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In Peskin & Schroeder section 9.2, they derive the two-point function in the path integral formalism:

$$\langle \Omega | \mathcal{T} \left\{ \hat{\phi}(x_1)\hat{\phi}(x_2)\right\} | \Omega \rangle = \frac{\int \mathcal{D}\phi \ \phi(x_1) \phi(x_2) e^{ i\int d^4 x\ \mathcal{L} }}{\int \mathcal{D}\phi \ e^{ i\int d^4 x\ \mathcal{L} }}. $$

The trick to derive this is to insert the identity

$$1 = \int \mathcal{D}\phi\ |\phi\rangle \langle \phi|$$

between the operators $\hat{\phi}(x_i)$. Then we can change the operator for regular functions using:

$$ \hat{\phi}(x_i) |\phi_i\rangle = \phi(x_i) |\phi_i\rangle.$$

My first question is: what are the states that form the complete orthogonal basis $|\phi\rangle$? The authors never seem to sepecify it. It cannot be just any complete orthogonal basis since these states seem to be eigenstates of the field operator

$$\hat{\phi}(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}}(\hat{a}_p e^{i p\cdot x} + \hat{a}_p^\dagger e^{-i p\cdot x}).$$ From what I understand, the states $|\phi\rangle$ represent all the possible classical field configurations (classical as in well-defined at all points in space at a given time, with no uncertainty) over which we integrate between two boundary states. But I don't see how these classical states are the eigenstates of $\hat{\phi}(x)$. Is there a simple expression for $|\phi\rangle$ in terms of e.g. creation/annihilation operators?

Actually what bothers me is that eigenstates of the field operator are supposed to be coherent states, which form an overcomplete set. Which means that if the $|\phi\rangle$'s are coherent states we cannot write the identity as the combination above since the states are not orthogonal (see section 8.1.3 of this document). My second question is: is it possible tha coherent states might be the eigenstates of another type of "field operator", not the one above? If so, what is this other operator? (Solved: see edit below) Note that in the given link they don't seem to define the operator for which the coherent state is an eigenstate.

(Related: 148200 and 109343. The answer in the first link doesn't really answer the question "what is $|\phi\rangle$?" and the second link mentions coherent states only, which as I mentioned are not orthogonal and therefore cannot be the states used in the derivation by Peskin & Schroeder)

EDIT: As @Mane.andrea suggested in the comments, I checked out section 4.1 of Condensed Matter Field Theory by Atland & Simons. It seems that they define the coherent state as the eigenstate of the annihilation operators $\hat{a}_i$ specifically, i.e. the positive-frequency part of the field above. So the answer to my 2nd question seems to be Yes, coherent states are the eigenstates of a different "field operator".

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    $\begingroup$ Is orthogonality necessary for the proof? Or only completeness? Because if so, having an over-complete basis is not a problem i think. Also, check out Section 4.1 of Altland Simons: Condensed Matter Field Theory. They discuss coherent states. $\endgroup$ – MannyC Jan 27 at 6:32
  • $\begingroup$ Orthogonality is necessary to obtain the standard path integral formulation above. Inserting the identity in terms of coherent states leads to additional factors and what is called the "coherent state path integral", which is important for fermionic fields. See my first link for reference. $\endgroup$ – Jasmeru Jan 27 at 13:52
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    $\begingroup$ Related and further. $\endgroup$ – Cosmas Zachos Jan 27 at 15:22
  • $\begingroup$ The answer to my first question seems to be the result given by the OP in the first link of @CosmasZachos, thanks. Do you want to write an answer so I can accept it? $\endgroup$ – Jasmeru Jan 27 at 19:16
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I think I found a solution with the help of @CosmasZachos's links. The state given by OP here didn't seem to match exactly my definition of the field operator. My guess is because it comes from the Schrödinger wavefunctional formalism. I'm not totally sure about my answer and I haven't found anything like it on the web so if anyone could double-check this, I'd appreciate it.

Let's define our operators carefully:

$$\hat{\phi}(x) = \hat{\phi}_+(x) + \hat{\phi}_-(x),$$

where $$ \hat{\phi}_+(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}}\hat{a}_p e^{-i p\cdot x};\quad \hat{\phi}_-(x) = \left(\hat{\phi}_+(x)\right)^\dagger.$$

It is useful to introduce the momentum:

$$\hat{\pi}(x) = \frac{\partial\hat{\phi}(x)}{\partial t } = \hat{\pi}_+(x)+\hat{\pi}_-(x),$$

where $$ \hat{\pi}_+(x) = -i \int \frac{d^3p}{(2\pi)^3} \sqrt{\frac{E_p}{2}}\hat{a}_p e^{-i p\cdot x};\quad \hat{\pi}_-(x) = \left(\hat{\pi}_+(x)\right)^\dagger.$$

The equal-time canonical commutation relations are:

$$[\hat{\phi}(\vec{x}),\hat{\pi}(\vec{y})] = i \delta^{(3)}(\vec{x}-\vec{y})\qquad [\hat{a}_\vec{p},\hat{a}_\vec{q}^\dagger] = (2\pi)^3 \delta^{(3)}(\vec{p}-\vec{q}).$$

From these we can find: $$ [\hat{\phi}(x)_+,\hat{\pi}_-(y)] = \frac{i}{2}\delta^{(3)}(\vec{x}-\vec{y}); \qquad [\hat{\phi}_+(\vec{x}),\hat{\phi}_-(\vec{y})] = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}e^{i\vec{p}\cdot(\vec{x}-\vec{y})}.$$

Note that the last relation is not a Dirac delta function (it is in fact the propagator of the field), which is precisely why the state given by OP in the link above is not an eigenstate of the field operator defined here.

In fact, through trials and errors, I think I've found a state $|\phi\rangle$ that satisfies $\hat{\phi}(x)|\phi\rangle = \phi(x)|\phi\rangle$ (again, feel free to double-check):

$$|\phi\rangle \equiv \mathcal{N} \exp\left\{i\int d^3y\ \left(\hat{\phi}_-(y) - 2 \phi(y)\right)\hat{\pi}_-(y)\right\}|0\rangle,$$

where $\mathcal{N}$ is a normalization factor (which I haven't computed). I'm not even sure if those states are orthogonal, but at least I know what an eigenstate of the field operator might look like.

Also to answer my 2nd question, I think coherent states are defined as eigenstates of the positive-frequency field operator $\hat{\phi}_+(x)$ only (the part containing the annihilation operators). One should be careful not to confuse the two operators. Also, through my researches, I noticed that some references only deal with coherent states or field eigenstate in a non-relativistic theory or the Schrödinger wavefunctional formalism, where the definitions might be different from the convention of Peskin & Schroeder.

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  • $\begingroup$ Yes, you are essentially on the right track. If you are not using coherent states (exponent linear in fields) but bona-fide eigenstates of the full fields, you need exponents quadratic in the fields. I cannot vouch for your normalizations (and the unconventional $\pm$-inversion of field pieces, but you are "warm", broadly. Section 14.2.3 of M Schwarz's standard book works it all out, especially eqn (14.19-14.22), and footnote: your last paragraph. I thought you were doing his problem 14.4, which is why I sent you to these refs, but you are evidently reconstructing it. $\endgroup$ – Cosmas Zachos Jan 28 at 21:52

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