The reason that I have this problem is that I'm trying to solve problem 14.4 of Schwartz's QFT book, which've confused me for a long time.

The problem is to construct the eigenstates of a quantum field $\hat{\phi}(\vec{x})$, such that $$ \hat{\phi}\left(\vec{x}\right)\left|\Phi\right\rangle =\Phi\left(\vec{x}\right)\left|\Phi\right\rangle . $$ I think the eigenstate should be

$$ \left|\Phi\right\rangle =e^{-\int d^{3}x\frac{1}{2}(\Phi(\vec{x})-\hat{\phi}_{+}(\vec{x}))^{2}}\left|0\right\rangle $$ where $\hat{\phi}_{+}\left(\vec{x}\right)$ is the part of $\hat{\phi}\left(\vec{x}\right)$ that only includes creation operators. I haven't found any book talking about this problem so I'm actually not sure whether this is the correct result.

And similarly the eigenstate of $\hat{\pi}\left(\vec{x}\right)$ such that $$ \hat{\pi}\left(\vec{x}\right)\left|\Pi\right\rangle =\Pi\left(\vec{x}\right)\left|\Pi\right\rangle $$ should be $$ \left|\Pi\right\rangle =e^{\int d^{3}x\frac{1}{2}(\Pi(\vec{x})+\hat{\phi}_{+}(\vec{x}))^{2}}\left|0\right\rangle . $$ Equation 14.21 and Equation 14.22 of Schwartz's book are $$ \left\langle \Pi|\Phi\right\rangle =\exp\left[-i\int d^{3}x\Pi\left(\vec{x}\right)\Phi\left(\vec{x}\right)\right] $$ $$ \left\langle \Phi'|\Phi\right\rangle =\int\mathcal{D}\Pi\left\langle \Phi'|\Pi\right\rangle \left\langle \Pi|\Phi\right\rangle =\int\mathcal{D}\Pi\exp\left(-i\int d^{3}x\Pi\left(\vec{x}\right)\left[\Phi\left(\vec{x}\right)-\Phi'\left(\vec{x}\right)\right]\right) $$ Now my question is how do I verify these relations? I've wasted a lot of time on this problem without any success.

And why doesn't any book talk about this problem? I mean the eigenstate of a quantum field? Isn't this kind of stuff very fundamental to QFT?

  • 1
    $\begingroup$ you will find some useful references in physics.stackexchange.com/a/231309/84967 Also, check out Weinberg's book on QFT, Vol I., chapter 9. $\endgroup$ Commented Feb 14, 2017 at 12:42
  • $\begingroup$ @AccidentalFourierTransform Weinberg only assume that such states exist and then he didn't talk about what they are or how to construct them. $\endgroup$
    – Nahc
    Commented Feb 14, 2017 at 15:36
  • $\begingroup$ The problem is formally identical to your 312004, and it is certainly easier to explain on the dumb oscillator, rather than perorating on quantum fields. Stringers and squeezed lighters have it for breakfast. Show that you recognize the SU(1,1) involved. $\endgroup$ Commented Feb 16, 2017 at 22:54
  • $\begingroup$ @Nahc: References: lss.fnal.gov/archive/other/nr-1-1995.pdf, Greiner - Field Quantization - page no. 365. $\endgroup$
    – rainman
    Commented Jul 31, 2017 at 3:50
  • $\begingroup$ you can interpret this language in an "intuitive" way if you have this "dictionary": physics.stackexchange.com/a/694149/226902 $\endgroup$
    – Quillo
    Commented Feb 14, 2022 at 9:49

4 Answers 4


So here is the thing. I post here my answer to 312004, since that one is closed; while also insisting this one here has its answer in one oscillator: as usual in QFT, the infinity of oscillators is only a smokescreen to test the student's understanding of the second quantization notation, just part b) of your text's problem. In an ideal world, this FT question above should be merged with the closed one, parts a) and c).

And no, field eigenstates are not an integral part of the QFT toolbox, given the output expected of QFT, at least in particle physics, the focus of your text. Nevertheless, Schroedinger wave functional theory (Jackiw, 1988 and Lüscher, 1985) still floats around.

(Coherent states are more commonly used, e.g. Itzykson and Zuber's QFT text, eqn (3-65) et seq.)

Essentially answering 312004 :

This is the obverse/completion of the twin question 292899. It hinges on a point not uncommon in squeezed-states and string discussions. I’ll just race to the answer suitably tweaking standard techniques, e.g., of Fischer, Nieto, & Sandberg, 1984, and leaving you to patch up numerical normalizations to your satisfaction.

Recall $[a , a^\dagger ]=1$. $$ |x\rangle\equiv n(x) ~ \exp\left ( -(a^\dagger -x\sqrt{2})^2 /2  \right )|0\rangle   \Longrightarrow  \hat{~x}~|x\rangle=x|x\rangle,\\ |p\rangle\equiv n(p) ~ \exp((a^\dagger +ip\sqrt{2})^2 /2   ) |0\rangle  \Longrightarrow  \hat{~p}~|p\rangle=p|p\rangle ~. $$ You may try to fix $n(x)=e^{x^2/2}/\pi^{1/4}$ from $\langle 0|x\rangle$, the Schrödinger ground state of the oscillator. (Yes, indeed, n is the inverse of the Gaussian!)

Your assignment is to normal-order $$ \langle x|p\rangle= n(x) n(p) \langle 0| e^{-(a-x\sqrt{2})^2/2}      e^{(a^\dagger +ip\sqrt{2})^2/2}   |0\rangle~. $$ That is to say, commute all $a$s to the right, where they annihilate the vacuum, and so disappear themselves, leaving the x’s, etc, and the same for the creation operators on the vacuum to the left, depositing their ps.

Actually, given this task, it is easier to normal order, instead, the new oscillators $$ b\equiv a-x\sqrt{2},  \qquad b^\dagger\equiv a^\dagger +ip\sqrt{2}, \qquad [b,b^\dagger]=1. $$ Thus, $$ b|0\rangle= -x\sqrt{2} |0\rangle, \qquad   \langle0|b^\dagger=\langle 0| ip\sqrt{2} , $$ so that the normal-ordered expression will net <0|0> times a function of x cancelling n(x), same for p, and the $e^{ixp}$ sought.

The answer (derived by standard moves in the Appendix below) is $$ e^{-b^2/2}    e^{b^{\dagger 2}/2}= e^{b^{\dagger 2}/4 }e^{-\ln 2 (1/2+b^\dagger b) }e^{-b^2/4}=\\ e^{b^{\dagger 2}/4 } \left ( \sum^\infty_{k=0}        \frac{(-1/2)^k}{k!} b^{\dagger k} b^k \right )e^{-b^2/4}   /\sqrt{2} . $$ Consequently, the v.e.v. of the above operator is $$ \langle p |x\rangle= e^{-p^2/2}  e^{ixp}e^{-x^2/2} n(x) n(p) /\sqrt{2\pi}= e^{ixp}/ \sqrt{2\pi}. $$

To prove orthogonality, e.g. of the position states, well, insert a complete set of momentum eigenstates in $\langle x|y\rangle$, and do the dp integral of the two plane waves you just found, netting you a  $\delta(x-y)$.

Appendix.  The crucial lemmata in the derivation are:

  • Defining $$ L_-\equiv -b^2/2, \qquad L_+ \equiv b^{\dagger 2 }/2,  \qquad L_0\equiv \tfrac{1}{2} ( 1/2 +b^\dagger b), $$ note they close into the algebra of SU(1,1) (underlying the Virasoro algebra),
    $$ [L_+, L_-]=2 L_0, \qquad [L_0, L_{\pm} ]=\pm L_{\pm}. $$

  • The key point: A group element identity (product of exponentials of generators) holds for all representations, but, conversely, an identity of such group elements in a faithful representation, such as the doublet (the Pauli matrices), cannot hold if the generic abstract one does not. (This is nontrivial: it requires Poincaré's exponential theorem, to the effect that the CBH expression found in the exponent is fully in the Lie algebra, and hence the representation is immaterial.) In this simplest rep, then, $$ L_+=  \begin{pmatrix}      0&1\\      0&0    \end{pmatrix} , \qquad      L_-=  \begin{pmatrix}      0&0\\      1&0    \end{pmatrix} , \qquad     2L_0=  \begin{pmatrix}      1&0\\      0&-1    \end{pmatrix} .      $$ It is then evident that $e^{L_-} e^{L_+} =e^{L_+/2} e^{-\ln 2 \cdot 2L_0} e^{L_-/2}  $ , since $$  \begin{pmatrix}      1&0\\      1&1    \end{pmatrix}   \begin{pmatrix}      1&1\\      0&1    \end{pmatrix}  =   \begin{pmatrix}      1&1/2\\      0&1    \end{pmatrix}    \begin{pmatrix}      1/2&0\\      0&2    \end{pmatrix}    \begin{pmatrix}      1&0\\      1/2&1    \end{pmatrix}  . $$

  • However, the middle group element, $ e^{-\ln 2 ~(1/2+b^\dagger b)} $, is not normal ordered yet, but straightforward to get there, given the identity $$ \bbox[yellow,5px]{e^{c b^\dagger b}= \sum^\infty_{k=0}    \frac{(e^c-1)^k}{k!} b^{\dagger k} b^k }. $$ (Just write out the first four powers of c in the Wick theorem ascending expansion in normal-ordered bs.  This is attributed to Wilcox (1967) JMP 8 962-982, and ultimately McCoy, 1932, but it is all but manifest. The oscillator propagator.)

The shortest way to see it is by exploiting the algebraic isomorphism $b^\dagger =x, ~~b=\partial$; so that the left-hand side acting on f(x) is just scaling, $\exp(c x\partial) ~f(x)= f(e^c x)$; while the right-hand one is the Taylor series around x, shifted by $x(e^c-1)$, and hence $f(x)\mapsto f(x+x(e^c-1))=f(e^c x)$.

Back to field theory: all you need to do now is consider an infinity of oscillators, whence, loosely, $\hat{x}\mapsto \hat{\phi}(x),\quad \hat{p}\mapsto \hat{\pi}(x)$, construct the eigenstates as in 292899, and generalize the above to $\langle \Pi|\Phi\rangle $, mutatis mutandis. As a straightforward consequence, observe the translationally invariant state in the kernel of the momentum is $$ \left|\Pi=0\right\rangle =e^{\int d^{3}x ~ \hat{\phi}_{+} (\vec{x})^{2}/2}\left|0 \right\rangle . $$

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    $\begingroup$ I tried doing the problem mutatis mutantdis. I think it is not as trivial as it sounds. The $1/E_p$ in the lorentz invariant integration measure leads to issues with doing the problem analogously, and the only solution for the field eigenstate had a product of operators in the exponent, which left me clueless how to prove the orthogonality relation. $\endgroup$ Commented Sep 1, 2019 at 19:04
  • $\begingroup$ Yes, the normalizations and the altered commutation relations of the modes is part of the mutandis, of course. You may track ti down in your text to reassure yourself how it mutates! As of separate modes, a product of exponentials yields an exponential of the sum of the respective exponents, no? $\endgroup$ Commented Sep 1, 2019 at 19:42
  • $\begingroup$ Do you have a reference for this? I'm not aware of any text where this is done and I don't know of a solutions manual for the Schwartz QFT book. Splitting the integrals into infinite products is an interesting idea I haven't tried yet, and will if I get a moment to. $\endgroup$ Commented Sep 1, 2019 at 21:17
  • $\begingroup$ Sorry, no. Lowell Brown’s book is very tasteful... $\endgroup$ Commented Sep 1, 2019 at 21:18
  • $\begingroup$ I hope this is not homework. You have checked this one and links thereof, no? $\endgroup$ Commented Feb 27, 2020 at 21:45

My solution refers to answers by Cosmas Zachos at this question and 292899. I will adopt notions by Swhartz's QFT book. Besides, finishing problem 14.3 will be helpful to solve this problem.

a) $$ |x\rangle = \frac{1}{\pi^{1/4}}\exp(x^2/2) \exp \left(-\left(a^{\dagger}-x \sqrt{2}\right)^{2} / 2\right)|0\rangle\\ |p\rangle = \frac{1}{\pi^{1/4}}\exp(p^2/2) \exp \left(\left(a^{\dagger}+i p \sqrt{2}\right)^{2} / 2\right)|0\rangle $$ b) Introducing convention to describe the "inner product" of field operators: $$ <\hat{A}(\vec{x}),\hat{B}(\vec{y})>_{f(\vec{x},\vec{y})}:=\int d^3x\,d^3y\,\hat{A}(\vec{x})\hat{B}(\vec{y})f(\vec{x},\vec{y}) $$ where $f$ is a function. Then the eigenstates of $\hat{\phi}(\vec{x}),\ \hat{\pi}(\vec{x})$ are: $$ |\Phi\rangle = \mathcal{N}\exp(<\Phi,\Phi>_{\mathcal{E}}/2) \exp \left(-<\hat{\phi}_+-\Phi,\hat{\phi}_+-\Phi>_{\mathcal{E}}\right)|0\rangle\\ |\Pi\rangle = \mathcal{N}\exp(<\Pi,\Pi>_{\mathcal{E}^{-1}}/2) \exp \left(-<\hat{\pi}_+-\Pi,\hat{\pi}_+-\Pi>_{\mathcal{E}^{-1}}\right)|0\rangle $$ where $\mathcal{N}$ is a normalization number, and $$ \mathcal{E}(\vec{x},\vec{y})=\int \frac{d^3 p}{(2\pi)^3}\omega_pe^{i\vec{p}\cdot(\vec{x}-\vec{y})}\\ \mathcal{E}^{-1}(\vec{x},\vec{y})=\int \frac{d^3 p}{(2\pi)^3}\frac{1}{\omega_p}e^{i\vec{p}\cdot(\vec{x}-\vec{y})} $$ To check this answers, first compute $$ [\hat{\phi}_-(\vec{z}),F(<\hat{\phi}_+-\Phi,\hat{\phi}_+-\Phi>_{\mathcal{E}})]=\frac{\delta F}{\delta<\hat{\phi}_+-\Phi,\hat{\phi}_+-\Phi>_{\mathcal{E}}}(\hat{\phi}_+(\vec{z})-\Phi(\vec{z})) $$ Then $$ \begin{align} \hat{\phi}(\vec{z})|\Phi\rangle &=\hat{\phi}_+(\vec{z})|\Phi\rangle+\hat{\phi}_-(\vec{z})|\Phi\rangle\\ &=\hat{\phi}_+(\vec{z})|\Phi\rangle+[\hat{\phi}_-(\vec{z}),\mathcal{N}\exp(<\Phi,\Phi>_{\mathcal{E}}/2) \exp \left(-<\hat{\phi}_+-\Phi,\hat{\phi}_+-\Phi>_{\mathcal{E}}\right)]|0\rangle\\ &=\hat{\phi}_+(\vec{z})|\Phi\rangle-\mathcal{N}\exp(<\Phi,\Phi>_{\mathcal{E}}/2)\exp \left(-<\hat{\phi}_+-\Phi,\hat{\phi}_+-\Phi>_{\mathcal{E}}\right)(\hat{\phi}_+(\vec{z})-\Phi(\vec{z}))|\Phi\rangle\\ &=\Phi(\vec{z})|\Phi\rangle \end{align} $$ Another eigenstate can be checked similarly.


Define argument of exponential as $$ \hat{L}_+:=<\hat{\phi}_--\Phi,\hat{\phi}_--\Phi>_{\mathcal{E}}\\ \hat{L}_-:=<\hat{\pi}_+-\Pi,\hat{\pi}_+-\Pi>_{\mathcal{E}^{-1}} $$ compute their commutator $$ [\hat{L}_+,\hat{L}_-]=i\int d^3x\,\{(\hat{\phi}_--\Phi),(\hat{\pi}_+-\Pi)\}(\vec{x})=:2\hat{L}_0 $$ NOTE $\{,\}$ is anti-commutator. Then $$ [\hat{L}_0,\hat{L}_{\pm }]=\pm\hat{L}_{\pm } $$ so we get Lie algebra $\mathfrak{su}(2)=\mathrm{span}_{\mathbb{C}}\{\hat{L}_0,\hat{L}_+,\hat{L}_-\}$.

According to Cosmas Zachos' answer above, we have $$ e^{-\hat{L}_+}e^{-\hat{L}_-}=e^{-\hat{L}_-/2}e^{2ln2\,\hat{L}_0}e^{-\hat{L}_+/2}\\ \exp[\frac{2}{i}\lambda<\hat{\pi}_+-\Phi,\hat{\phi}_--\Phi>_1]=\mathrm{N}\{\exp[\frac{2}{i}(e^\lambda-1)<\hat{\pi}_+-\Pi,\hat{\phi}_--\Phi>_1]\} $$ where $\mathrm{N}$ is regular order operator to arrange $(\hat{\pi}_+-\Pi)$ left to $(\hat{\phi}_--\Phi)$.

Finally, $$ \begin{align} \langle\Phi|\Pi\rangle&=\mathcal{N}^2\exp(<\Phi,\Phi>_{\mathcal{E}}/2)\exp(<\Pi,\Pi>_{\mathcal{E}^{-1}}/2)\langle0|e^{-\hat{L}_+}e^{-\hat{L}_-}|0\rangle\\ &=\mathcal{N}^2\langle0|e^{2ln2\,\hat{L}_0}|0\rangle\\ &=\mathcal{N}^2\langle0|\exp\{ln2\,i\int d^3x[\frac{i}{2}\delta^3(0)+2<\hat{\pi}_+-\Phi,\hat{\phi}_--\Phi>_1]|0\rangle\\ &=\mathcal{N}^2\exp\left(-\frac{1n2}{2}\int d^3x\,\delta^3(0)\right)\langle0|\mathrm{N}\{\exp[\frac{2}{i}(e^{ln(1/2)}-1)<\hat{\pi}_+-\Pi,\hat{\phi}_--\Phi>_1]\}|0\rangle\\\\ &=\exp[i\int d^3x\,\Pi(\vec{x})\Phi(\vec{x})] \end{align} $$ where we set $\mathcal{N}=\exp\left(\frac{1n2}{4}\int d^3x\,\delta^3(0)\right)$.


I also tried exercise 14.4 in Schwartz's book. (I use the same conventions and notations as in the book) My guess for the eigenstate reads $$ |\Phi\rangle = \exp\Big{[}-\int d^3x \int d^3y (\Phi(x)-\hat{\phi}_+(x))\mathcal{E}(x,y)(\Phi(y)-\hat{\phi}_+(y))\Big{]}| 0 \rangle $$ where $$ \mathcal{E}(x,y)= \int\frac{d^3p}{(2\pi)^3} e^{ip(x-y)}\omega_p. $$ I could have made some error in calculation but I think the Ansatz should be correct. Likewise for momentum eigenstates I get $$ |\Pi\rangle = \exp\Big{[}-\int d^3x \int d^3y (\Pi(x)-\hat{\pi}_+(x))\mathcal{E}^{-1}(x,y)(\Pi(y)-\hat{\pi}_+(y))\Big{]}| 0 \rangle $$ where $$ \mathcal{E}^{-1}(x,y) = \int\frac{d^3p}{(2\pi)^3} e^{ip(x-y)}\frac{1}{\omega_p} $$ is just the inverse of $\mathcal{E}$. However I still have problems verifying $$ \langle \Phi | \Pi \rangle = \exp\Big{[}-i\int d^3x \Phi(x) \Pi(x) \Big{]}. $$ (This and the completeness relation immediately implies the orthogonality). I want to use https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula. If I define $A$ as the daggered exponent in $|\Phi\rangle$ and $B$ as the exponent in $|\Pi \rangle$. It get that $$ [A,B] = -2i\int d^3x \Big{[} (\Phi- \hat{\phi}_-) (\Pi - \hat{\pi}_+) + (\Pi - \hat{\pi}_+)(\Phi- \hat{\phi}_-) \Big{]}. $$ Notice that $$ \langle 0 | [A,B] | 0 \rangle = -4i \int d^3x \Pi(x) \Phi(x) + \text{ infinite constant} $$ so I am almost there (factors $e^A$ and $e^B$ become scalars when sandwiched between vaccum states and can be pulled out and absorbed by normalization). However in the Baker-Campbell-Hausdorff formula there appear more terms in the exponent. As far as I can see these terms (after being sandwiched) either can be absorbed by normalization or are proportional to $[A,B]$ (since $[A,[A,B]] \propto A,~[B,[A,B]] \propto B$), so that only leaves me with a descripancy of an unknown prefactor in the exponential (maybe after the summation of the infinite series the prefactor becomes $1$ as stated?).
It is also probable that I have made a mistake and I would be thankful if one could point it out and/or show me the solution.

  • $\begingroup$ Hi, why did you include the $\mathcal{E}$ in your ansatz? $\endgroup$ Commented Jan 19, 2021 at 15:58
  • $\begingroup$ @TheQuantumMan There has been some time since I did this calculation. If I recall correctly one can also $\mathcal{E}(x,y) = \delta(x-y)$ and it is still an eigenfunction. But I found this form more convenient for showing the other relations (if they even fulfill them?). Also I remember that I saw Eqs (14.65) and (14.66) in Schwartz, which suggest more the form that I have. I also remember that I was quite unsatisfied with my solution, so if you know a solution, let me know :) $\endgroup$
    – jkb1603
    Commented Jan 19, 2021 at 16:21
  • $\begingroup$ As of now I'm unsatisfied with my own solution as well! If I have anything worthwhile to share, I'll leave a comment :) $\endgroup$ Commented Jan 20, 2021 at 9:06
  • 1
    $\begingroup$ I have verified everything. The only difference in my calculation is $[A,B] = i \int \dots $ instead of $[A,B] = -2i \int \dots $. BTW, $\mathcal{E}(x,y)=\delta(x-y)$ would not work, because we need to cancel terms like $\sqrt{\omega_p/2}$ or $1/\sqrt{2\omega_p}$ in $\hat{\pi}$ or $\hat{\phi}$. $\endgroup$ Commented Aug 3, 2022 at 2:18

The problem essentially boils down to the commutation of complicated operators. If a state is represented by $$ |\psi\rangle = \exp(f(\hat{a},\hat{a}^{\dagger})) |\text{vac}\rangle \equiv \hat{K} |\text{vac}\rangle , $$ where $f(\hat{a},\hat{a}^{\dagger})$ is some function of the ladder operators, then the generic inner product is of the form $$ \langle\psi_1|\psi_2\rangle = \langle\text{vac}|\hat{K}_1^{\dagger} \hat{K}_2 |\text{vac}\rangle . $$ To solve the problem one needs to get the product $\hat{K}_1^{\dagger} \hat{K}_2$ into normal order. Hence the complicated commutation.

There is a general procedure to compute complicated commutations of operators, provided the operators can be represented in exponential form with arguments that are most second order in the ladder operators. The problem in the current situation can be expressed as $$ \exp(f_1(\hat{a},\hat{a}^{\dagger})) \exp(f_2(\hat{a},\hat{a}^{\dagger})) = A \exp(g_1 \hat{a}^2) \exp(g_2\hat{a}^{\dagger}\hat{a}) \exp(g_3\hat{a}^{\dagger 2}) , $$ where $A$, $g_1$, $g_2$, and $g_3$ are the unknowns to be solved. Note that the second operator on the right-hand side with the number operators is not in normal order, but since the vacuum is an eigenstate of the number operators, this is not a problem.

The general procedure is now to insert a dummy variable into the arguments of the exponents and make all the unknowns functions of this dummy variable: $$ \exp(t f_1(\hat{a},\hat{a}^{\dagger})) \exp(t f_2(\hat{a},\hat{a}^{\dagger})) = A(t) \exp(g_1(t) \hat{a}^2) \exp(g_2(t)\hat{a}^{\dagger}\hat{a}) \exp(g_3(t) \hat{a}^{\dagger 2}) , $$ where $t$ is the dummy variable. Then compute the derivative of the equation with respect to $t$. Next you remove as many of the operators by apply the respective adjoint operators. Then you use the identity for $\exp(\hat{X})\hat{Y}\exp(-\hat{X})$ to simplify the expression. The result can be separated into a number of differential equations that can be solved to get solutions for the unknowns.

I have worked out this problem in full for the eigenstates of the quadrature operators in an article: Phys. Rev. A 98, 043841 (2018) - arXiv:1810.04396. Also see the Erratum: Phys. Rev. A 101, 019903(E) (2020). The analysis provides expressions with which one can obtain the result for all the required inner products.


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