# QM - Time derivative of the expected value of position

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

• 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...)! Apr 1 at 15:41

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

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

• Thanks! Now everything is clear Apr 1 at 16:56
• Normalization does not pose any restriction on the behavior at infinity. There are square-integrable functions that are not even defined at infinity. Apr 1 at 20:55
• @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. Apr 1 at 22:03

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.