Interpreting the expectation values of $\langle p \rangle$ and $\langle p^2 \rangle$

Why the expectation value of momentum $$\langle p \rangle$$ is zero for the one dimensional ground-state wave function of an infinite square well? And why $$\langle p^2 \rangle = \frac{\hbar^2 \pi^2}{L^2}$$? I am not asking for a proof. I am trying to interpret physical their meanings. I understand $$\langle p \rangle = 0$$ does not imply that momentum itself is zero, though at first I thought that way. I would really appreciate if you can help me build a good intuition. I have seen other questions related to this matter here, but they have not given the intuition I am looking for.

• An infinite square well is symmetrical. Why would the particle have an average momentum to the right or the left? – G. Smith Apr 5 at 3:30
• I don't know that is why I am asking this, and if the average momentum is zero why $\langle p^2 \rangle \neq 0$? – Rob Apr 5 at 3:32
• What is the average value of the numbers -1 and +1? What is the average value of their squares? – Prahar Apr 5 at 3:33
• Think about the bell curve of a normal distribution around zero. Its average is zero. Its mean square average is not. – G. Smith Apr 5 at 3:35
• @Prahar I was hoping to see a physical explanation. I knew that. – Rob Apr 5 at 3:35

To be $$0$$ on average, a quantity must either be always $$0$$ as you suggest or else it must have positive or negative outcomes, for instance $$\{-2,-1,3\}$$. Of course it is possible for a particle in $$1d$$ to have positive or negative values of momentum, with negative values corresponding to motion in the negative direction.

In the case of steady state solution to the Schrödinger equation (as the ground state of the infinite well), the distribution of positions does not change with time, so $$\langle x\rangle$$ does not depend on $$t$$; as a result, the average momentum $$\langle p\rangle= m\frac{d}{dt}\langle x\rangle=0$$. If you solution is not a steady-state solution then $$\langle p\rangle\ne 0$$ at all times.

On the other hand, $$\langle p^2\rangle$$ is the average of non-negative quantities since $$p^2\ge 0$$; this average cannot be negative, v.g. the average of $$\{4,1,9\}$$ is certainly not $$0$$. The precise actual value of $$\langle p^2\rangle$$ just happens to work out to $$\hbar^2\pi^2/L^2$$ and is basically related to the energy of your system: the ground state energy is $$E_1= \pi^2\hbar^2/(2mL^2)$$; in an infinite well the energy is completely kinetic. Since $$E_1=\langle p^2\rangle/(2m)$$, it's easy matter to find $$\langle p^2\rangle= \pi^2\hbar^2/L^2$$.

So when trying to build intuition about a subject, ideally you want something experiential to help you understand it. By the statement of your question you seem to get why the math woks out the way it does (please correct me if this is not the case) but don't get what it's telling you about the system. Let's see if we can construct some intuition from experiences that a lot of have, driving. Let's say you live 5 miles away from work, which for the sake of this example takes you 20 minutes to drive to work and 20 minutes to drive back (obviously this is hardly ever the case, but we're idealizing here). On average you're driving at 15 mph to and from work. Though because you spend equal amounts of time driving to work as you do driving home. I can't say I expect you to be do one more than the other, so my expectation value would have to be equally weighted between both possibilities. However, that doesn't mean I can't have an idea on how fast you're going (15 mph) which is definite and positive.

In terms of the particle in the box, classically the particle can be viewed as traveling from one side of the box just as much as it does to the other. Therefore, if we had any nonzero value we would be claiming that it would be spending more time traveling to one side than the other (though in reality "time" isn't the right way to think about it, but it's a good starting point). However, it does have a nonzero speed so if we average that and square it, it will still be nonzero (multiple by mass squared to get $$\langle p ^{2} \rangle$$. The statements of the first quantity $$\langle p \rangle$$ are more a value about where the particle is on average, while the latter $$\langle p^{2} \rangle$$ are more about how fast the particle is going.

This all of course is simplifying quite a bite, but hopefully it's enough to get the gears moving.

• That was more or less what I needed. As you pointed out, I understand why the math works. This gave me intuition. The only part I was bothered with was using miles in your answer (Haha I've used metric system my entire life). Jokes aside, it certainly got the gears moving. Thanks! – Rob Apr 5 at 4:26

Expectation value is nothing but an average value ,So

is the average of many p values in that state , there can even be positive momentum's or negative momentum's as reference to the direction (in mathematical perspective),So

=0 but for p square it's no 0