Chapter 8 The Path Integral Formulation of Quantum Theory

8.1 The Path Integral Recipe

8.2 Analysis of the Recipe

8.3 An Approximation to $U(t)$ for the Free Particle

8.4 Path Integral Evaluation of the Free-Particle Propagator

8.5 Equivalence to the Schrödinger Equation

8.6 Potentials of the Form $V=a+bx+cx^{2}+d\dot{x}+ex\dot{x}$

Exercise 8.6.1 Verify that

U(x,t;x^{\prime},0)=A(t)\exp(\mathrm{i}S_{\text{cl}}/\hbar),~A(t)=\left(\dfrac{m}{2\pi\hbar\mathrm{i}t}\right)^{1/2}

agrees with the exact result, Eq. (5.4.31), for $V(x)=-fx$ . Hint: Start with $x_{\text{cl}}(t^{\prime\prime})=x_{0}+v_{0}t^{\prime\prime}+\dfrac{1}{2}(f/m)t^{\prime\prime 2}$ and find the constants $x_{0}$ and $v_{0}$ from the requirement that $x_{\text{cl}}(0)=x^{\prime}$ and $x_{\text{cl}}(t)=x$ .

Exercise 8.6.2 Show that for the harmonic oscillator with

\mathscr{L}=\dfrac{1}{2}m\dot{x}^{2}-\dfrac{1}{2}m\omega^{2}x^{2}

U(x,t;x^{\prime})=A(t)\exp\left\{\dfrac{\mathrm{i}m\omega}{2\hbar\sin\omega t}[(x^{2}+x^{\prime 2})\cos\omega t-2xx^{\prime}]\right\}

where $A(t)$ is an unknown function. (Recall Exercise 2.8.7.)

Exercise 8.6.3 We know that given the eigenfunctions and the eigenvalues we can construct the propagator: \begin{equation}\label{8.6.15} U(x,t;x^{\prime},t^{\prime})=\sum_{n}\psi_{n}(x)\psi_{n}^{*}(x^{\prime})\mathrm{e}^{-\mathrm{i}E_{n}(t-t^{\prime})/\hbar}\tag{8.6.15} \end{equation} Consider the reverse process (since the path integral approach gives $U$ directly), for the case of the oscillator.

(1) Set $x=x^{\prime}=t^{\prime}=0$ . Assume that $A(t)=(m\omega/2\pi\mathrm{i}\hbar\sin\omega t)^{1/2}$ for the oscillator. By expanding both sides of Eq. (\ref{8.6.15}), you should find that $E=\hbar\omega/2$ , $5\hbar\omega/2$ , $9\hbar\omega/2$ , $\ldots$ , etc. What happened to the levels in between?

(2) Now consider the extraction of the eigenfunctions. Let $x=x^{\prime}$ and $t^{\prime}=0$ . Find $E_{0}$ , $E_{1}$ , $|\psi_{0}(x)|^{2}$ , and $|\psi_{1}(x)|^{2}$ by expanding in powers of $\alpha=\exp(\mathrm{i}\omega t)$ .

Exercise 8.6.4 Recall the derivation of the Schrödinger equation (8.5.8) starting from Eq. (8.5.4). Note that although we chose the argument of $V$ to be the midpoint $x+x^{\prime} / 2$ , it did not matter very much: any choice $x+\alpha \eta$ , (where $\eta=x^{\prime}-x$ ) for $0\leqslant \alpha \leqslant 1$ would have given the same result since the difference between the choices is of order $\eta \varepsilon \simeq \varepsilon^{3 / 2}$ . All this was thanks to the factor $\varepsilon$ multiplying $V$ in Eq. (8.5.4) and the fact that $|\eta| \simeq \varepsilon^{1 / 2}$ , as per Eq. (8.6.5).

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