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Quantum Mechanics$\Rightarrow$}Schrodinger Equation

Recall the Time-independent Schrodinger Equation,

$H\psi=E\psi\Rightarrow -\frac{\hbar^2}{2m}\psi '' + V(x) \psi = E\psi,
$

where $H=-\frac{\hbar^2}{2m}\psi '' + V(x) \psi = E\psi$.

The classical Hamiltonian is $H=\frac{p^2}{2m} + V(x)$, where one sees the only difference is $p \rightarrow \frac{\hbar}{i}\frac{d}{dx}$. From the classical Hami, one can directly reach the Hami operator in the TISE via substituting a differential operator for momentum, as in choice (B).

Alternate Solutions
 sunnysunny2019-08-07 04:43:51 The information you share is very useful. It is closely related to my work and has helped me grow. Thank you! subway surfers\r\nReply to this comment
 ernest212019-08-23 02:02:18 I thought of it in terms of the selection rules , . If the initial state has , which corresponds to spherical symmetry in the wavefunction, then according to these the final state must have , which is not spherically symmetric. candle in the tombReply to this comment joshuaprice1532019-08-09 06:41:39 I completely agree with you. I really like this article. It contains a lot of useful information. I can set up my new idea from this post. It gives in depth information. sprinkler repair MiamiReply to this comment sunnysunny2019-08-07 04:43:51 The information you share is very useful. It is closely related to my work and has helped me grow. Thank you! subway surfers\r\nReply to this comment UNKNOWNUMBER2018-10-15 01:52:20 Easy free points!Reply to this comment

I thought of it in terms of the selection rules , . If the initial state has , which corresponds to spherical symmetry in the wavefunction, then according to these the final state must have , which is not spherically symmetric. candle in the tomb

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