Question 7.1: Derivation of the Transition Amplitude (7.5) Derive the tran......
Derivation of the Transition Amplitude (7.5)
Derive the transition amplitude (7.5) from the concept of path integrals.
\left\langle x_{\mathrm{f}}\left(t_{\mathrm{f}}\right) \mid x_{\mathrm{i}}\left(t_{\mathrm{i}}\right)\right\rangle=\mathcal{N} \underset{N \rightarrow \infty}{\lim} \int \prod\limits_{n=1}^{N-1} \mathrm{~d} x\left(t_n\right) \exp \left(\mathrm{i} S\left(\left\{x\left(t_n\right), \dot{x}\left(t_n\right)\right\}\right)\right) (7.5)
We extract the time dependence of the state vectors
T \equiv\left\langle x_{\mathrm{f}}\left(t_{\mathrm{f}}\right) \mid x\left(t_{\mathrm{i}}\right)\right\rangle=\langle x_{\mathrm{f}}|\exp \left(-\mathrm{i} \hat{H}\left(t_{\mathrm{i}}-t_{\mathrm{f}}\right)\right)| x\rangle . (1)
In the next step we divide the time interval \left(t_{\mathrm{f}}-t_{\mathrm{i}}\right) into N-1 intermediate time steps with \Delta t=\left(t_{\mathrm{f}}-t_{\mathrm{i}}\right) / N as done before in (7.3):
T=\langle x_{\mathrm{f}} \equiv x_{N}|\underbrace{\exp (-\mathrm{i} \hat{H} \Delta t) \exp (-\mathrm{i} \hat{H} \Delta t) \cdots}_{N \text { times }}| x_{\mathrm{i}} \equiv x_{0}\rangle (2)
and insert complete sets of eigenstates of the positions between each step
\begin{aligned} T= & \left\langle x_{N}\left|\exp (-\mathrm{i} \hat{H} \Delta t) \int \mathrm{d} x_{N-1}\right| x_{N-1}\right\rangle\left\langle x_{N-1}\right| \exp (-\mathrm{i} \hat{H} \Delta t) \\ & \times \int \mathrm{d} x_{N-2}\left|x_{N-2}\right\rangle\left\langle x_{N-2}|\cdots \exp (-\mathrm{i} \hat{H} \Delta t)| x_{0}\right\rangle . & (3) \end{aligned}
In addition we also insert complete sets of momentum eigenstates into (3)
\begin{aligned} T= & \int \mathrm{d} x_{1} \mathrm{~d} x_{2} \mathrm{~d} x_{3} \cdots \mathrm{d} x_{N-1} \frac{\mathrm{d} p_{0}}{2 \pi} \cdots \frac{\mathrm{d} p_{N-1}}{2 \pi} \\ & \times\left\langle x_{N} \mid p_{N-1}\right\rangle\left\langle p_{N-1}|\exp (-\mathrm{i} \hat{H} \Delta t)| x_{N-1}\right\rangle \\ & \times\left\langle x_{N-1} \mid p_{N-2}\right\rangle\left\langle p_{N-2}|\exp (-\mathrm{i} \hat{H} \Delta t)| x_{N-2}\right\rangle \\ & \times \cdots \times\left\langle p_{0} \mid x_{0}\right\rangle . & (4) \end{aligned}
The main purpose of the whole procedure is that for sufficiently large N and small \Delta t one can expand the exponential
\exp (-\mathrm{i} \hat{H} \Delta t) \sim 1-\mathrm{i} \Delta t \hat{H}. (5)
In (4) the operator \hat{H} is sandwiched between eigenstates of the position and momentum operator. Thus, using \hat{x}|x\rangle=x|x\rangle and \hat{p}|p\rangle=p|p\rangle we can replace the operators in \hat{H} by their eigenvalues: { }^{4}\langle p|\hat{H}(\hat{p}, \hat{x})| x\rangle=\langle p|H(p, x)| x\rangle= \langle p \mid x\rangle H(p, x). This is the main trick used in the path integral formalism: operators are replaced by c numbers at the cost of (infinitely) many sets of basis states representing the possible paths of the particle. Using this results and \langle x \mid p\rangle=\exp (\mathrm{i} p x) we obtain the expression
\begin{aligned} T= & \int \prod_{n=1}^{N-1} \mathrm{~d} x_{n} \prod_{m=0}^{N-1} \frac{\mathrm{d} p_{m}}{2 \pi} \exp \left(\mathrm{i} \sum_{n=0}^{N-1} p_{n}\left(x_{n+1}-x_{n}\right)\right) \\ & \times \prod_{n=0}^{N-1}\left(1-\mathrm{i} \Delta t H\left(p_{n}, x_{n}\right)\right) \\ \underset{N \to ∞}{=} & \mathcal{N} \int[\mathrm{d} x][\mathrm{d} p] \exp \left(\mathrm{i} \int\limits_{t}^{t^{\prime}} \mathrm{d} \tilde{t}(p \dot{x}-H(p, x))\right) . & (6) \end{aligned}
Equation (6) is the path integral in phase space. Depending on the specific structure of H it is rather straightforward to get rid of the momentum integration if H is quadratic in the momentum, e.g.
H=\frac{p^{2}}{2 m}+U(x). (7)
Using this expression we see that we can complete the square with respect to the momentum variable in (6):
T=\mathcal{N} \int[\mathrm{d} x][\mathrm{d} p] \exp \left(-\mathrm{i} \int\limits_{t}^{t^{\prime}} \mathrm{d} \tilde{t}\left(\frac{1}{2 m}(p-m \dot{x})^{2}-\frac{m \dot{x}^{2}}{2}+U(x)\right)\right) (8)
Now one can transform the variable p^{\prime}=p-m \dot{x}, which amounts to a Legendre transformation, and integrate the Gaussian integral of p^{\prime}, which generates a constant that can be absorbed in the normalization factor:
\begin{aligned} & =\mathcal{N}^{\prime} \int[\mathrm{d} x] \exp \left(\mathrm{i} \int\limits_{t}^{t^{\prime}}\left[\mathrm{d} \tilde{t} \frac{m \dot{x}^{2}}{2}-U(x)\right]\right) \\ & =\mathcal{N}^{\prime} \int[\mathrm{d} x] \exp (\mathrm{i} S) . & (9) \end{aligned}
Equation (9) is the path integral formula in the Lagrange formalism (7.5), here in Minkowski space-time.
{ }^{4} Some care has to be taken with the ordering of the operators when the Hamiltonian contains mixed products of position and momentum operators.