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
    Commented 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
    Commented Apr 27, 2017 at 12:38

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.