# What is the overlap $\langle \phi | 0 \rangle$ for a scalar field?

Consider a massive free real scalar field $$\hat{\Phi}$$ (with $$\mathcal{L}[\Phi] = \partial_{\mu}\Phi\partial^\mu \Phi - \tfrac{1}{2} m^2 \Phi^2$$). I was wondering what is the overlap for the following states? $$\langle \phi | 0 \rangle$$ where $$|0\rangle$$ is the vacuum state, and $$| \phi \rangle$$ is the eigenstate for the operator $$\hat{\Phi}$$ where $$\hat{\Phi}(\mathbf{x}) | \phi \rangle = \phi(\mathbf{x}) | \phi \rangle$$ where $$\hat{\Phi}(\mathbf{x})$$ is the Schrodinger picture field operator.

I have read online that $$| \phi \rangle$$ is supposed to be a coherent state, so I am thinking like something along the lines of $$| \phi \rangle = \exp\left(i\int d^3\mathbf{y}\ \phi(\mathbf{y}) \hat{\Pi}(\mathbf{y})\right) | 0 \rangle$$ where $$\hat{\Pi} = \partial_0 \hat{\Phi}$$ is the conjugate momentum operator to the field. But I am not sure of this, and even if this is true I am unsure of how to use this result to evaluate $$\langle \phi | 0 \rangle$$.

• I'm almost positive this is calculated in Weinberg, Vol.1, chapter 9. – AccidentalFourierTransform Nov 14 '18 at 4:48
• Related. The state you wrote is an eigenstate of the annihilator part of the field, Only. – Cosmas Zachos Nov 14 '18 at 22:36
• – Cosmas Zachos Nov 15 '18 at 0:45
• Indeed, the overlap is in Weinberg v1, (9.2.9), as @AccidentalFourierTransform points out, but it is for the real McCoy, and not for the coherent state the OP picked up on the internet. That state is not an eigenstate of the full field. So the OP is mixing apples and oranges, a distinction parsed out in linked questions and links thereto. – Cosmas Zachos Nov 15 '18 at 15:27

However, if the (free) field operator is written in terms of creation and annilation operators in the usual way, then we can define an eigenstate of the annihilation-operator part of the (smeared) field operator, and this is a coherent state. This is analogous to the usual single-mode coherent states $$|\alpha\rangle\equiv\exp(\alpha a^\dagger)|0\rangle \tag{1}$$ where $$\alpha$$ is a complex number. In the free-scalar-field case, we have $$|\alpha\rangle \equiv \exp\left(\int \frac{d^3p}{(2\pi)^3}\ \alpha(\mathbf{p})a^\dagger(\mathbf{p})\right) |0\rangle \tag{2}$$ instead, where $$\alpha(\mathbf{p})$$ is a suitable complex-valued function. This is an eigenstate of the smeared annihilation operator $$a(\beta)\equiv\int \frac{d^3p}{(2\pi)^3}\ \beta^*(\mathbf{p})a(\mathbf{p}). \tag{3}$$ Use $$\big[a(\mathbf{p}),\,a^\dagger(\mathbf{q})\big] =(2\pi)^3\delta(\mathbf{p}-\mathbf{q}) \hskip2cm a(\mathbf{p})|0\rangle=0. \tag{4}$$ to get $$a(\beta)\,|\alpha\rangle = \left(\frac{d^3p}{(2\pi)^3}\ \beta^*(\mathbf{p})\alpha(\mathbf{p})\right)\,|\alpha\rangle. \tag{5}$$ The inner product $$\langle\alpha|0\rangle$$ can be calculated by adapting the techniques used in the single-mode case.
By the way, notice that the annihilation-operator part of $$\phi(\mathbf{x})$$ is a non-local operator. If we write $$\phi^\pm(\mathbf{x})$$ for the annihilation- and creation-parts, then the commutator of $$\phi^-(\mathbf{x})$$ with $$\phi^+(\mathbf{y})$$ is non-zero even when $$\mathbf{x}\neq \mathbf{y}$$. This is unavoidable, because the Reeh-Schlieder theorem implies that an operator that is strictly localized in any bounded region of spacetime cannot annihilate the vacuum state (section 2 in "Notes on Some Entanglement Properties of Quantum Field Theory", http://arxiv.org/abs/1803.04993).