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Expectation of Momentum in a Classical (Infinite) Potential Well

"Look at" the particle at many different randomly-chosen times.
Thank you for this. Classically the system will have two values, but there magnitudes are equal.
The only thing I am now concerned about is lacking some kind of pdf to bring about the variances for momentum and position in a way that both values are found using the same pdf

But I have to say this was very illuminating.
 
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kuruman

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Thank you for this. Classically the system will have two values, but there magnitudes are equal.
The only thing I am now concerned about is lacking some kind of pdf to bring about the variances for momentum and position in a way that both values are found using the same pdf

But I have to say this was very illuminating.
What's a "pdf" in the context of your last post?
 
What's a "pdf" in the context of your last post?
The probability density function so I can find <value>
 

Orodruin

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The expectation value in the classical setting is a time-average over a full period of the classical system.
 

jtbell

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some kind of pdf to bring about the variances for momentum and position in a way that both values are found using the same pdf
I don't think you can do that for the classical case, because the pdf's for momentum and position are in terms of different variables, ##p## and ##x##: ##P_p(p)## and ##P_x(x)##. It's not obvious to me that there's a general relationship between them.

The pdf's for position and momentum are different functions in QM, too. For position: $$P_x(x) = \psi^*(x)\psi(x)$$ For momentum, $$P_p(p) = \phi^*(p)\phi(p)$$ where $$\phi(p) = \frac 1 {\sqrt{2\pi\hbar}} \int^{+\infty}_{-\infty} {\psi(x)e^{-ipx/\hbar}\,dx}$$ The two pdf's are related, but definitely different.

Are you perhaps thinking of the fact that in the QM case you can calculate the expectation value of momentum from the probability amplitude function for position, ##\psi(x)##: $$\langle p \rangle = \int^{+\infty}_{-\infty} {\psi^*(x) \left( -i\hbar \frac {\partial}{\partial x} \right) \psi(x)\,dx}$$ as well as from the probability distribution function for momentum: $$\langle p \rangle = \int^{+\infty}_{-\infty} {p P_p(p)\,dp}$$ for a continuous pdf, or $$\langle p \rangle = \sum_i {p_i P_p(p_i)}$$ for a discrete pdf?
 

kuruman

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To elaborate on @jtbell #24. The expectation value, as you know, is an average.

Suppose you performed ##N## measurements of the momentum at random intervals and you got ##N_1## values at ##+p## and ##N_2## at ##-p##, where ##N=N_1+N_2##. What is the average value of the momentum?

Suppose you flip a fair coin ##N## times and you get ##N_1## tails and ##N_2## heads. You assign +1 point to tails and -1 point to heads. How would you express the average of the flips as a number? What do you think the a priori average could be before you do any measurements?
 

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