The Wave Function

1.5 Momentum

Eqn 1.29 - 1.31

Referring to Equation 1.25 and 1.28, we see that

$$ \begin{aligned} \frac{d\braket{x}}{dt} &= \int x {\color{red}\frac{\partial }{\partial t} |\psi|^{2}} dx \\ &= \text{Substituting equation 1.25 on red colored equation} \\ &= {\color{red}\frac{i \hbar}{2m}} \int x {\color{red}\frac{\partial }{\partial x} \left( \psi^{*} \frac{\partial \psi}{\partial x} - \frac{\partial \psi^{*}}{\partial x} \psi \right)} dx \quad (1.29)\\ &= \frac{i \hbar}{2m} \left[ \int x \frac{\partial }{\partial x}\left( \psi^{*}\frac{\partial \psi}{\partial x} \right) dx - \int x \frac{\partial }{\partial x} \left( \frac{\partial \psi^{*}}{\partial x} \psi \right) dx \right] \\ &= \text{Using integration by-parts, } \\ & \quad \int uv dx = u \int v ~dx - \int \left( \frac{du}{dx} \int v ~dx \right) dx \\ &= \frac{i \hbar}{2m} \Bigg[ x ~ \int\frac{\partial }{\cancel{\partial x}}\left( \psi^{*}\frac{\partial \psi}{\partial x} \right) \cancel{dx} - \int \left\{\frac{\cancel{dx}}{\cancel{dx}} \int \frac{\partial }{\cancel{\partial x}}\left( \psi^{*}\frac{\partial \psi}{\partial x} \right) \cancel{dx} \right\} dx \\ &\quad - \left\{ x \int \frac{\partial }{\cancel{\partial x}} \left( \frac{\partial \psi^{*}}{\partial x} \psi \right) \cancel{dx} - \int \left\{ \frac{\cancel{dx}}{\cancel{dx}} \int \frac{\partial }{\cancel{\partial x}} \left( \frac{\partial \psi^{*}}{\partial x} \psi \right) \cancel{dx} \right\} dx \right\} \Bigg] \\ &= \frac{i \hbar}{2m} \Bigg[ x ~ \int \partial\left( \psi^{*}\frac{\partial \psi}{\partial x} \right) - \int \left\{\int \partial \left( \psi^{*}\frac{\partial \psi}{\partial x} \right) \right\} dx \\ &\quad - \left\{ x \int \partial \left( \frac{\partial \psi^{*}}{\partial x} \psi \right) - \int \left\{ \int \partial \left( \frac{\partial \psi^{*}}{\partial x} \psi \right) \right\} dx \right\} \Bigg] \\ &= \text{Since, } \int \partial(y) = y + \text{integration constant}.\\ &= \frac{i \hbar}{2m} \Bigg[ x ~ \left( \psi^{*}\frac{\partial \psi}{\partial x} \right) \Bigg|_{-\infty}^{\infty} - \int \left( \psi^{*}\frac{\partial \psi}{\partial x} \right) dx\\ & \quad - \left\{ x \left( \frac{\partial \psi^{*}}{\partial x} \psi \right)\Bigg|_{-\infty}^{\infty} - \int \left( \frac{\partial \psi^{*}}{\partial x} \psi \right) dx \right\} \Bigg] \\ &= \text{We assume wave function vanishes at infinity (as eqn 1.26).} \\ &= - \frac{i\hbar}{2m} \int \left( \psi^{*}\frac{\partial \psi}{\partial x} - \frac{\partial \psi^{*}}{\partial x} \psi \right) dx. \quad (1.30) \\ &= \text{Again, apply by-parts on second term. i.e. } \int \frac{\partial \psi^{*}}{\partial x} \psi~dx \\ &= - \frac{i\hbar}{2m} \left[ \int \psi^{*} \frac{\partial \psi}{\partial x} ~dx - \psi^{*}\psi \Big|_{-\infty}^{\infty} + \int \psi^{*} \frac{\partial \psi}{\partial x} ~dx \right] \\ &= \text{Wave function vanishes.}\\ &= - \frac{i \hbar}{m} \int \psi^{*} \frac{\partial \psi}{\partial x} ~dx. \quad (1.31) \end{aligned} $$

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Published on Aug 25, 2021

Last revised on Jan 26, 2023