Schroedinger Equation from Variational Principle

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SUMMARY

The forum discussion centers on the derivation of the Schrödinger equation from the variational principle as presented in Landau's nonrelativistic quantum mechanics. The key equation discussed is $$\delta \int \psi^{\ast} (\hat{H} - E) \psi dq = 0$$, which leads to the conclusion that varying the wave function ##\psi## and its conjugate ##\psi^{\ast}## independently results in the Schrödinger equation ##\hat{H} \psi = E \psi##. The discussion raises questions about the minimization of the difference between the expectation value of the Hamiltonian and energy, suggesting a potential analogy to classical mechanics. Additionally, it highlights the utility of this action in constructing a path integral for quantum observables.

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Sturk200
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Landau's nonrelativistic quantum mechanics has a "derivation" of Schroedinger's equation using what he calls "the variational principle". Apparently such a principle implies that:

$$\delta \int \psi^{\ast} (\hat{H} - E) \psi dq = 0$$

From here I can see that varying ##\psi## and ##\psi^{\ast}## independently gives rise to the equation ##\hat{H} \psi = E \psi##. But where does that first equation come from? I think it is saying that the the allowed energies are those for which the difference between the expectation value of the hamiltonian and the energy is minimized. I'm not sure why that should be so. Is there some analogue to classical mechanics that I'm missing?

(Note: this is section 20 of the book).
 
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If you have a classically motivated differential equation of motion for some degree(s) of freedom, you can usually hack your way towards a Lagrangian that yields that equation. A reasonable starting point is to multiply the equation on the left by your DOF (here, \psi) and start integrating by parts.

In classical mechanics this would usually be pretty pointless. In quantum theory this action can be used to construct a path integral, providing a different way of computing observables.
 

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