I am reading "QFT in a Nutshell", and the beginning of the book progresses like this:

  • Show how $\langle q_F|e^{-iHt}|q_I\rangle=\int Dq\ e^{iS}$
  • Says that we are more interested in $\langle F|e^{-iHt}|I\rangle$ (where $I$ and $F$ are initial/final states) than $\langle q_F|e^{-iHt}|q_I\rangle$, and then says we want $\langle F|=\langle 0|$ and $|I\rangle=|0\rangle$
  • Generalize to fields obtaining the path integral for fields.

I could not find an explanation on why we want the final and initial states to be the ground states. So my question is this: why do we want the vacuum expectation value?

  • 1
    $\begingroup$ related. $\endgroup$ – AccidentalFourierTransform Oct 25 '18 at 2:38
  • $\begingroup$ Isn't the LSZ formula relevant only for the canonical formalism, not path integrals? $\endgroup$ – Alex1994 Oct 25 '18 at 2:44
  • $\begingroup$ Nope. It is a key element of QFT, regardless of the formalism you use. $\endgroup$ – AccidentalFourierTransform Oct 25 '18 at 2:45
  • $\begingroup$ Surprisingly, the LSZ formula was not mentioned all the way from the expectation value to the derivation of the Feynman rules. How that can be? $\endgroup$ – Alex1994 Oct 25 '18 at 2:48
  • $\begingroup$ @Alex1994 Just Zee's style. He focuses more on the physical picture than mathematical rigor. $\endgroup$ – Avantgarde Oct 25 '18 at 4:08

For comparison, consider a simple harmonic oscilllator. In this system, we have operators $a$ and $a^\dagger$ satisfying $[a,a^\dagger]=1$, and the Hamiltonian is $H=\omega a^\dagger a$. We can define the vacuum state $|0\rangle$ to be the state of lowest energy, and we can describe this state more explicitly as the one that satisfies $a|0\rangle=0$.

Again for comparison, consider the model of a free scalar field. Schematically, the Hamiltonian is $H\sim \int dx\ \big(\nabla\phi(x)\big)^2+m^2\phi^2(x)$, and the equal-time commutation relation is $[\phi(x),\dot\phi(y)]\sim\delta(x-y)$. If we again define the vacuum state $|0\rangle$ to be the state of lowest energy, then we can define creation/annihilation operators $a^\dagger(p)$ and $a(p)$, expressed explicitly in terms of $\phi(x)$, in such a way that the vacuum state satisfies $a(p)|0\rangle=0$. We can also construct states with any specified number of particles by acting on the vacuum state with that number of creation operators, which in turn may be explicitly expressed in terms of the field operators that were used to define the model in the first place.

In most interesting QFTs, we don't know how to do this. We still define the model in terms of field operators by specifying their commutation relations and specifying the Hamiltonian, and we can still define the vacuum state to be the state of lowest energy, but we don't know how to characterize the vacuum state (much less states with any given number of particles) in any more explicit way using the field operators.

What we can do is consider expressions like $\langle 0|\phi(x)\phi(y)\cdots|0\rangle$ and make some general arguments (like LSZ) about how these vacuum expectation values are related to things of more direct physical interest. With the help of Wick rotation, we can use the path-integral formulation to define expressions like $\langle 0|\phi(x)\phi(y)\cdots|0\rangle$ without knowing anything more about $|0\rangle$ than the fact that it is the state of lowest energy. Then we can extract information about inner products between states with various numbers of particles by indirect arguments like LSZ. I think this is ultimately why we typically want the initial and final states to be ground states in QFT.


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