Complex phase of the path integral in QM?

Question

The square modulus of an amplitude must be real. Given that, I am having some trouble understanding the square modulus of a path integral being absolutely real. Given \begin{equation} \int\!Dq(t)\equiv \lim\limits_{N\to\infty}\left( \frac{m}{2\pi i\hbar \delta t} \right)^{\!\frac{N}{2}}\left( \prod^{N-1}_{k=1} \int dq_k \right) , \end{equation}

and \begin{equation} \big\langle \psi_2 \big| e^{-i\hat H T/\hbar}\big| \psi_1 \big\rangle=\int\!Dq(t)\,\exp\left\{\frac{i}{\hbar}\int_0^t\!dt'\,\frac{1}{2}m \dot q_k^2\right\}, \end{equation}

introduce concise notation such that \begin{equation}\label{eq:od7diyh} \int\!Dq(t)\equiv \lim\limits_{N\to\infty}\left( i\beta\right)^{\!\frac{N}{2}}\big[ F(q) \big] ,\quad\text{and}\quad\big\langle \psi_2 \big| e^{-i\hat H T/\hbar}\big| \psi_1 \big\rangle=\int\!Dq(t)\,e^{i\gamma}. \end{equation}

Quantum mechanics requires that \begin{equation} \big\|\big\langle \psi_2 \big| e^{-i\hat H T/\hbar}\big| \psi_1 \big\rangle\big\|^2\in\mathbb{R}~~,\nonumber \end{equation}

so it follows that we must have \begin{equation} \left\{\lim\limits_{N\to\infty}\left( i\beta\right)^{\!\frac{N}{2}}\big[ F(q) \big]e^{i\gamma}\right\}\left\{\lim\limits_{N\to\infty}\left( -i\beta\right)^{\!\frac{N}{2}}\big[ F(q) \big]e^{-i\gamma}\right\}\in\mathbb{R}~~.\nonumber \end{equation}

This simplifies as \begin{equation} \lim\limits_{N\to\infty}\beta^N\big[ F(q) \big]^2 e^{i\gamma}e^{-i\gamma}~~.\nonumber \end{equation}

If I cancel the two exponentials, that will mean that the probability (squared amplitude) doesn't depend on the action at all. If I don't cancel them (for some reason that is not clear to me), then how can I guarantee that the expression is real? What is going on here?

Answer 1

I assume that your $\gamma$ is the action. You can't cancel the two exponentials $e^{i\gamma}$ because they are being integrated. The basic object in the path integral formalism is the propagator \begin{equation} \left\langle q_f | e^{-i H t} |q_i \right\rangle = \int^{q(t_f)=q_f}_{q(t_i)=q_i} \mathcal{D}q \, \exp{ \Big( i \gamma[q] \Big) } = \int^{q(t_f)=q_f}_{q(t_i)=q_i} \mathcal{D}q \, \exp{ \Big( i \int^{t_f}_{t_i} dt \, L\left( q(t),\dot{q}(t)\right) \Big) }, \end{equation} where the integration is only over all of the paths such that the endpoints are $q_i$ and $q_f$. You question is about a transition amplitude between arbitrary states $\left| \psi_1 \right\rangle$ and $\left| \psi_2 \right\rangle$. To write this in terms of the path integral, you can use the position space completeness relation to get \begin{equation} \left\langle \psi_2 | e^{-i H t} |\psi_ 1 \right\rangle = \int dq_f \, dq_i \left\langle \psi_2 | q_f \right\rangle \left\langle q_f | e^{-i H t} | q_i \right\rangle \left\langle q_i | \psi_1 \right\rangle. \end{equation} The matrix element between the wavefunctions in the middle is what you compute with the path integral. Usually if you try to compute the amplitude squared $||\left\langle q_f | e^{-i H t} | q_i \right\rangle||^2$ you will run into problems because the position eigenstates are not normalizable, but that does not matter here so we will ignore this point. We have \begin{equation} ||\left\langle q_f | e^{-i H t} | q_i \right\rangle||^2 = \left\langle q_f | e^{-i H t} | q_i \right\rangle \left\langle q_i | e^{i H t} | q_f \right\rangle = \int \mathcal{D}q \, \mathcal{D}q^\prime \exp{ i\Big( \gamma[q] - \gamma[q^\prime] \Big) }, \end{equation} with the appropriate limits. You can't cancel $\gamma[q]$ with $-\gamma[q^\prime]$ because these objects are being integrated. Of course, if after doing the path integrals you get a constant phase times some function, this phase cancels in all amplitudes, but the action under the path integral is not constant.

Answer 2

Your notation is a bit confusing. The indice in $\dot{q}$ is not necessary if you already write the classical action as a time integral.

Let's discretize the time $t$ to have a more clear expression: $$ \int Dq(t)\exp\left(i \int_0^t dt' \frac{1}{2}m\dot{q}_k^2\right) =\lim_{N\rightarrow \infty} (i\beta)^{N/2}\prod_{k=1}^{N-1} \int dq_{k} \exp\left[i\frac{1}{2}m\left(\frac{q_k - q_{k-1}}{t_{k}-t_{k-1}}\right)^2(t_{k}-t_{k-1})\right]\\ = \lim_{N\rightarrow \infty} (i\beta^{N/2})\prod_{k=1}^{N-1} \int dq_{k} \exp\left[\frac{im}{2\Delta t}(q_k - q_{k-1})^2\right]. $$

Now if we take te amplitude, we will get $$ \lim_{N\rightarrow \infty} \beta^{N}\prod_{k,k'=1}^{N-1} \int dq_{k}\int dq_{k'} \exp\left[\frac{im}{2\Delta t}(q_k - q_{k-1})^2\right] \exp\left[\frac{-im}{2\Delta t}(q_{k'} - q_{k'-1})^2\right]. $$ When $k=k'$, the exponentials cancel each other, as you expect. When $k\neq k'$, this not necessary happen. However, we can solve every possible integral in index $k$ and $k'$ separately, since they are independent (It would not possibly if something like $(q_k-q_{k'})^2$ appeared in the final expression, but there's nothing like that). So, integrating over all $q_k$ and $q_{k'}$, we get $$ \lim_{N\rightarrow \infty} \beta^{N}|A(t)|^2\exp\left[\frac{im}{2t}(q(t) - q(0)^2\right] \exp\left[\frac{-im}{2t}(q(t) - q(0)^2\right] = \lim_{N\rightarrow \infty} \beta^{N}|A(t)|^2 $$ which is something real. What is happening here? You missed one point: First, your $\gamma$ is dependent of the dummy index $k$, so if you have a product in $k$ and $k'$, they will cancel each other before integration only if $k=k'$. However, we can solve all the integrals in each dummy index separately and take the product of the results, as showed above. The amplitude $A(t)$ is just the normalization of the free kernel, and the final expression only depends of it's modulus, so it's a real thing.