# Complex phase of the path integral in QM?

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 $$$$\int\!Dq(t)\equiv \lim\limits_{N\to\infty}\left( \frac{m}{2\pi i\hbar \delta t} \right)^{\!\frac{N}{2}}\left( \prod^{N-1}_{k=1} \int dq_k \right) ,$$$$

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

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

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

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

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

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?

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