# Expectation values in a quantum field theory

Consider a quantum field $\phi(x)$ (say, a scalar) at a spacetime location $x$. It turns out that $$\langle\phi(x)\rangle_0\equiv\langle 0|\phi(x)|0\rangle=0.\tag{1}$$ This is easily obtaied by using the Fourier exapansion $$\phi(x)=\int\frac{d^3\textbf{p}}{(2\pi)^{3/2}\sqrt{2E_{\textbf{p}}}}(a_{\textbf{p}}e^{-ip\cdot x}+a^\dagger_{\textbf{p}} e^{ip\cdot x}).\tag{2}$$

Now in quantum mechanics, any expectation value is computed w.r.t a probability distribution $p(x)$. For example, $$\langle x\rangle_\psi\equiv\langle \psi|x|\psi\rangle=\int\limits_{-\infty}^{\infty}xp(x)dx\tag{3}$$ where $p(x)=\psi^*(x)\psi(x)$.

Which probability distribution $p[\phi(x)]$ is used to compute (1)? Does the path-integral quantization provide any answer?

The idea is exactly the same as in single particle QM. Take the expectation value $\langle 0 | \hat\phi(x) | 0 \rangle$ and insert a complete set of field eigenstates $|\varphi\rangle$; these are the states that satisfy $\hat\phi(x) |\varphi\rangle = \varphi(x)|\varphi\rangle$. We do this with a path integral:

$$\langle 0 | \hat\phi(x) | 0 \rangle = \int \mathcal{D}\varphi\ \langle 0 | \hat\phi(x)|\varphi\rangle \langle\varphi | 0 \rangle = \int \mathcal{D}\varphi\ \langle 0 |\varphi\rangle \langle\varphi | 0 \rangle \varphi(x).$$

So your probability distribution is $p[\varphi] = |\langle\varphi | 0 \rangle|^2$, just like in regular QM. I'm not sure yet how to calculate this, but by analogy with the harmonic oscillator it would probably be along the lines of $\exp(-\varphi^2)$.

Edit: I think I've managed to calculate it, though I make no guarantees of this being correct. In particular I'm pretty sure there should be some $2\pi$'s around.

We proceed by analogy with the harmonic oscillator. You can find the ground state wavefunction by using the fact that $\langle x | \hat a | 0 \rangle = 0$; this gives you a differential equation for $\langle x | 0 \rangle = 0$. In much the same way, we write $\langle \varphi | \hat a_\mathbf{p} | 0 \rangle = 0$. We then express $\hat a_\mathbf{p}$ in terms of $\hat \phi$ and $\hat \pi$, and use the fact that acting on a wavefunctional $\hat \pi$ is $-i \delta / \delta \varphi(x)$. I did it following Peskin and Schroeder's conventions for normalizations and working in the Schrödinger picture, and arrived at the following equation (kx is the dot product $\mathbf{k} \cdot \mathbf{x}$):

$$\int \mathrm{d}^3x\ e^{-ikx} \left( \omega_k \varphi(x) \langle\varphi | 0 \rangle + \frac{\delta}{\delta \varphi(x)}\langle\varphi | 0 \rangle \right) = 0$$

We will eventually express everything in terms of the Fourier transform $\hat \varphi$ of $\varphi$ (who knew, right?), so we will need to use the fact that

$$\int \mathrm{d}^3x\ e^{-ikx} \frac{\delta}{\delta \varphi(x)}\langle\varphi | 0 \rangle = \frac{\delta}{\delta \hat \varphi} \langle\varphi | 0 \rangle \bigg|_{-k}$$

so our equation becomes

$$\frac{\delta}{\delta \hat \varphi (k)} \langle\varphi | 0 \rangle = - \omega_k \hat \varphi (-k) \langle\varphi | 0 \rangle$$

and we can guess the solution

$$\langle\varphi | 0 \rangle = \exp \left( -\frac12 \int \mathrm{d}^3k\ \omega_k \hat \varphi(k) \hat \varphi(-k) \right).$$

You could rewrite this in position space but the Fourier transform of $\omega_k$ involves Bessel functions so I don't know if it's worth it.

I think the path integral approach provides an answer. In terms of the normalized generating functional $\mathcal{Z}[J]=\frac{Z[J]}{Z[0]}$, the expectation value is given by $$\langle 0|\phi(x)|0\rangle=\frac{\delta \mathcal{Z}[J]}{\delta J(x)}\Bigg|_{J=0}=\int\mathcal{D}\phi\phi(x)\exp\Big[\int i(\mathcal{L}+j\phi)d^4x\Big]\Big|_{J=0}\tag{1}$$ which when compared with the average of any function w.r.t a normalized probability distribution given by $$\overline{f(u)}=\int \limits_{a}^{b}duf(u)p(u)$$ tells that the relevant probability distribution in this case (1) is $$p[\phi(x)]=\exp(iS)=\text{complex}!!!$$ But what bugs me here is that $p[\phi(x)]$ is complex, and therefore, the name probability distribution may not be justified.

• QFT is not necessarily relativistic. If you look at the Euclidean theory, then you get a perfectly nice $\mathrm{e}^{-S}$, which is the main reason for the better convergence behaviour there. – ACuriousMind May 29 '17 at 17:17
• Minor detail: there should be an integral inside the exponential, and you forgot to set $J=0$. Your "complex probability distribution" is then $\exp(iS)$. – Javier May 30 '17 at 18:18

You may as well ask which $|\psi\rangle$ to use. We simply use the vacuum state, viz. $\langle 0|a_\mathbf{p}|0\rangle =0$. Thus $p=|\langle 0|x\rangle |^2$.

• What do you mean by $|x\rangle$ in quantum field theory? – SRS May 29 '17 at 14:37
• The label $x$ couldn't denote a single particle's position as it does in QM, as you've probably noticed. It would have to be another label of a complete basis, e.g. a $4$-momentum. – J.G. May 29 '17 at 14:50

## protected by Qmechanic♦May 30 '17 at 4:40

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