Lagrangian multiplier and ground state search

Question

Hi I am trying to understand the paper of Schollwoeck (arxiv version: https://arxiv.org/pdf/1008.3477.pdf)

There on p.64 formular 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 \(\langle \psi | \psi \rangle = 0\) .

See for example https://en.wikipedia.org/wiki/Lagrange_multiplier.

My question now 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?

Comments

You must first consider $|\psi\rangle$ as unknown variable, find the solution to the extremum problem, and only after that apply the constraint, whatever it is.

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The stationary points are the same whether you add your term or the term in Schollwoeck; he is just less careful than you in writing the details. The difference is just a constant $\lambda$.

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@ArnoldNeumaier what do you mean with stationary points?
I only know Lagrange multiplier in context of optimization of functions over real valued vector spaces.  So what does it actually mean if I want to find the minimal \(| \psi \rangle\)? Does my intuition translate to Hilbert spaces? Since the optimal solution point in a real valued vector space can look very different for different constraints, even if the difference is just a constant. Or are there some properties of the hamiltonian which were implicitly used in the argument?

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The infinite-dimensional case works the same, at the physical level of rigor. You minimize a quadratic subject to a norm constraint, by forming the Lagrangian and finding its stationary points (zero variation = zero gradient). The constant does not affect the variation.

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@ArnoldNeumaier For a real valued Lagrangian I have to calculate the gradient wrt. x but also wrt. \(\lambda\) . This is were the constant will appear and matter. Then I'll have to solve a system of equations where the constant does not simply vanishes. What is the analog in the infinite-dim. case? What is the search term I have to use to  find more details on this subject?

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No. Independent of the dimension, the stationary points are at the gradient of the Lagrangian with respect to the original variables only. In addition, the original constraint must be imposed. The derivation of the technique is the same in all dimensions.