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In the book Many-Particle Physics by Gerald D. Mahan, he points out that the Schrodinger equation in the form $$i\hbar\frac{\partial\psi}{\partial t}=\Big[-\frac{\hbar^2\nabla^2}{2m}+U(\textbf{r})\Big]\psi(\textbf{r},t)$$ can be obtained as the Euler-Lagrange equation corresponding to a Lagrangian of the form $$L=i\hbar\psi^*\dot{\psi}-\frac{\hbar^2}{2m}\nabla\psi^*\cdot\nabla\psi-U(\textbf{r})\psi^*\psi.$$

I have a discomfort with this derivation. As fas as I know a Lagrangian is a classical object. Is it justified in constructing a Lagrangian that has $\hbar$ built into it?

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  • $\begingroup$ This really seems to be deriving a classical field equation which happens to look like the Schrödinger equation for a quantum particle. The interpretations are completely different. $\endgroup$ – knzhou Feb 10 '18 at 10:28
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Firstly, one may think of this as a mathematical rather than physical procedure. In the end one is simply constructing a functional,

$$S = \int \mathrm dt \, L$$

whose extremisation, $\delta S = 0$ leads to the Schrodinger equation. However, Lagrangians containing $\hbar$ are not uncommon. In quantum field theory, one can construct effective actions from computing Feynman diagrams, which may have factors of $\hbar$, outside of natural units.

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  1. As user JamalS correctly points out in his answer:

    • Quantum actions in QFT are allowed to have $\hbar$-dependence.
    • If we merely want a stationary action principle for the TDSE, and view the action functional as just a mathematical tool without physical consequences beyond the EL equations, then the $\hbar$-dependence doesn't matter.
  2. However, perhaps OP's discomfort with the TDSE derivation is spurred by the following deeper question:

    How we can get the correct semiclassical limit$^1$ and loop expansion of a second-quantized path integral $$Z~=~\int\! {\cal D}\psi{\cal D}\psi^{\ast} \exp\left(\frac{i}{\hbar} S\right),$$ if the Schroedinger action $S$ depends on $\hbar$, so that various parts of the actions $S$ scales/are suppressed inhomogeneously in the semiclassical limit $\hbar\to 0$?

    That's a good question. The answer is that there are implicit hidden $\hbar$-dependence, i.e. one should rescale the variables $$\psi~=~\frac{\psi_0}{\sqrt{\hbar}},\qquad m~=~\hbar m_0,\qquad U~=~\hbar U_0,$$ to obtain a classical ($\hbar$-independent) action $S$, and to restore a correction loop expansion.

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$^1$ See e.g. this Phys.SE post.

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  • $\begingroup$ ℏ or not, however, what would one make of non-extremizing, "off-shell" ψ configurations? Surely they are not "quantum corrections", to the extent that all the quantum corrections are already in the ψ ("on shell") SE solution? $\endgroup$ – Cosmas Zachos Mar 8 '18 at 19:59

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