# Time-independent Schrödinger equation Lagrangian derivation

Recently I was taking a calculus of variations class and our professor casually obtained the time-independent Schrödinger equation for a free particle from the integral (constants dropped) and it's constraint to normalise the distribution: $$S=\int_{-\infty}^{\infty}(\psi')^2dx - E\int_{-\infty}^{\infty}\psi^2dx$$ where $$E$$ is the Lagrange multiplier that turns out to be the energy.

Does the Lagrangian $$(\psi')^2$$ have a special name or any significance in actual quantum mechanics or is this just a coincidence?

With some further tinkering to the Lagrangian, I managed to derive the full time independent Schrödinger equation by replacing $$(\psi')^2$$ with $$(\psi')^2+V(x)\psi^2$$.

• Is that second term supposed to be $E \int \lvert \psi \rvert^2 \, dx$? If so, then I can answer this. – DanielSank Mar 18 at 7:12
• – Qmechanic Mar 18 at 7:20
• @DanielSank Yes (in this case we assumed $\psi$ is real valued. I know that $|\psi|^2$ is the probability distribution my main curiosity is about the main Lagrangian. – Jepsilon Mar 18 at 7:34
• @Qmechanic I saw that question but I am having a hard time relating it to my query. In particular, the other question deals with time dependence. If it can be shown that these two are equivalent I am all ears because I cannot make the connections on my own. – Jepsilon Mar 18 at 7:38