Defining the ground state energy of a QFT

Question

• I would like to hear of some general discussion on how is the ground state and its energy defined in QFT and how does one go about finding it. (..at least in some simple cases I have seen the use of Bogulibov transformation to make the theory look like a free-field theory to be able to identify the vacuum - but does every theory have a generalized Bogulibov transformation?..)

To initiate let me point out one specific formalism which I am confused about.

If one takes the Lagrangian as $L = \frac{1}{2}\left(\partial _\mu \phi\right)^2 - V(\phi)$ then in an expansion in powers of $\hbar$ gives the first term of the partition function as, $Z \approx \frac{1}{\sqrt{\mathrm{det}\left[-\partial^2 - V^{\prime\prime}\left(\phi_c\right)\right]}} e^{iS\left[\phi_c\right]}$ where $S$ is the action of the theory and $\phi _c$ is the classical extrema of that action.

Then one sees that if $E_0$ is the energy of the classical solution and the action is evaluated for a time interval $T$ then one has the relation $S\left[\phi_c\right] = -T\,E_0$

In the footsteps of the above identity one wants to define $E_G$ as the true quantum groundstate energy as $E_G = \lim_{T\rightarrow \infty} \frac{i \ln Z}{T}$.

If one used the above definition to the same order of perturbation as $Z$ is given and one replaces the value of $E_0$ before one takes the limit of $T \rightarrow \infty$ then one one has the relation,

$E_G \approx E_0 - \lim_{T \rightarrow \infty} \frac{i}{2T} \ln \left\{ \det \left[-\partial ^2 - V^{\prime\prime}\left(\phi_c\right)\right]\right\}$

• Now I am not sure as to how to interprete the second term of the above expression since naively that seems to go to $0$ since the determinant doesn't depend on the time-interval. But I guess I am wrong and I would want to know how at least this way of thinking works out.

Answer 1

My answer is based on the book "Quantum field theory in a nutshell, by A. Zee, Chapter IV.3 : Effective Potential"

Take a scalar field theory $(\phi)$, when you have a classical potential $V(\phi)$, you can find the state V.E.V. (vacuum expectation value) $\phi_0$ by minimizing this potential $V(\phi)$. But quantum fluctuations change this potential, so to calculate the modifyed state V.E.V. $\phi_{eff}$, you have first to calculate the modifyed potential $V_{eff}(\phi)$.

Define $W(J) = i log Z(J)$, You have V.E.V. $\phi(x) = \frac{\delta W(J)}{\delta J(x)}$ (a particular example of Green function obtained by diffentiating $W$ by the source $J$ repeatedly). By using a Legendre transformation, define $\Gamma(\phi) = W(J) - \int d^4x J(x) \phi(x)$. Note that there is an implicite dependence of $J$ on $\phi$, and you have $-J(x) = \frac{\delta \Gamma(\phi)}{\delta \phi(x)}$.

You may define $V_{eff}$ by : $\Gamma(\phi) = \int d^4x (- V_{eff}(\phi) + Z(\phi) (\partial\phi)^2) + ...$, which is justifyed because you can show that $V^\prime_{eff}(\phi) = J$ (rigourously when $\phi$ and $J$ do not depend on x). This means, that without source, $V^\prime_{eff}(\phi) = 0$, so $V_{eff}$ is the correct potential which, minimized, gives the ground state.

With step descent approximation applyied to $Z$, around the step descent point field $\phi_s$, where $V^{\prime}(\phi_s) = 0$ you will find, with explicit $\hbar$: $$Z(J) = \int D\phi\;e^{(i/\hbar)[S(\phi)+ J\phi]}$$ $$Z(J) \approx e^{(i/\hbar)[S(\phi_s)+ J\phi_s]} \int D\phi\;e^{(i/\hbar)\int d^4x\,\frac{1}{2} [(\partial\phi)^2) - V^{\prime\prime}(\phi_s)\phi^2]}$$ In the previous expression, the new integration variable is $\phi= \phi-\phi_s$. $$Z(J) \approx e^{(i/\hbar)[S(\phi_s)+ J\phi_s] - \frac{1}{2} (Tr\, log (\partial^2 +V^{\prime\prime}(\phi_s)) - Tr\, log (\partial^2))}$$ Note that the appearance of the term $Tr\, log (\partial^2))$ may be seen as a renormalization factor.

$$W(J) = S(\phi_s) + J\phi_s + i\frac{\hbar}{2} (Tr\, log (\partial^2 +V^{\prime\prime}(\phi_s)) - Tr\, log (\partial^2)) +O(\hbar^2)$$ $$\Gamma(\phi) = S(\phi) + i\frac{\hbar}{2} (Tr\, log (\partial^2 +V^{\prime\prime}(\phi)) - Tr\, log (\partial^2)) +O(\hbar^2)$$

With $\phi$ independant of x, you can write : $$Tr\, log (\partial^2 +V^{\prime\prime}(\phi)) = \frac{1}{(2\pi)^4}\int d^4x \int d^4k \,\,log \frac{(-k^2 + V^{\prime\prime}(\phi))}{-k^2}$$

So, finally, by the definition of $V_{eff}(\phi)$ from $\Gamma(\phi)$ $$V_{eff}(\phi) = V(\phi) - i\frac{\hbar}{2} \frac{1}{(2\pi)^4} \int d^4k \,\,log \frac{(k^2 - V^{\prime\prime}(\phi))}{k^2} + O(\hbar^2)$$

In practice, this integral is quadratically divergent, so you have to put a cut-off $\Lambda$, and you get : $$V_{eff}(\phi) = V(\phi) + \frac{\Lambda^2}{32\pi^2}V^{\prime\prime}(\phi) - \frac{1}{64\pi^2}[V^{\prime\prime}(\phi)]^2 log\frac{e^{1/2}\Lambda^2}{V^{\prime\prime}(\phi)}$$ So, by minimizing this effective potential $V_{eff}$, you can find the new state V.E.V. $\phi_{eff}$.

Now, in 3+1 dimensions, a scalar like $\phi_{eff}$ has the dimension of a mass (or energy), so by some Higgs mechanism, it can give mass to spin 1 gauge bosons, but the minimum of the effective potential $V_{eff}$ and the effective V.E.V. field $\phi_{eff}$ are 2 different things.