For non-relativistic quantum mechanics, the boundary conditions are rather simple to deal with, they are just

\begin{equation} \langle x_1, t_1 \vert x_2, t_2\rangle = \int_{x_1(t_1)}^{x_2(t_2)} \mathcal{D}x(t)e^{\frac{i}{\hbar}S[x(t)]} \end{equation}

And then we can solve the path integral by hand with the boundary conditions that the position must be $x_1$ at time $t_1$ and $x_2$ at time $t_2$, with whatever method is used to solve the integral (since they usually can be solved exactly in some manner).

For quantum field theory, though, boundary conditions seem to usually ignored, and only the vacuum expectation value is calculated

\begin{equation} \langle \Omega \vert \hat{A}\vert\Omega\rangle = \int \mathcal{D}\varphi(x)A(\varphi(x))e^{\frac{i}{\hbar}S[\varphi(x)]} \end{equation}

What would be the generic method to compute an arbitrary transition amplitude from one state to the other? It is usually written as

\begin{equation} \langle \varphi_1\vert \hat{A}\vert\varphi_2\rangle = \int_{\varphi_1(x_1)}^{\varphi_2(x_2)} \mathcal{D}\varphi(x)A(\varphi(x))e^{\frac{i}{\hbar}S[\varphi(x)]} \end{equation}

But the indications on how to do it are rather scarce. I suppose the method that suggests itself the most naturally would be to just express the wavefunction as a function of the vacuum of the theory, $\vert \varphi_1 \rangle = \hat{B} \vert \Omega \rangle$, but this seems to be both a bit contrary to the spirit of the formalism (ideally something that does not refer to wavefunctions) and also possibly problematic with regard to operator ordering (I know that operator ordering can be dealt with stochastic integrals in the action, but I'm not sure how it applies outside of it).

Is there a way to deal with boundary conditions without solving by hand the path integral for a quantum field for some boundary conditions of that field (which, I see, is extremely rarely done)? What would be a simple example for it, for instance for a boundary condition of a free field such that

\begin{equation} \varphi_i(\vec x, t_i) \propto e^{-\frac{(\vec{x} - \vec{x}_i)^2}{2\sigma^2}} \end{equation}

to compute the transition amplitude between two gaussian wavepackets, ideally without referring to wavefunctions?


Hello handsome poster (why thank you kind stranger). The answer to your question it turns out is in Rovelli's "Quantum Gravity", at least insofar as the free scalar field is concerned. This is done in the following way. As you may recall (from Feynman and Hibbs), through various arguments about doing the path integral as a perturbation on the classical path, the path integral of a Gaussian lagrangian for a point particle

$$L = a(t) \dot x^2 + b(t) \dot x x + c(t) x^2 + d(t) \dot x + e(t) x + f(t)$$

will be

$$K(x_a, t_a; x_b, t_b) = e^{\frac{i}{\hbar} S_{cl} [x_b, x_a]} \int_0^0 \exp\{ \frac{i}{\hbar} \int_{t_a}^{t_b}[a(t) \dot y^2 + b(t) \dot y y + c(t) y^2]dt\}\mathcal Dy(t)$$

Where the second term will be some function only depending on the beginning and end time, as it does not depend on the position.

Through a rather tedious calculation, you can work out that the classical action for the harmonic oscillator is

$$S_{cl} = \frac{m\omega}{2\sin(\omega T)} ((x_a^2 + x_b^2) \cos (\omega T) - 2 x_a x_b)$$

with the remaining term of the kernel

$$\int_0^0 \exp\{ \frac{i}{\hbar} \int_{t_a}^{t_b} \frac{m}{2}[\dot y^2 - \omega y^2]dt\}\mathcal Dy(t)$$

As the function $y(t)$ goes from $0$ to $0$ in the interval $T = t_b - t_a$, it can be written as the Fourier series $$y(t) = \sum_n a_n \sin(\frac{n\pi t}{T})$$

The action can thus be transformed into

$$\int_{t_a}^{t_b} \frac{m}{2}[\dot y^2 - \omega y^2]dt = \frac{mT}{4} \sum_n [(\frac{n\pi}{T})^2 - \omega^2] a_n^2$$

With a path integral being a simple infinite product of the individual gaussians for each $a_n$, which becomes (cf Feynman Hibbs for more details)

$$F(T) = \sqrt{\frac{m\omega}{2 \pi i \hbar \sin(\omega T)}}$$

Now the free scalar field can be decomposed as an infinite number of harmonic oscillators via some Fourier transform

\begin{eqnarray} S &=& \int_{t_a}^{t_b} dt \int d^3x \frac{1}{2}[\partial_\mu\varphi \partial^\mu \varphi - m^2 \varphi^2]\\ &=& \int \frac{d^3k}{(2\pi)^3} \{\int_{t_a}^{t_b} dt \frac{1}{2}[\dot \varphi(k)^2 -(\vec k^2 + m^2)\vert \varphi(k) \vert^2]\} \end{eqnarray}

(with some use of Parseval's theorem) which is just the action of the harmonic oscillator. Then, via some generous physicist magic, we can write

$$\exp[\frac{i}{\hbar}\int \frac{d^3k}{(2\pi)^3} S_{SHO}[\varphi(\vec k)]] \approx \prod_k \exp[\frac{i}{\hbar (2\pi)^3} S_{SHO}[\varphi(\vec k)]]$$

For every mode, the transition amplitude will be

$$\exp[\frac{i}{\hbar(2\pi)^3} \frac{\omega}{2\sin(\omega T)} ((\varphi_a^2(k) + \varphi_b^2(k)) \cos (\omega T) - 2 \bar \varphi_a(k) \varphi_b(k)) \sqrt{\frac{\omega}{2 \pi i \hbar \sin(\omega T)}}$$

Which will give us, once properly multiplied back,

$$\exp[\frac{i}{\hbar} \int \frac{d^3 k}{(2\pi)^3} \frac{\omega}{2\sin(\omega T)} ((\varphi_a^2(k) + \varphi_b^2(k)) \cos (\omega T) - 2 \bar \varphi_a(k) \varphi_b(k)) \prod_k [\sqrt{\frac{\omega}{2 \pi i \hbar \sin(\omega T)}}]$$

$\prod_k [\sqrt{\frac{\omega}{2 \pi i \hbar \sin(\omega T)}}]$ corresponds to a normalization factor, which Rovelli writes as being

$$\mathcal N \approx \prod_k [\sqrt{\frac{m\omega}\hbar}] \exp[-\frac{V}{2} \int \frac{d^3 k}{(2\pi)^3} \ln [\sin (\omega T)]]$$

which is apparently formally divergent (I assume due to the volume term $V$ - Rovelli does not seem to mention what it is tho), but I suppose this is similar to the Hamiltonian of the Hilbert space method also being divergent.

| cite | improve this answer | |

It depends a little bit on what you wish to compute, but the method almost universally adopted in numerical simulations is to use periodic boundary conditions in imaginary time, $\varphi(x,0)=\varphi(x,iT)$. This corresponds to thermal expectation values $$ \langle \Omega_T|O|\Omega_T\rangle = \frac{1}{Z}\sum_n \langle n|Oe^{-\beta H}|n\rangle $$ where $\beta=1/T$. We have $$ \langle \Omega_T|O|\Omega_T\rangle = \int_{\varphi(x,0)=\varphi(x,T)}{\cal D}\varphi\, O(\varphi(x,t))e^{-S_E}\, , $$ where $S_E$ is the euclidean action. In the limit $T\to\infty$ this gives the Euclidean correlation functions in the ground state (for $O=\varphi(x_1,\tau_1)\ldots\varphi(x_n,\tau_n)$), which is the standard observable in lattice field theory.

You can work harder, generate a certain state using an appropriate operator, measure its wave function, and then impose this wave function as a boundary condition. However, this is rarely done, and would be impractical in field theory.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.