# Introduction to Operators

1. Aug 31, 2015

### paradoxymoron

First, I have a question regarding the conservation of probability. The book shows (quite elegantly) that

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

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?

2. Aug 31, 2015

### Orodruin

Staff Emeritus
If the limits are finite, then your wave function is going to be identically zero at the boundaries (this occurs, e.g., for the infinite potential well).

No, $x$ is a coordinate and it does not depend on time. What depends on time is the wave function, which in turn depends on $x$ and $t$. You can deduce the average position which does in general depend on time, but this is not the coordinate $x$, but a statement of a property of the wave function.

It is a definition here and a way of making sense of the term "velocity" at the quantum level.

In general, you should try to limit your questions to one question (or at least one concept) per thread. Posting several topics in one thread will only make your responses scattered among each other.

3. Aug 31, 2015

### blue_leaf77

Is it not already guaranteed by the Ehrenfest theorem?

4. Aug 31, 2015

### Orodruin

Staff Emeritus
The Ehrenfest theorem tells you that the expectation value should follow the classical equation of motion. It is unrelated to the interpretation of this to the "velocity".

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