- #1
redtree
- 285
- 13
I note the following:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\delta(\vec{x}_n-\vec{x}_0)e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}
\end{split}
\end{equation}I divide the time interval as follows:
\begin{equation}
\begin{split}
t_n-t_0 &= \sum_{j=1}^{n} t_j - t_{j-1}
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\delta(\vec{x}_n-\vec{x}_0) e^{-i \frac{\mathcal{H}_n}{\hbar} \sum_{j=1}^{n} t_j - t_{j-1}}
\\
&=\delta(\vec{x}_n-\vec{x}_0) \prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\end{split}
\end{equation}I note the following:
\begin{equation}
\begin{split}
\delta(\vec{x}_n-\vec{x}_0)&=\int_{-\infty}^{\infty} d\vec{x}_{n-1} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{0})
\\
&=\int_{-\infty}^{\infty} d\vec{x}_{n-1}\int_{-\infty}^{\infty} d\vec{x}_{n-2} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{n-2})\delta(\vec{x}_{n-2}-\vec{x}_{0})
\\
&=\int_{-\infty}^{\infty} d\vec{x}_{n-1}\int_{-\infty}^{\infty} d\vec{x}_{n-2}\int_{-\infty}^{\infty} d\vec{x}_{n-3} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{n-2})\delta(\vec{x}_{n-2}-\vec{x}_{n-3})\delta(\vec{x}_{n-3}-\vec{x}_{0})
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1})
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1}) \prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1}) e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\delta(\vec{x}_{j}-\vec{x}_{j-1}) e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\end{split}
\end{equation}Given:
\begin{equation}
\begin{split}
\delta(\vec{x}_{j}-\vec{x}_{j-1})&=\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar} e^{i \frac{\vec{p}_j}{\hbar}(\vec{x}_{j}-\vec{x}_{j-1})}
\end{split}
\end{equation}Thus:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{i \frac{\vec{p}_j}{\hbar}(\vec{x}_{j}-\vec{x}_{j-1})} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{i (t_j - t_{j-1})\frac{\vec{p}_j}{\hbar}\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}-\mathcal{H}_n\right)}
\end{split}
\end{equation}Given ##\vec{v}_j=\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}##:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j \vec{v}_j-\mathcal{H}_n\right)}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j \vec{v}_j\frac{m}{m}-\mathcal{H}_n\right)}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\frac{\vec{p}_j^2}{m}-\mathcal{H}_n\right)}
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
\mathcal{H}_n&=T_n + V_n
\\
&=\frac{\vec{p}_n^2}{2 m}+ V_n
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n \right)}
\end{split}
\end{equation}However, ##\vec{p}_j \neq \vec{p}_n##. Therefore, where ##\mathcal{L}## denotes the Lagrangian:
\begin{equation}
\begin{split}
\frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n &\neq \frac{\vec{p}_j^2}{2m}- V_n
\\
&\neq \mathcal{L}
\end{split}
\end{equation}What am I missing?
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\delta(\vec{x}_n-\vec{x}_0)e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}
\end{split}
\end{equation}I divide the time interval as follows:
\begin{equation}
\begin{split}
t_n-t_0 &= \sum_{j=1}^{n} t_j - t_{j-1}
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\delta(\vec{x}_n-\vec{x}_0) e^{-i \frac{\mathcal{H}_n}{\hbar} \sum_{j=1}^{n} t_j - t_{j-1}}
\\
&=\delta(\vec{x}_n-\vec{x}_0) \prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\end{split}
\end{equation}I note the following:
\begin{equation}
\begin{split}
\delta(\vec{x}_n-\vec{x}_0)&=\int_{-\infty}^{\infty} d\vec{x}_{n-1} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{0})
\\
&=\int_{-\infty}^{\infty} d\vec{x}_{n-1}\int_{-\infty}^{\infty} d\vec{x}_{n-2} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{n-2})\delta(\vec{x}_{n-2}-\vec{x}_{0})
\\
&=\int_{-\infty}^{\infty} d\vec{x}_{n-1}\int_{-\infty}^{\infty} d\vec{x}_{n-2}\int_{-\infty}^{\infty} d\vec{x}_{n-3} \delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{n-1}-\vec{x}_{n-2})\delta(\vec{x}_{n-2}-\vec{x}_{n-3})\delta(\vec{x}_{n-3}-\vec{x}_{0})
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1})
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right]\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1}) \prod_{j=1}^{n} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\delta(\vec{x}_{n}-\vec{x}_{n-1})\delta(\vec{x}_{l}-\vec{x}_{l-1}) e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\delta(\vec{x}_{j}-\vec{x}_{j-1}) e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\end{split}
\end{equation}Given:
\begin{equation}
\begin{split}
\delta(\vec{x}_{j}-\vec{x}_{j-1})&=\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar} e^{i \frac{\vec{p}_j}{\hbar}(\vec{x}_{j}-\vec{x}_{j-1})}
\end{split}
\end{equation}Thus:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{i \frac{\vec{p}_j}{\hbar}(\vec{x}_{j}-\vec{x}_{j-1})} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{i (t_j - t_{j-1})\frac{\vec{p}_j}{\hbar}\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}} e^{-i \frac{\mathcal{H}_n}{\hbar} (t_j - t_{j-1})}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}-\mathcal{H}_n\right)}
\end{split}
\end{equation}Given ##\vec{v}_j=\frac{(\vec{x}_{j}-\vec{x}_{j-1})}{(t_j - t_{j-1})}##:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j \vec{v}_j-\mathcal{H}_n\right)}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\vec{p}_j \vec{v}_j\frac{m}{m}-\mathcal{H}_n\right)}
\\
&=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\frac{\vec{p}_j^2}{m}-\mathcal{H}_n\right)}
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
\mathcal{H}_n&=T_n + V_n
\\
&=\frac{\vec{p}_n^2}{2 m}+ V_n
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\langle\vec{x}_n|e^{-i \frac{\mathcal{H}_n}{\hbar} (t_n-t_0)}|\vec{x}_{0}\rangle &=\prod_{l=1}^{n-1}\left[\int_{-\infty}^{\infty} d\vec{x}_l\right] \prod_{j=1}^{n}\left[\int_{-\infty}^{\infty} \frac{d\vec{p}_j}{2 \pi \hbar}\right] e^{\frac{i (t_j - t_{j-1})}{\hbar} \left(\frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n \right)}
\end{split}
\end{equation}However, ##\vec{p}_j \neq \vec{p}_n##. Therefore, where ##\mathcal{L}## denotes the Lagrangian:
\begin{equation}
\begin{split}
\frac{\vec{p}_j^2}{m}-\frac{\vec{p}_n^2}{2 m}- V_n &\neq \frac{\vec{p}_j^2}{2m}- V_n
\\
&\neq \mathcal{L}
\end{split}
\end{equation}What am I missing?