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 October 15th, 2017, 06:02 PM #1 Senior Member     Joined: Nov 2015 From: United States of America Posts: 188 Thanks: 25 Math Focus: Calculus and Physics Quantum Mechanics: Momentum Expectation Value Hello forum, I was doing some quantum mechanics earlier and I was finding expectation values for a given wave function $\psi(x,t)$ specifically for momentum $\langle p \rangle_\psi$ I got $\langle p \rangle_\psi = 0$. What does this solution tell me? That the particle is most likely remaining in place? I would love some insight on this solution from anyone comfortable with quantum mechanics. Thanks!
October 16th, 2017, 03:09 AM   #2
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 Originally Posted by SenatorArmstrong Hello forum, I was doing some quantum mechanics earlier and I was finding expectation values for a given wave function $\psi(x,t)$ specifically for momentum $\langle p \rangle_\psi$ I got $\langle p \rangle_\psi = 0$. What does this solution tell me? That the particle is most likely remaining in place? I would love some insight on this solution from anyone comfortable with quantum mechanics. Thanks!
The expectation value can be considered a little bit like a "mean" value. Basically, it's the typical mean you would expect for a given a number of trials on that statistical distribution.

Consider a die for example. There is a 1/6 probability of getting 1, 2, 3, 4, 5 or 6. The expectation value can be calculated from this distribution, giving 3.5. That means that for, say, several rolls of the die, you can expect the mean to be around 3.5. It doesn't mean that it will be, just that that's what you'd expect given the probabilities.

If the expectation value is 0, it means that the typical mean you can expect for a set of particles in the distribution (like a gas, for example) is zero.
If there are particles with positive momentum in your distribution, there are going to be particles with negative momentum also.

Expectation value should not be confused with "the typical value obtained for a single trial" because you might have a double-peak distribution or something unusual. In those cases, the mode (i.e. locations of peaks in the PDF) is probably a better indicator of the typical value obtained for a single trial.

October 16th, 2017, 11:00 AM   #3
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 Originally Posted by Benit13 The expectation value can be considered a little bit like a "mean" value. Basically, it's the typical mean you would expect for a given a number of trials on that statistical distribution. Consider a die for example. There is a 1/6 probability of getting 1, 2, 3, 4, 5 or 6. The expectation value can be calculated from this distribution, giving 3.5. That means that for, say, several rolls of the die, you can expect the mean to be around 3.5. It doesn't mean that it will be, just that that's what you'd expect given the probabilities. If the expectation value is 0, it means that the typical mean you can expect for a set of particles in the distribution (like a gas, for example) is zero. If there are particles with positive momentum in your distribution, there are going to be particles with negative momentum also. Expectation value should not be confused with "the typical value obtained for a single trial" because you might have a double-peak distribution or something unusual. In those cases, the mode (i.e. locations of peaks in the PDF) is probably a better indicator of the typical value obtained for a single trial.
Thanks for clearing things up!

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