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I'm currently reading Griffiths' book about Quantum Mechanics but I cannot understand how he derives the formula for the time derivative of the expected value of position in 1 dimension.

He writes:

$$\frac{\mathrm d\langle x \rangle}{ \mathrm dt} = \int x\,\frac{\partial}{\partial t}\left(|\psi|^2\right) \mathrm dx =\frac{i\hbar}{2m}\int x\,\frac{\partial}{\partial x}\left(\psi^*\frac{\partial\psi}{\partial x}+\psi \frac{\partial\psi^*}{\partial x}\right)\mathrm dx \quad . \tag 1$$

Now he integrates by parts: He integrates $$\frac{\partial}{\partial x}\left(\psi^*\frac{\partial\psi}{\partial x}+\psi \frac{\partial\psi^*}{\partial x}\right) \tag 2 $$

and he differentiates $x$. Explicitly the integral above becomes:

$$x\left(\psi^*\frac{\partial\psi}{\partial x}+\psi\frac{\partial\psi^*}{\partial x}\right)\big\rvert_\mathbb{R} + \int\psi^*\frac{\partial\psi}{\partial x}+\psi\frac{\partial \psi^*}{\partial x} \mathrm d x\quad . \tag 3$$

The author concludes that the boundary term evaluates to $0$, but I can't see how. $x$ goes to $\infty$ and both $\psi$ and $\psi^*$ go to $0$ when $x$ approaches $\infty$, so in principle we should have an indeterminate form of the type $\infty * 0$. How can I make some progress here?

Reference: David J. Griffiths - Introduction to Quantum Mechanics 2nd Ed.

Chapter 1, Section 1.5, Pages 15-16

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  • $\begingroup$ Please use MathJax instead of images/screenshots. $\endgroup$ Commented Apr 1, 2023 at 14:16
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    $\begingroup$ I've edited your post. Feel free to undo if necessary. For example, you can use \langle and \rangle to get $\langle$ and $\rangle$, respectively. In any case, you should include a full citation (i.e. edition of the book, chapter, page number...)! $\endgroup$ Commented Apr 1, 2023 at 15:41

2 Answers 2

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Normalization requires $\Psi(x, t)$ must go to zero faster than $1/\sqrt{|x|}$. Even at this lower bound (insufficient for proper normalization) the boundary term in the integration by parts will vanish.

$$\Psi(x,t) \sim \frac{1}{\sqrt{x}} \Rightarrow \frac{\partial \Psi}{\partial x} \sim \frac{1}{x^{3/2}}$$

So

$$\lim_{x\rightarrow \infty}\left[x\left(\Psi^*\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^*}{\partial x}\Psi\right)\right] \sim \lim_{x\rightarrow\infty}\left[x\cdot\frac{1}{\sqrt{x}}\cdot\frac{1}{x^{3/2}}\right]\sim\lim_{x\rightarrow\infty}\frac{1}{x}=0$$

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  • $\begingroup$ Thanks! Now everything is clear $\endgroup$
    – Tom Avery
    Commented Apr 1, 2023 at 16:56
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    $\begingroup$ Normalization does not pose any restriction on the behavior at infinity. There are square-integrable functions that are not even defined at infinity. $\endgroup$
    – DanielC
    Commented Apr 1, 2023 at 20:55
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    $\begingroup$ @DanielC Of course you're right. However as this is chapter 1 of Griffiths, I thought it best to restrict the Hilbert space to well-defined wavefunctions over the entire 1D domain (as Griffiths himself does) for conceptual ease. $\endgroup$
    – Aiden
    Commented Apr 1, 2023 at 22:03
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The magnitude of x grows linearly. The probability density must decay much faster than that if we are to satisfy the normalization condition over the entire domain.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Apr 1, 2023 at 19:27

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