Overlap $\langle \phi|\pi \rangle$ in quantum field theory

Question

I was reading through Kapusta & Gale, "Finite temperature Field theory Principles and applications". In chapter 2, they derive a partition function for a normal field theory (0 temperature case). I see the following argument:

Let $\hat{\phi}({\bf x},0)$ and $\hat{\pi}({\bf x},0)$ be Schroedinger operators.

Then $$\hat{\phi}({\bf x},0)|\phi\rangle = \phi({\bf x})|\phi \rangle\tag{2.1}$$ and similarly, for the conjugate momenta field, $$\hat{\pi}({\bf x},0)|\pi\rangle = \pi({\bf x})|\pi \rangle.\tag{2.4}$$ Here, $\phi({\bf x})$ and $\pi({\bf x})$ are the eigenfunctions to the Schroedinger operators; while $|\phi\rangle$ and $|\pi\rangle$ are the eigenstates.

What they then say in Eq. (2.7) and Eq. (2.8) is the following:

• In quantum theory, one has $$\langle x|p \rangle = e^{ipx}. \tag{2.7}$$

• On similar lines, in field theory one should have $$\langle \phi|\pi \rangle = \exp \left (i\int d^3x \pi({\bf x})\phi({\bf x}) \right ). \tag{2.8}$$

Is there any way to prove the relation (2.8)?

To me, $\langle \pi|\phi \rangle$ is just a normal inner product space, and hence it should simply be $$\langle \phi|\pi \rangle = i\int d^3x \pi(\bf{x})\phi(\bf{x}),$$ i.e. there should be no exponentiation.

Also, I believe, $$\langle \pi|\phi \rangle\langle \phi|\pi \rangle = |\langle \pi|\phi \rangle|^2 .$$

But, if I use (2.8), I get, $$\langle \pi|\phi \rangle\langle \phi|\pi \rangle = 1.$$ This does not sound right.

Is there a mistake I am making somewhere?

Answer 1

The QFT formulas follows from the corresponding QM formulas via the usual heuristic discretization rules, e.g.,

$$ \text{index }\{1,\ldots,n\!\equiv\! N^3\}~\ni~j \qquad\longrightarrow\qquad {\bf x}~\in~[0,L]^3 \text{ spatial position}, \tag{A}$$

$$\text{position } \mathbb{R}~\ni~q^j~ \qquad\longrightarrow\qquad \phi({\bf x})~\in~\mathbb{R} \text{ field} , \tag{B}$$

$$\text{momentum } p_j~ \qquad\longrightarrow\qquad V \pi({\bf x})~=~ \text{ unit volume}\times \text{ momentum density} , \tag{C}$$

$$\text{sum } \sum_{j=1}^n~ \qquad\longrightarrow\qquad \int_{[0,L]^3} \!\frac{d^3{\bf x}}{V} \text{ integral} , \tag{D}$$

$$ \hat{q}^j(t)|q\rangle ~=~ q^j|q\rangle \qquad\longrightarrow\qquad \hat{\phi}({\bf x},t)|\phi\rangle ~=~ \phi({\bf x})|\phi\rangle, \tag{2.1}$$

$$ \int \!d^nq ~|q\rangle \langle q |~=~{\bf 1} \qquad\longrightarrow\qquad \int \!\left[ \prod_{\bf x} d\phi({\bf x})\right] |\phi\rangle \langle \phi |~=~{\bf 1}, \tag{2.2}$$

$$ \langle q |q^{\prime}\rangle ~=~\delta^n(q\!-\!q^{\prime}) \qquad\longrightarrow\qquad \langle \phi |\phi^{\prime}\rangle ~=~\prod_{\bf x} \delta(\phi({\bf x})\!-\!\phi^{\prime}({\bf x})), \tag{2.3}$$

$$ \hat{p}_j(t)|p\rangle ~=~ p_j|p\rangle \qquad\longrightarrow\qquad \hat{\pi}({\bf x},t)|\pi\rangle ~=~ \pi({\bf x})|\pi\rangle, \tag{2.4}$$

$$ \int \!\frac{d^np}{(2\pi\hbar)^n} ~|p\rangle \langle p |~=~{\bf 1} \qquad\longrightarrow\qquad \int \!\left[ \prod_{\bf x} \frac{d\pi({\bf x})}{2\pi\hbar}\right] |\pi\rangle \langle \pi |~=~{\bf 1}, \tag{2.5}$$

$$ \langle p |p^{\prime}\rangle ~=~(2\pi\hbar)^n\delta^n(p\!-\!p^{\prime}) \qquad\longrightarrow\qquad \langle \pi |\pi^{\prime}\rangle ~=~\prod_{\bf x} 2\pi\hbar\delta(\pi({\bf x})\!-\!\pi^{\prime}({\bf x})), \tag{2.6}$$

$$ \langle q |p\rangle ~=~\exp\left\{ \frac{i}{\hbar}\sum_{j=1}^nq^jp_j\right\}\tag{2.7}$$

$$\qquad\longrightarrow\qquad \langle \phi |\pi\rangle ~=~\exp\left\{ \frac{i}{\hbar}\int_{[0,L]^3} \!d^3{\bf x}~\phi({\bf x})\pi({\bf x})\right\}. \tag{2.8}$$

Concerning how to prove the overlap (2.7) from the CCR, see e.g. this Phys.SE post. OP's sought-for proof of eq. (2.8) follows from transcribing eq. (2.7) via above QM$\to$ QFT dictionary.

Answer 2

Consider $f(x) = \langle p | x \rangle$. Now consider $$ f'(x) = \frac{d}{dx} \langle p | x \rangle = \langle p | \frac{d}{dx} | x \rangle . $$ Next, we use the fact that ${\hat p} | x \rangle = i \hbar \frac{d}{dx} | x \rangle$. Then, $$ f'(x) = \frac{1}{i\hbar} \langle p | {\hat p} | x \rangle = \frac{p}{i\hbar} \langle p | x \rangle = \frac{p}{i\hbar} f(x) $$ We can easily solve this differential equation $$ f(x) = \langle p | x \rangle = A e^{- \frac{i}{\hbar} p x } $$ To fix the constant $A$, we use $$ \delta ( x - x') = \langle x' | x \rangle = \int \frac{dp}{2\pi\hbar}\langle x' | p \rangle \langle p | x \rangle = \int \frac{dp}{2\pi\hbar} |A|^2 e^{- \frac{i}{\hbar} p ( x - x' ) } = |A|^2 \delta(x-x') $$ It follows that $A=1$. Here, we have chosen a particular phase which we can do by simply rescaling the phase of either $| x \rangle$ or $|p\rangle$.

Now use the exact same proof and apply it to the QFT case.

Answer 3

There does not seem to be a straight forward simple derivation. Well in that case, the best approach is probably to do it the hard way:

a) solve for the eigenstates of the two field operators

b) show that they are orthogonal and complete

c) compute the overlap between elements of the different bases

These field operators can be represented as linear combinations of the quadrature operators $\hat{q}_s(\mathbf{k})$ and $\hat{p}_s(\mathbf{k})$. Therefore, one can also follow the above procedure for these quadrature operators to show that $$ \langle p|q\rangle = \exp\left(i\int q_s(\mathbf{k})p_s(\mathbf{k})d^3k \right) . $$ The full derivation of this result is provided in one of my publications [Phys. Rev. A 98, 043841 (2018)], available here.