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I've been told that in QFT, everything is turned into their continuum type, i.e:

$$q_i \to \phi(x)$$ $$p_i \to \pi(x)$$ $$[q_i, p_j] = i\delta_{ij} \to [\phi(x), \pi(y)] = i\delta(x-y)$$

etc.

Now I've been wondering how one would represent states in this theory. Knowing that in QM, a state of n-variables can be represented as:

$$|\Psi\rangle = \int dq_1\dots dq_n \psi(q_1,\dots, q_n) |q_1, \dots, q_n\rangle$$

I suspect in QFT, a state would be represented something like:

$$|\Psi\rangle = \int \mathcal{D}[\phi] \psi(\phi)|\phi\rangle$$

Is that the case? Moreover, in some sense, is every concept in QM updated to fit into QFT? i.e.:

$$i\frac{\partial}{\partial t} \psi(\phi, t) = \mathcal{H}\psi(\phi,t)$$

If so, how would then generalize the momentum operator? $$-i\frac{\partial}{\partial q_i} \to \frac{\delta}{\delta\phi(x)}$$

Cosmas Zachos
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Habouz
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    You may find this article interesting: https://en.wikipedia.org/wiki/Schr%C3%B6dinger_functional. The short answer to your questions is "yes", although you need to think of $\psi$ as a functional and your $\partial/\partial \phi(x)$ should actually be a functional derivative $\delta/\delta\phi(x)$. – Andrew Aug 21 '23 at 19:50
  • Linked. Also this. – Cosmas Zachos Aug 21 '23 at 20:13

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Yes, the states in quantum field theory are what are called "wave functionals." The momentum operator is represented by a functional derivative, $\pi(\mathbf{x}) \rightarrow -i\frac{\delta}{\delta \phi(\mathbf{x})}$. Note that the minus sign comes from the fact that, in 1-d a right-going wave has the form $\psi_p(x) = e^{ipx/\hbar}$. To get $p\psi_p(x)$ you need $-i\hbar \frac{\partial}{\partial x}$.

Sean E. Lake
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