Understanding the Conservation of Probability and Operators in Quantum Mechanics

paradoxymoron
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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?
 
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paradoxymoron said:
What if the limits were finite?

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).

paradoxymoron said:
Is x not a function of time?

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.

paradoxymoron said:
The book then "postulates" that the rate of change of the average position is the average velocity, without justification. How is that true?
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.
 
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Orodruin said:
It is a definition here and a way of making sense of the term "velocity" at the quantum level.
Is it not already guaranteed by the Ehrenfest theorem?
 
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".
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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