I Checking My Understanding: Lagrangian & Path Integral Formulation

[redtree]

I note the following:

\begin{equation}

\begin{split}

\langle \vec{x}| \hat{U}(t-t_0) | \vec{x}_0 \rangle&=\langle \vec{x}| e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} | \vec{x}_0 \rangle

\\

&=e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} \delta(\vec{x}-\vec{x}_0)

\end{split}

\end{equation}Given:

\begin{equation}

\begin{split}

\delta(\vec{x}-\vec{x}_0) &= \int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i \vec{k}(\vec{x}-\vec{x}_0)}

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\langle \vec{x}| e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} | \vec{x}_0 \rangle&=e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} \int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i \vec{k}(\vec{x}-\vec{x}_0)}

\\

&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i \vec{k}(\vec{x}-\vec{x}_0)} e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)}

\\

&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i \left(\vec{k}(\vec{x}-\vec{x}_0)- \frac{\mathcal{H}}{\hbar} (t-t_0)\right)}

\\

&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i (t-t_0)\left(\vec{k}\frac{(\vec{x}-\vec{x}_0)}{(t-t_0)}- \frac{\mathcal{H}}{\hbar} \right)}

\end{split}

\end{equation}Given $\vec{v}=\frac{(\vec{x}-\vec{x}_0)}{(t-t_0)}$:

\begin{equation}

\begin{split}

\langle \vec{x}| e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} | \vec{x}_0 \rangle&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i (t-t_0)\left(\vec{k}\vec{v}- \frac{\mathcal{H}}{\hbar} \right)}

\end{split}

\end{equation}Given:

\begin{equation}

\begin{split}

\frac{\mathcal{H}}{\hbar}&=\frac{T}{\hbar} + \frac{V}{\hbar}

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

\frac{T}{\hbar}&=\frac{\vec{p}^2}{2 m \hbar}

\end{split}

\end{equation}Given $\vec{p}=\hbar\vec{k}$:

\begin{equation}

\begin{split}

\frac{T}{\hbar}&=\frac{\hbar}{2 m }\vec{k}^2

\end{split}

\end{equation}Thus:

\begin{equation}

\begin{split}

\frac{\mathcal{H}}{\hbar}&=\frac{\hbar}{2 m }\vec{k}^2 + \frac{V}{\hbar}

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\langle \vec{x}| e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} | \vec{x}_0 \rangle&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i (t-t_0)\left(\vec{k}\vec{v}- \frac{\hbar}{2 m }\vec{k}^2 - \frac{V}{\hbar} \right)}

\end{split}

\end{equation}Where:

\begin{equation}

\begin{split}

\vec{k}\vec{v}&=\frac{m}{m}\vec{k}\vec{v}

\\

&=\frac{\vec{k}\vec{p}}{m}

\\

&=\frac{\hbar \vec{k}^2}{m}

\end{split}

\end{equation}Such that:

\begin{equation}

\begin{split}

\langle \vec{x}| e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} | \vec{x}_0 \rangle&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i (t-t_0)\left(\frac{\hbar \vec{k}^2}{m}- \frac{\hbar}{2 m }\vec{k}^2 - \frac{V}{\hbar} \right)}

\\

&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i (t-t_0)\left(\frac{\hbar \vec{k}^2}{2m} - \frac{V}{\hbar} \right)}

\\

&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i (t-t_0)\left(\frac{T}{\hbar} - \frac{V}{\hbar} \right)}

\end{split}

\end{equation}Given the Lagrangian $\mathcal{L}= T- V$:

\begin{equation}

\begin{split}

\langle \vec{x}| e^{-2 \pi i \frac{\mathcal{H}}{\hbar} (t-t_0)} | \vec{x}_0 \rangle&=\int_{-\infty}^{\infty} d\vec{k} e^{2 \pi i (t-t_0)\left(\frac{\mathcal{L}}{\hbar} \right)}

\end{split}

\end{equation}

I want to make sure that I am understanding this correctly. Is this correct? If not, where is my mistake?

 

[blue_leaf77]

First I would like to say that the Hamiltonian operators appearing in the first and second line in eq. (1) are not really the same. In the first line the Hamiltonian is still in operator form while in the second the it's already expressed in position space. In equation (3), I don't think you can arbitrarily swap the order of the exponential containing $x-x_0$ and that containing $H$. If the one with $x-x0$ follows the one with $H$ then it's supposed to be understood as $\exp \left( 2\pi i H(t-t_0)/\hbar \right)$ acting on $\exp(ik(x-x_0))$.

 

[vanhees71]

It's not as simple since $V$ is a function of $\hat{x}$. For the free particle you can of course simply insert a unit operator in terms of the completeness relation of momentum eigenstates (I set $\hbar=h/2 \pi=1$ for simplicity):
$$\langle \vec{x}|\exp(-\mathrm{i} t \hat{\vec{p}^2}/(2m)|\vec{x}' \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \langle \vec{x}|\exp(-\mathrm{i} t \hat{\vec{p}^2}/(2m)|\vec{p} \rangle \langle p|\vec{x}' \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^3} \exp[-\vec{p}^2 t /(2m)] \exp(\mathrm{i} \vec{p}(\vec{x}-\vec{x}').$$
Now you regularize the integral by making $t \rightarrow t-\mathrm{i} \epsilon$ with $\epsilon>0$. Then you get a Gaussian integral which can be solved easily.

For $V \neq 0$ this trick doesn't work anymore, and you have to use other techniques. To make contact with the action in Lagrange form the Feynman path-integral approach is a good one. You find a simple introduction in my QFT manuscript (also for non-relativistic QT in Chpt. 1):

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[redtree]

It's not as simple since $V$ is a function of $\hat{x}$.

I apologize for not seeing, but how does $V = V(\vec{x})$ change things?

 

[vanhees71]

You'd need to insert the more complicated complete set of energy eigenstates of the complete Hamiltonian including the potential, and it's usually not so simple to evaluate the corresponding sum/integral. It can be done, e.g., for the harmonic oscillator.

 

[redtree]

Thank you for your reply. Your manuscript is very good.From your manuscript, equation 1.56, I extract the following term:

\begin{equation}

\begin{split}

\text{exp}\left[-i\Delta t \sum_{k=1}^{N} \frac{\vec{p}_{k}^2}{2 m}+ V(\vec{x}_{k}) + i \sum_{k=1}^{N} \vec{p}_{k}(\vec{x}_{k+1}-\vec{x}_{k})\right]

\end{split}

\end{equation}For simplicity and the purposes of this discussion, I set $N=1$, such that:

\begin{equation}

\begin{split}

\text{exp}\left[-i\Delta t \left(\frac{\vec{p}_{1}^2}{2 m}+ V(\vec{x}_{1})\right) + i \vec{p}_{1}(\vec{x}_{2}-\vec{x}_{1})\right]&=\text{exp}\left[-i\Delta t \left(\frac{\vec{p}_{1}^2}{2 m}+ V(\vec{x}_{1})\right) + i \vec{p}_{1}\Delta t \frac{(\vec{x}_{2}-\vec{x}_{1})}{\Delta t}\right]

\end{split}

\end{equation}Given $\vec{v}=\frac{(\vec{x}_{2}-\vec{x}_{1})}{\Delta t}$:

\begin{equation}

\begin{split}

\text{exp}\left[-i\Delta t \left(\frac{\vec{p}_{1}^2}{2 m}+ V(\vec{x}_{1})\right) + i \vec{p}_{1}(\vec{x}_{2}-\vec{x}_{1})\right]&=\text{exp}\left[-i\Delta t \left(\frac{\vec{p}_{1}^2}{2 m}+ V(\vec{x}_{1})\right) + i \Delta t \vec{p}_{1} \vec{v}\right]

\\

&=\text{exp}\left[i\Delta t \left(\vec{p}_{1} \vec{v}-\frac{\vec{p}_{1}^2}{2 m}- V(\vec{x}_{1})\right) \right]

\\

&=\text{exp}\left[i\Delta t \left(\vec{p}_{1} \vec{v}\frac{m}{m}-\frac{\vec{p}_{1}^2}{2 m}- V(\vec{x}_{1})\right) \right]

\\

&=\text{exp}\left[i\Delta t \left(\frac{\vec{p}_{1}^2}{m}-\frac{\vec{p}_{1}^2}{2 m}- V(\vec{x}_{1})\right) \right]

\\

&=\text{exp}\left[i\Delta t \left(\frac{\vec{p}_{1}^2}{2 m}- V(\vec{x}_{1})\right) \right]

\\

&=\text{exp}\left[i\Delta t \left(\mathcal{L}\right) \right]

\end{split}

\end{equation}Which seems essentially the same as my derivation above.

 

[vanhees71]

Yes, it's essentially integrating out the momentum in the Hamiltonian version of the path integral (which must always be the starting point). Then, indeed, if the Hamiltonian is of this specific form, you get the Lagrangian version with the classical Lagrangian. If you have interactions involving the momenta (or velocities in the Lagrangian), you get something different when integrating out the momentum.