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

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.

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 $$\pi = 0.$$

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?

• The first part of your Hamiltonian density is the kinetic energy of fields, and the rest if potential energy. So what your lecturer did is basically finding the configuration that minimizes the potential energy. Apr 7, 2022 at 15:36
• But why were those values of $\sigma$ and of $\pi$ specifically chosen? I understand that we get $(4)$ when choosing them, but I don't know why those were the values we attributed to them. Apr 7, 2022 at 16:08
• PS you have also messed up signs in the potential, big time,and your solution does not satisfy the minimum cubic equation. Try solving for it perturbatively in g, like most. Apr 7, 2022 at 18:41
• You must be right, but how can I have done so if $\frac{\partial \mathcal{L}}{\partial \sigma} = -\dot{\sigma}$? and similarly for $\pi$ ? Apr 7, 2022 at 19:07
• ?? It is the canonical procedure of deriving the hamiltonian from the Lagrangian. Yo must have misunderstood something. are you talking about the definition of canonical momenta? Show your work! Apr 7, 2022 at 19:56

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 $$-g^2/(4\lambda m^2)$$.

• The sign of the kinetic part is conventional. It looks like the convention was $g_{00}=-1$ where you use $g_{00}=+1$. Apr 7, 2022 at 20:08
• The first $\sigma$ has dropped out of your Lagrangian, too. Apr 7, 2022 at 20:09
• @my2cts Possibly, but if so he still needs a positive semidefinite kinetic term in his hamiltonian. Apr 7, 2022 at 22:22