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Why can't the expectation value of momentum be computed over some finite interval of space? Something like, $$ \int_a^b \psi^* \hat{p}\psi ~\mathrm{d}x.\tag{1}$$ I understand that usually we compute expectation value over all space, but does the above quantity mean anything? Also, I am not assuming that $[a,b]$ is the width of some box that the particle is confined to.

Qmechanic
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    Because that isn't the definition of expectation value? By definition, the expectation value of an operator is the value that one would obtain if one repeated the same experiment on infinite identical copies of the same system and took the average of all the results. Given this, $\langle p \rangle$ is defined as integral over all space. Now, if the wave-function itself is non-zero only in the range $(b,a)$, then it would be what you said above. – Prahar Feb 18 '17 at 05:10

2 Answers2

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  1. Technically, OP's eq. (1) is the expectation value $$ \langle \hat{A} \rangle ~:=~ \int_{\mathbb{R}} \!\mathrm{d}x~\psi^{\ast} \hat{A}\psi \tag{i}$$ of a non-Hermitian operator $$ \hat{A}~:=~ 1_{[a,b]}(\hat{x})\hat{p}.\tag{ii}$$ Here $x\mapsto 1_{[a,b]}(x)$ denotes the characteristic/indicator function for the interval $[a,b]\subseteq \mathbb{R}$. We assume $a<b$.

  2. $\hat{A}$ is non-Hermitian, since $\hat{x}$ and $\hat{p}$ do not commute $$ [\hat{x},\hat{p}]~=~i\hbar~{\bf 1}.\tag{iii} $$

  3. Formally, to get a Hermitian operator/observable, consider e.g. instead the symmetrized operator $$ \hat{B}~:=~\frac{1}{2}\left\{1_{[a,b]}(\hat{x}),~\hat{p}\right\}_+ ~:=~\frac{1}{2}\left(1_{[a,b]}(\hat{x}) \hat{p}+\hat{p}1_{[a,b]}(\hat{x})\right) ~=~\hat{A} +\frac{i\hbar}{2}\left(\delta(\hat{x}\!-\!b)-\delta(\hat{x}\!-\!a)\right).\tag{iv} $$

  4. If there exist some boundary conditions such that $$\psi(a)~=~0~=~\psi(b), \tag{v}$$ (e.g. because of an infinite potential at $x=a$ and $x=b$), then the operators $\hat{A}$ and $\hat{B}$ are effectively the same. \psi

  5. Fun fact: If the wave function $\psi\in\mathbb{R}$ is differentiable, real, and vanishes $\psi(\pm\infty)=0$ at infinity, then the expectation value vanishes $$ \langle \hat{B} \rangle~\stackrel{(\text{iv})}{=}~ \frac{1}{2}\int_a^b \!\mathrm{d}x\left(\psi^{\ast} (\hat{p}\psi) + (\hat{p}\psi)^{\ast}\psi\right)~\stackrel{(\text{vii})}{=}~\int_a^b \!\mathrm{d}x ~mj~=~0, \tag{vi}$$ where $$j~:=~\frac{1}{2m}\left(\psi^{\ast} \hat{p}\psi-\psi \hat{p}\psi^{\ast}\right)~=~0\tag{vii}$$ is the probability current.
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Qmechanic
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  • Thanks so much for your answer. Could you please tell me what $1_{[a,b]}(\hat{x})$? Is it $\Theta(\hat{x} -a) - \Theta(\hat{x}-b)$? Where $\Theta$ is the step function. – play Feb 18 '17 at 14:54
  • $\uparrow$ Yes. – Qmechanic Feb 18 '17 at 15:12
  • If I compute $\langle \hat{B} \rangle$ for $\psi(x,0)=e^{-x^2}$ I get zero* for any $[a,b]$ (Of course this is a wave packet with zero mean momentum). Does $\hat{B}$ have any physical interpretation, for e.g., measuring momentum only in a fixed region of space and then calculating the momentum average over many measurements.

    *Hopefully I did the calculation correctly.

     – play Feb 18 '17 at 18:26
  • Okay, thinking about this a bit, $\hat{B}$ will always give zero because of integration by parts, at least when $\psi$ is real. – play Feb 18 '17 at 22:56
  • $\uparrow$ Right, when $\psi$ is real – Qmechanic Feb 19 '17 at 00:20
  • Okay I feel like I have come around in a circle because the integrand in (vi) is the probability current of a unit mass particle. – play Feb 19 '17 at 01:51
  • Apologies for making this thread so long. There is an extra factor of 2 in (vi) after the equality (vii). – play Feb 19 '17 at 17:52
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A wavefunction gives you the probability of a particle to be seen somewhere. If it's not confined to be in some region, like a potential well or something, then it has a probability to be seen anywhere in the space. That's why the integration runs from $-\infty$ to $+\infty$ (remember that you could never pinpoint a quantum mechanical particle in space). That can be converted to a finite volume of space if the confinement is strong. For example, in the case of a particle in an infinite potential well, the boundary condition is that the wavefunction should vanish at the potential well boundaries.

But, in reality, there are no such infinite potential traps where you can confine a particle to. All potentials available are finite. So, the condition that the wavefunction should vanish at the boundary is no longer valid, which means that, the wavefunction could spread out of the boundary (called quantum tunneling). Such wavefunctions are called unnormalized. That's why you integrate over the entire space for getting the probability and hence the expectation value of the particle.

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