How to find the ground state of a system via its Hamiltonian density?

Question

I am trying to find the ground state of the following Lagrangian (with $\lambda> 0 , g > 0$):

$$\tag{1} \mathcal{L}= -\frac{1}{2}(\partial_\mu \partial^\mu \sigma + \partial_\mu \pi \partial^\mu \pi ) + \frac{1}{2} m^2 (\sigma^2 + \pi^2) - \frac{1}{4} \lambda ( \sigma^2 + \pi^2)^2 - g\sigma$$

This gives me the following Hamiltonian density:

$$\tag{2} \mathcal{H}= - (\frac{1}{2} \dot{\sigma}^2 + \nabla^2 \sigma + \frac{1}{2} \dot{\pi} + \nabla ^2 \pi) - \frac{1}{2} m^2 (\sigma^2 + \pi^2) + \frac{1}{4} \lambda (\sigma^2 + \pi^2)^2 -g\sigma$$

$\pi$ and $\sigma$ are scalar fields.

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I have seen it on my lecturer's notes that:

$$\tag{3} \mathcal{H_{min}} = \left[- \frac{1}{2} m^2 (\sigma^2 + \pi^2) + \frac{1}{4} \lambda (\sigma^2 + \pi^2)^2 -g\sigma \right]_{min}\\= \frac{1}{4} \lambda \left[(\sigma^2 + \pi^2 ) - \frac{m^2}{\lambda}\right]^2 - \frac{m^4}{4\lambda} + g\sigma$$

and therefore that this means that:

$$\tag{4} \mathcal{H}_{min} = -\frac{m^4}{4 \lambda} - \frac{gm}{\sqrt{\lambda}}$$

for $\sigma = -m /\sqrt{\lambda}$ and $\pi = 0.$

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Why is the first part of $(2)$ ignored for equation $(3)$?

I understand that choosing those values for $\sigma$ and $\pi$ lead to equation $4$? But what defines equation $4$ as the minimum that we are searching for?

Answer 1

You would be able to answer both your questions if you had not thoroughly messed up your signs, which I fix, $$\tag{1} \mathcal{L}= \frac{1}{2}(\partial_\mu \sigma\partial^\mu \sigma + \partial_\mu \pi \partial^\mu \pi ) + \frac{1}{2} m^2 (\sigma^2 + \pi^2) - \frac{1}{4} \lambda ( \sigma^2 + \pi^2)^2 - g\sigma$$

The canonically resulting hamiltonian density then has two pieces, a kinetic and a potential term, $$\tag{2} \mathcal{H}= \frac{1}{2} \Bigl (\dot{\sigma}^2 + (\nabla \sigma)^2 + \dot{\pi}^2 + (\nabla \pi)^2\Bigr ) +V\\ V=- \frac{1}{2} m^2 (\sigma^2 + \pi^2) + \frac{1}{4} \lambda (\sigma^2 + \pi^2)^2 -g\sigma.$$

The kinetic term in the parenthesis is positive semidefinite, minimized for constant fields.

Thus, the remaining term is the potential for constant fields, $$\tag{3} V= \frac{1}{4} \lambda \left[(\sigma^2 + \pi^2 ) - \frac{m^2}{\lambda}\right]^2 - \frac{m^4}{4\lambda} - g\sigma.$$

To determine the minimizing values for constant π and σ, vary w.r.t. both of them. The first vanishing derivative is satisfied by π =0. But the second not, by contrast, for vanishing σ. Instead, you need to satisfy $$ \lambda \sigma (\sigma^2 -m^2/\lambda) -g=0. \tag{5} $$ So your solution, as it stands, is untenable nonsense! But solving a (depressed) cubic equation is messy. (If you were ambitious, Taylor expand Viète's formula for small g.)

Still, for g =0, $\sigma = m/\sqrt{\lambda}$ would do. So perturb around that, for small g, i.e., neglect terms of $O(g^2)$, in $$ \sigma= m/\sqrt{\lambda}+ c g/m^2, $$ where I introduced masses in the perturbative correction for dimensional consistency, striving to determine the dimensionless constant c.

Plugging into (5), confirm that c=1/2 solves the condition to $O(g^2)$. Plugging this corrected solution into (3) finally confirm that $$\tag{4} V_{min} = -\frac{m^4}{4 \lambda} - \frac{gm}{\sqrt{\lambda}} +O(g^2).$$

NB _{The next term in the expansion of the minimum energy is $-g^2/(4\lambda m^2)$.}