In Peskin p.282, it is said "The general functional integral formula (9.12) derived in the last section holds for any quantum system, so it should hold for a quantum field theory." (9.12) is the formula
$U(q_a,q_b,T)=(\prod_i\int Dq(t)Dp(t))\exp[i\int_0^T dt(\sum_ip^i\dot{q}^i-H(q,p)]$

In deriving this formula, the identity operators $\int dq|q\rangle\langle q|$ and $\int dp|p\rangle\langle p|$ are inserted, and the equalities $\langle q|q'\rangle=\delta(q-q')$, $\langle q|p\rangle=e^{ipq}$ and $\int dp e^{ip(q-q')}=2\pi \delta (q-q')$ are used.

In a parallel derivation in Klein-Gordon field, I should first assume that there exist eigenstates of both the field operator $\hat \phi(\boldsymbol r)$ and the momentum operator $\hat \pi(\boldsymbol r)$:
$\hat \phi(\boldsymbol r)|\phi\rangle = \phi(\boldsymbol r)|\phi\rangle $
$\hat \pi(\boldsymbol r)|\phi\rangle = \pi(\boldsymbol r)|\phi\rangle $
and insert the identity operators $\int D\phi|\phi\rangle\langle \phi|$ and $\int D\pi|\pi\rangle\langle \pi|$ into $\langle \phi_b|e^{-iHT}|\phi_a\rangle$, and use the equalities $\langle \phi|\phi'\rangle=\delta(\phi-\phi')$, $\langle \phi|\pi\rangle=e^{i\int d^3\boldsymbol r\pi(\boldsymbol r)\phi(\boldsymbol r)}$ and $\int D\pi e^{i\int d^3\boldsymbol r \pi(\boldsymbol r)(\phi(\boldsymbol r)-\phi'(\boldsymbol r))}=2\pi \delta (\phi-\phi')$. There seems to be no problem here.

Now consider the same procedure for a Schrodinger boson field (QFT itself does not require relativity). I need the eigenstates $|\psi\rangle$ of the firld operator $\hat \psi(\boldsymbol r)$:
$\hat \psi(\boldsymbol r)|\psi\rangle = \psi(\boldsymbol r)|\psi\rangle$
and also the orthogonality relation $\langle\psi|\psi'\rangle=\delta(\psi-\psi')$. So by inserting the identity operator I get a representation of $\hat \psi (\boldsymbol r)$:
$\hat \psi (\boldsymbol r)=\int D\psi^*D\psi\hat\psi(\boldsymbol r)|\psi\rangle\langle\psi|=\int D\psi^*D\psi\psi(\boldsymbol r)|\psi\rangle\langle\psi|$
Make Hermitian conjugation,
$\hat \psi^\dagger (\boldsymbol r)=\int D\psi^*D\psi\psi^*(\boldsymbol r)|\psi\rangle\langle\psi|$
Here comes the problem: with these two equalities, I will always get
$[\psi(\boldsymbol r), \psi^\dagger(\boldsymbol r')]=\int D\psi^*D\psi(\psi(\boldsymbol r)\psi^*(\boldsymbol r')-\psi^*(\boldsymbol r')\psi(\boldsymbol r))|\psi\rangle\langle\psi|=0$
in contradiction to the quantization condition $[\psi(\boldsymbol r), \psi^\dagger(\boldsymbol r')]=i\delta(\boldsymbol r-\boldsymbol r')$

What's wrong with that? If some of the above conditions should be relaxed, then how to derive the corresponding path integral formula for a quantum field?


1 Answer 1


The Schrödinger field $\psi$ is not Hermitian, it's an annihilation operator, so it does not necessarily have a basis of eigenstates, and the eigenstates that it does have are field versions of coherent states. Since nothing in your question really depends on the field-theoretical aspect, I will in the following just talk about ordinary creation/annihilation operators $a,a^\dagger$ and their coherent states $\lvert z \rangle$.

The resolution of the identity in terms of coherent states $\lvert z\rangle$ (with complex eigenvalues!) is not given by $\int \mathrm{d}z\mathrm{d}z^\ast \lvert z\rangle\langle z\rvert$ but by $$ \mathbf{1} = \int \frac{\mathrm{d}z\mathrm{d}z^\ast}{2\pi\mathrm{i}}\mathrm{e}^{\lvert z\rvert^2}\lvert z \rangle\langle z\rvert.$$ Furthermore, the set of coherent states is overcomplete - it is not a basis. None of this is the source of the error, but it bears mention nevertheless.

Your particular error lies in writing

$$[\psi(\boldsymbol r), \psi^\dagger(\boldsymbol r')]=\int D\psi^*D\psi(\psi(\boldsymbol r)\psi^*(\boldsymbol r')-\psi^*(\boldsymbol r')\psi(\boldsymbol r))|\psi\rangle\langle\psi|, $$

this is not true, as I will demonstrate for the analogous case of $a,a^\dagger$: $$ [a,a^\dagger] = \int \frac{\mathrm{d}z\mathrm{d}z^\ast}{2\pi\mathrm{i}}\mathrm{e}^{\lvert z\rvert^2}a \lvert z \rangle\langle z\rvert a^\dagger - \int \frac{\mathrm{d}z\mathrm{d}z^\ast}{2\pi\mathrm{i}}\mathrm{e}^{\lvert z\rvert^2}a^\dagger \lvert z \rangle\langle z\rvert a,$$ but $a^\dagger \lvert z\rangle \neq z^\ast \lvert z\rangle$. There is just no reason for that to be true, and in fact, it is not. Considering that $[a,a^\dagger]$ is pretty much the same relation as the standard relation of $[x,p]$, we should much rather expect that if $a$ acts as multiplication on $\lvert z\rangle$, then $a^\dagger$ acts by differentiation!

In any case, the crucial point is that you cannot derive the $=0$ in this way. The coherent state path integral does not contradict its own quantization assumption.

  • $\begingroup$ ♦: Thank you for answering! What you really mean is to relax the condition $\langle\psi|\psi'\rangle=\delta(\psi-\psi')$ in my derivation, so $\hat \psi(\boldsymbol r)|\psi\rangle = \psi(\boldsymbol r)|\psi\rangle$ does not imply $\hat \psi^\dagger(\boldsymbol r)|\psi\rangle = \psi(\boldsymbol r)^*|\psi\rangle$. Then comes the next problem: how to derive the path integral of Schrodinger field without this orthogonality relation? $\endgroup$
    – StupidBird
    Apr 27, 2017 at 12:30
  • $\begingroup$ @StupidBird Indeed, we have $\langle z \vert z'\rangle = \mathrm{e}^{z^\ast z'}\neq 1$. As for how the coherent state path integral works, that's a different question, and I'd just advise you to search for that term with your favourite search engine, there's plenty of explanations for that out there. $\endgroup$
    – ACuriousMind
    Apr 27, 2017 at 12:38

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