${\displaystyle H(t)\vert \psi (t)\rangle =i\hbar {\frac {\partial }{\partial t}}\vert \psi (t)\rangle }$

In your article, replace the image with:

<math>H(t)\vert\psi(t)\rangle = i\hbar\frac{\partial}{\partial t}\vert\psi(t)\rangle</math>

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