- #1

ConjugatedClanger

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*The Physics off Quantum Mechanics by*James Binney and David Skinner. On page 45, when discussing the

*probability current*(in the wave mechanics formalism) in calculating it they state:

I.e.$$ i \hbar \left( \psi^{*} \frac{\partial}{\partial t} \psi + \psi \frac{\partial}{\partial t} \psi^{*} \right) = \psi^{*} \hat{H} \psi - \psi \hat{H}^{*} \psi^{*} \tag{1}$$The intent is obviously to derive a"We multiply the TDSE by ## \psi^{*} ## and subtract it from the conjugate of of the TDSE multiplied with ## \psi ##"

*continuity equation*for the TDSE and I've seen this approach in a few other places, but I've always wondered:

**what is the intuition, or motivation, behind calculating ##(1)##, straight off the bat**? What does the quantity ## \psi^{*}\hat{H}\psi ## mean -- an "expectation density"?

To me the more intuitive approach (and one shown in David Tong's lecture notes (page 11)) is to say

$$ \frac{\partial}{\partial t} \rho = \psi^{*} \frac{\partial}{\partial t} \psi + \psi \frac{\partial}{\partial t} \psi^{*} \tag{2}$$ where ## \rho = |\psi|^{2} = \psi^{*} \psi ## is the probability density (as defined by the postulates of wave mechanics), and then rearrange the TDSE and its conjugate for ## \partial/\partial t \psi ## and ## \partial/\partial t \psi^{*} ## respectively and substitute back into ##(2)##, which will give us ##(1)##.

It might be that I'm looking for significance where there is none, and the approach shown in ##(1)## is from the insight we gained by first doing ##(2)## and lecturers and authors just choose this way to jump to an answer and then work backwards. But if anyone has some insight please let me know!