.

Why is the wavefunction expanded keeping the terms to the first order in $$\varepsilon$$ while to the second order in $$\eta$$?

• Anyone without the book is completely lost and it is not clear what you are asking. Note that questions (and answers) must be self-contained and useful for other users, a broader audience, too. You should a) give a more detailed reference (book title, authors, edition, chapter, page, equations numbers) and b) provide context, and type the relevant passage (using MathJax, not screenshots). Commented Jul 5 at 20:50

Briefly speaking, it follows from dimensional analysis that higher-order terms $${\cal O}(\eta^{n\geq 3})$$ will [after the Gaussian $$\eta$$-integration (4.5)] only produce higher-orders terms $${\cal O}(\epsilon^{n/2})$$, which are not important as $$\epsilon\to 0$$.