# Lagrangian multiplier and ground state search

I'm trying to understand the paper of Schollwoeck. On page 64, equation 203 he states:

In order to solve this problem, we introduce a Lagrangian multiplier λ, and extremize $$\langle \psi | H | \psi \rangle - \lambda \langle \psi|\psi \rangle$$

If I remember correctly this however translates into an optimization problem where one wants to min/maximize $\langle \psi | H | \psi \rangle$ subject to the constraint $\langle \psi|\psi \rangle$. See for example here.

My question is, how can I understand the constraint? If it would be something like $\lambda (\langle \psi |\psi \rangle -1 )$ then I could read it as the normalization constraint but in this form I don't know how to make sense of it. May some brighter person please enlighten me?

• This is a guess made without reminding myself of exactly how Lagrange multipliers work: could it be that the point is to limit the norm of $|\psi\rangle$, but that it doesn't matter what the norm actually is because you can always rescale it? – DanielSank Apr 3 '18 at 16:29
• @DanielSank If the actual value of the norm wouldn't matter I would rewrite the constraint as $\lambda (\langle \psi | \psi \rangle - c )$ with some constant $c$. If I omit this c this means I explicitly enforce the norm to be 0 which would result in an invalid qm state. Also scaling 0 with a factor doesn't change much. The only way I could maybe understand this is from an numerical point of view where one introduces the $\lambda$ term as a penalty to achieve 'small' states in the optimization process and then later rescales them to 1, since the numerical optimization never really reaches 0 – v.tralala Apr 4 '18 at 12:08

You're right that the full constraint should in principle be written $$\lambda \left( \langle \psi|\psi\rangle -1\right)$$ to describe normalization. Schollwöck makes this clearer in this book chapter (p. 16.23). If the $$-\lambda$$ term is included we can take a variation/derivative with respect to $$\lambda$$ and recover the constraint. However, if we want to find the ground state energy, we'll want to do a variation with respect to the state. In such a case, the $$-\lambda$$ term does not enter into the variation. Another way of putting it is that minimizing $$E'=\langle \psi | H |\psi\rangle - \lambda \langle \psi|\psi\rangle$$ is equivalent to minimizing $$E'-\lambda$$.
All in all, I think Schollwöck just suppressed the $$-\lambda$$ term for convenience. Note that there is no left-hand side in the equation either, which could be a hint that he doesn't consider this the full energy functional. Instead it's the minimal expression to minimize to obtain the ground state energy.