4
$\begingroup$

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?

$\endgroup$
4
  • 4
    $\begingroup$ A lot to unpack. You might want to understand how $\langle p| x\rangle=e^{ipx}$ is proved before going on to the QFT case. Also, in QFT the fields $\phi(x)$ and $\pi(x)$ are not wave functions, they are operators for which states can have definite values of. $\endgroup$
    – fewfew4
    Commented Feb 10, 2022 at 9:38
  • 1
    $\begingroup$ +1 $\uparrow$. $\langle \Psi_1 | \Psi_2 \rangle = \int d^3 x \Psi_1^*(x) \Psi_2(x)$ applies only to wave-functions $\Psi_i(x)$. I don't see why you think this formula should hold for $\phi(x)$ and $\pi(x)$ which are not wave-functions. $\endgroup$
    – Prahar
    Commented Feb 10, 2022 at 11:11
  • $\begingroup$ The fact that $|\langle\pi|\phi\rangle|^2=1$ simply tells you that the two bases are mutually unbiased. $\endgroup$ Commented Feb 10, 2022 at 13:09
  • $\begingroup$ related and useful (nice question with a lot of good links): physics.stackexchange.com/q/312006/226902 $\endgroup$
    – Quillo
    Commented Feb 14, 2022 at 9:47

3 Answers 3

6
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Thanks. What I was looking for is this: In QM, One can start with the basic de Broglie equation as an axiom, with $\lambda = h/p$. This gives the wave $e^{ipx}$. From here everything else follows. In QFT, the basic axiom is the commutator $[\phi(x),\pi(y)] = i\delta^3(x-y)$, from which the generalized fourier transform with $\phi$ and $\pi$ as the conjugate fields should follow. I'm trying to see how the generalized fourier integral $\exp(i\int d^3x \pi(x)\phi(x)$ follows from the commutator axiom. $\endgroup$
    – Angela
    Commented Feb 14, 2022 at 4:51
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Feb 14, 2022 at 9:17
  • $\begingroup$ The discretization is done twice, for both the position and momentum domain, which makes a finite dimensional representation. How does one know that the limit to the continuous representation is valid? $\endgroup$ Commented Feb 14, 2022 at 10:53
  • $\begingroup$ @Qmechanic , All : thanks it helps. $\endgroup$
    – Angela
    Commented Feb 14, 2022 at 15:59
3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ to do the exact same proof, one would need i$\frac{\delta |\phi \rangle}{\delta \phi(x)} = \pi(x)|\phi \rangle$. How would one arrive at this relation? In fact, I can express the schroedinger operator $\hat{\phi}(x,0)$ as a sum of creation and annihilation operators. I believe the creation operator does not have any normalizable eigenvector. In that case, what does the eigenket $|\phi \rangle$ mean? How does it exist? $\endgroup$
    – Angela
    Commented Feb 10, 2022 at 13:45
  • $\begingroup$ How do you derive the same thing in QM? Literally the same derivation goes over to QFT. What does the eigenket $| x \rangle$ mean in QM? It is not normalizable either. Why is still OK to use that state? $\endgroup$
    – Prahar
    Commented Feb 10, 2022 at 14:38
  • $\begingroup$ I also hope you wouldn't mind me saying that it seems to me that the questions you have so far can all be answered by learning QM a bit more carefully and I really suggest you do that first. There are far too many other confusing things in QFT. $\endgroup$
    – Prahar
    Commented Feb 10, 2022 at 14:40
  • $\begingroup$ The derivation relies on the assumption that the momentum operator can be expressed as the derivative with respect to $x$. However, this assumption is already based on the Fourier relationship between the bases. Hence, the argument for the derivation is somewhat circular. $\endgroup$ Commented Feb 14, 2022 at 10:50
0
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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