First, I have a question regarding the conservation of probability. The book shows (quite elegantly) that(adsbygoogle = window.adsbygoogle || []).push({});

$$ \frac {d}{dt} \int_{-\infty}^{\infty} |\Psi (x, t)|^2dx = \frac {i\hbar}{2m} \Big{(}\Psi ^* \frac{\partial \Psi}{\partial x} - \frac{\partial \Psi ^*}{\partial x} \Big{)} \Big |_{-\infty}^{\infty}$$

And it goes on to say that the derivative is zero, on the grounds that ##\Psi## approaches zero as ##x## approaches infinity. What if the limits were finite? Would the probability density still be normalizable for all ##t##?

We also recently started going over operators of classical dynamical variables. After stating the simple operator for position, the textbook goes on to derive an expression for the rate of change of the average position, i.e

##

\begin{align}

\frac{d \ \overline{x}}{dt}&=\frac{d}{dt} \int_{-\infty}^{\infty} x|\Psi (x, t)|^2dx \\

&=\int_{-\infty}^{\infty} x\frac{\partial |\Psi |^2}{\partial t}dx

\end{align}

##

The rest of the derivation - which uses the above result, and then double use of integration by parts - is easy to follow.

However, I'm confused as to why, when taking the derivative under the equal sign, it doesn't affect the ##x## and use the product rule. Is ##x## not a function of time?

The book then "postulates" that the rate of change of the average position is the average velocity, without justification. How is that true?

One more question. How is it that any measurable quantity ##Q## can be written as a function of position and momentum ##p##, i.e ##Q=Q(x, p)##? Can this be proven?

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# Introduction to Operators

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