# Transitioning from Path integral in QM to QFT

I've been recently introduce in the Path Integral formulation of Quantum Mechanics. The typical place to start is where Feynman himself, I believe, started. That is, the one dimensional case where a particle with Hamiltonian $$\hat{H}=\hat{T}(p)+\hat{V}(x)$$ starts from $$\left|x_i,t_i\right>$$ and transitions to $$\left|x_f,t_f\right>$$. Then we calculate the propagator $$\left$$ which is: $$\left=\int_{x(t_i)=x_i}^{x(t_f)=x_f}\mathcal{D}x\int\mathcal{D}p\ e^{iS[p,x]}$$ where $$\int\mathcal{D}x=\lim_{N\rightarrow\infty}\int \prod_{n=1}^Ndx_n$$ and also some constants whave been integrating in the definition of measures $$\mathcal{D}x$$ and $$\mathcal{D}p$$. If we also assume that the Hamiltonian describing the system is quadratic in its momentum, then we can perform the Gaussian integral and end up with $$\left=\int_{x(t_i)=x_i}^{x(t_f)=x_f}\mathcal{D}x\ e^{iS[x]}$$ where again, some constants have been included in $$\mathcal{D}x$$. The $$n$$-dimensional case seems to me that is a trivial generalization that gives us the same result. Am I wrong?

Then we have to translate the path integral method to quantum field theory. For that I first read Srednicki's book, section "The path integral for free-field theory". He makes the transition form QM to QFT just by saying that $$x\rightarrow \phi(x)$$ and uses the same formula essentially to calculate the generating function $$Z_0(J)=\left<0|0\right>_J$$. (I'm skipping some intermediate steps for brevity). However, at some point, in order to calculate the generating function, he takes the reverse process and discretises the path integral in order to perform a Gaussian integral. The way he does that is not clear to me, so I searched further and I ended up reading Schwartz's book, subsection "Path integral in quantum field theory".

There, first he also mentions that the QFT analog to $$\hat{x}$$ is $$\hat{\phi}(x)$$ which is defined as (here $$x$$ and $$p$$ are three-vectors): $$\hat{\phi}(\boldsymbol{x})=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_pe^{i\boldsymbol{p}\boldsymbol{x}}+a_p^\dagger e^{-i\boldsymbol{p}\boldsymbol{x}})$$ The Hamiltonian is also a function of the conjugate operator $$\pi(\boldsymbol{x})\equiv -i\int\frac{d^3p}{(2\pi)^3}\sqrt{\frac{\omega_p}{2}}(a_pe^{i\boldsymbol{p}\boldsymbol{x}}-a_p^\dagger e^{-i\boldsymbol{p}\boldsymbol{x}})$$ The two operators satisfy the cannonical commutation relation $$\left[\hat{\phi}(\boldsymbol{x}),\hat{\pi}(\boldsymbol{x})\right]=i\delta^{(3)}(\boldsymbol{x}-\boldsymbol{y})$$

Then, and this is the beginning of my problems to follow, he makes analogies to the QM case introducing the eigenstates of $$\phi$$ and $$\pi$$ $$\hat{\phi}(\boldsymbol{x})\left|{\Phi}\right>=\Phi(\boldsymbol{x})\left|{\Phi}\right>$$ and $$\hat{\pi}(\boldsymbol{x})\left|{\Pi}\right>=\Pi(\boldsymbol{x})\left|{\Pi}\right>$$ where $$\left<\Pi|\Phi\right>=\exp\left(-i\int d^3x\ \Pi(\boldsymbol{x})\Phi(\boldsymbol{x})\right)$$ Also, he states that it is true that $$\left<\Phi_1|\Phi_2\right>=\int \mathcal{D}\Pi\ \left<\Phi_1|\Pi\right>\left<\Pi|\Phi_2\right>=\int \mathcal{D}\Pi\ \exp\left(-i\int d^3x\ \Pi(\boldsymbol{x})[\Phi_2(\boldsymbol{x})-\Phi_1(\boldsymbol{x})]\right)$$ and then continues on like the Quantum mechanical case. At least in this case I get how the discretisation of fields takes place.

So my questions are:

1. Is the Schwartz's approach necessary or at least a sound transition from QM to QFT?
2. What are some alternative approaches for this transition?
3. How do we go from defining the eigenstates of the field operators to their inner product?
4. Why is $$\int\mathcal{D}\Pi\left|\Pi\right>\left<\Pi\right| = 1$$, so why is $$\mathcal{D}$$ used?

My problem is I have a vague idea of everything but not yet crystallised.

• At the typical physicist's level of (non-)rigor, the inner product $$\langle \Pi\vert \Phi\rangle$$ is precisely analogous to the inner product $$\langle x \vert p\rangle = \mathrm{e}^{\mathrm{i}xp}$$, just "in field space", since $$\phi(x)$$ and $$\pi(x)$$ fullfil a continuous version of the canonical commutation relations. Qmechanic lays out the argument for the value of this inner product in this excellent answer, the field theory version is supposed to work the same way, just with some additional $$\delta$$s and integrals floating around.
• $$\int \lvert \Pi\rangle\langle \Pi\rvert \mathcal{D}\Pi = 1$$ is just the analog of the usual completeness relation $$\int \lvert p\rangle\langle p\rvert \mathrm{d}p = 1$$ for the momentum operator. The $$\mathcal{D}$$ is just the suggestive notation that the "integral" is supposed to be over all possible field configurations $$\Pi : \mathbb{R}^{3,1}\to\mathbb{R}$$, not just over some real number as in the QM case. Again, this is extremely heuristic - we cannot make precise how the integral over that space is supposed to work, $$\hat{\pi}(x)$$ probably isn't actually an operator but an operator-valued distribution so it having eigenstates makes even less sense that the mathematically also rather thorny $$\lvert p\rangle$$ for the momentum operator, and so every piece of this formula is really just that - purely formal. It represents less a well-defined computation and more an idea or concept of the eigenstates of the momentum field operator being complete.