Why is the momentum of the wave function in a 1-D particle-in-a-box = 0?

Is this implying that a particle in the box for 1-dimension has no momentum? Or is the momentum average out to $0$ on throughout the region $0\leq x\leq a$? For reference, I was told the state was $n = 1$, and assuming the wave equation is therefore
$$\psi(x) = \sin\left(\frac{\pi}{a}x\right).$$


It just doesn't intuitively make sense to me. I'm thinking it has to be more in line with my latter thought that the momentum throughout the box would end up equaling $0$. Does that mean inside any box, whether it be 1, 2, or 3 dimensional, the momentum would always average out to $0$, since it is closed in a system?


I am calculating $\langle\hat{p}\rangle$.





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> Why is the momentum of the wave function in a 1-D particle-in-a-box = 0?
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> 
> 


There are a couple ways of looking at this. One is to churn through the maths with the integrals, and you find that the expectation value comes out to zero:


$$\int\_0^a \psi^\* \left(-\mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}x}\right)\psi \,\mathrm{d}x = 0$$


Another is to search for a more meaningful physical interpretation of this, which we will try to do so now. First of all, note that when you say *the* wavefunction, you are actually only referring to one, specific, wavefunction for the system.


In introductory QM courses, you are typically exposed only to *stationary states*, which are states that satisfy the time-**independent** Schrödinger equation (TISE):


$$\hat{H}\psi = E\psi$$


Only a certain set of states, let's denote them by $\{\psi\_1, \psi\_2, \ldots, \psi\_n, \ldots \}$, obey this equation. You know already that these states are given by:


$$\psi\_n = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)$$


However, these are not the only possible states of the system. These are actually the set of **stationary states** of the system: stationary in the sense that they do not change with time, or in other words, they are time-independent (hence TISE).


Formally speaking, the expectation values of all operators are time-independent. The proof for this is not too hard and can be found in standard QM textbooks such as Griffiths (see also the second box in <https://en.wikipedia.org/wiki/Ehrenfest_theorem>).


For all stationary states, it is thus impossible to have a nonzero momentum because that would imply that the particle is moving, i.e. the *position* of the particle is changing with time. This would directly contradict the fact that such states are time-independent.


Thus, it turns out that for *all* the states $\psi\_n$ (regardless of what number $n$ is), the expectation value $\langle \hat{p}\rangle$ is always zero. This applies to any stationary state of any system, regardless of its dimension.




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> 
> [How can a system have nonzero $\langle \hat{p}\rangle$, then?]
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The system would need to be in a non-stationary state, i.e. a wavefunction that is *not* one of the $\psi\_n$'s. In general, such a wavefunction can be constructed by linear combination of the $\psi\_n$'s:


$$\psi = c\_1 \psi\_1 + c\_2 \psi\_2 + \cdots + c\_n\psi\_n + \cdots$$


Such a state no longer obeys the TISE, but that's not a problem, as we never set out to claim that it was a stationary state. These states will evolve in time: in particular, the coefficients $c\_i$ are time-dependent, and therefore the expectation value of various operators can also vary with time.


The time evolution of these states is governed by the time-**dependent** Schrödinger equation, instead:†


$$\mathrm{i}\hbar\frac{\partial \psi}{\partial t} = \hat{H}\psi.$$


I wouldn't worry about this too much for now, but I just wanted to point out that the stationary states are *special* states which almost by definition have $\langle\hat{p}\rangle = 0$, and that it is totally possible to construct states which do not have this property.




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> 
> Is this implying that a particle in the box for 1-dimension has no momentum?
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> 
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Having a zero *expectation value* for momentum is not the same as having zero momentum. When you say zero momentum, what you are suggesting is that whenever the momentum of the particle is measured, it will always return the value zero.


However, that's not the case here, actually. The truth is that whenever you measure the momentum of the particle, it can return pretty much *any* value. However, if you prepare many, many copies of the same system and measure each of their momenta, the average of these will come out to zero.


One way of expressing this is using the standard deviation as explained in Hans Wurst's answer. A nonzero standard deviation (or "uncertainty", as it's often called in QM) means that you will in general get nonzero momentum measurements.


This probabilistic nature of quantum mechanics can be difficult to grasp, but I think the best solution is to (re)read a textbook: virtually all textbooks spend a good amount of time explaining this.


We could think of this somewhat classically (but don't try to stretch this analogy too much). If you roll a six-sided die, what is the *expectation value*? Well, you can roll a 1, 2, 3, 4, 5, or 6 with equal probability, so the expectation value is


$$E = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$$


However, that doesn't mean that every time you roll a die, you'll get 3.5 (that's obviously impossible). Likewise, with a particle in a box (in a stationary state $\psi\_n$), each time you measure the momentum, it very well may not be zero; but if you do it lots of times, it will average to 0.




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**Footnotes**


\* Except for a global phase factor, $\exp(-\mathrm{i}E\_nt/\hbar)$, which is meaningless when it comes to expectation values because it cancels out in the $\psi^\*$ and $\psi$ terms whenever they are integrated.


† The time evolution of the stationary states is also governed by the TDSE, so it's not that it suddenly fails to apply. It's just that when you work it out, the stationary states have *no* time evolution.


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You can calculate the variance of an observable to quantify the spread of measurements around the expectation value. Only if the variance is zero, every measurement will return the same value equal to the expectation value.


We can calculate the variance via
$$\text{Var}(\hat p)\_{|\psi}= \langle \hat p^2\rangle \_{|\psi} -\langle \hat p\rangle^2\_{|\psi}$$
In your case we get for the properly normalized $\psi(x)=\sqrt{\frac{2}{a}}\sin(\frac{\pi}{a}x), 0\leq x \leq a$,
$$
\langle \hat p\rangle = \int ^a\_0 \psi^\*(x)\frac{\hbar }{i }\frac{d}{dx}\psi(x)dx=0\\
\langle \hat p^2\rangle = \int ^a\_0 \psi^\*(x)\left(-\hbar ^2\frac{d^2}{dx^2}\right)\psi(x)dx=\frac{2\pi^2\hbar^2}{a^2}
$$
The variance for this state is then
$$\begin{aligned}
\text{Var}(\hat p)\_{|\psi} &= \langle \hat p^2\rangle \_{|\psi} -\langle \hat p\rangle^2\_{|\psi}\\
\text{Var}(\hat p)\_{|\psi} &=\frac{2\pi^2\hbar^2}{a^2}-0 = \frac{2\pi^2\hbar^2}{a^2}\\
\sigma(\hat p)\_{|\psi} &= \sqrt{\text{Var}} = \frac{\pi\hbar }{a}\sqrt 2
\end{aligned}$$


where I've also calcualted the standard deviation $\sigma$.


As you can see the variance isn't zero. This means that a single momentum measurement doesn't have to be zero, even if the expectation value is zero.


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