If a particle in a stationary state is bound, the expectation value of its momentum must be 0.

(a). In words, why?

(b) Prove it.

Starting from the general expression(5-31) with $\hat{\mathbf{p}}$in the place of $\overbrace{\mathbf{Q}}$, integrate by parts, then argue that the result is identically 0. Be careful that your argument is somehow based on the particle being bound: a free particle certainly may have a non zero momentum. (Note: Without loss of generality,$\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$ may be chosen to be real.)

The expectation value of momentum is 0.

A bound state is a special quantum state of a particle in which the particle has tendency to remain localized in one dimensional.

A stationary state is a state which is independent of time.

A stationary state is a state in which the particle's motion is independent of time. The particle moves independently with time. Stationary states have a particular value of energy for a particular state.

A bound state is a state in which a particle possesses a definite value of potential. The particle in the bound state moves in one-dimensional space. That means the accuracy in the position is maximum. The accuracy in the measurements for momentum will be 0 according to the uncertainty principle.

The particle moving in a stationary state possesses zero value of expectation value of momentum. If the expectation value is not zero, it will not possess the characteristics of stationary states. The motion of particles will not be constrained.

Thus, the given statement for the stationary state has been proved.

The obtained expectation value of the momentum is 0.

The expression for expectation value of momentum.

$\overline{p}={\int }_{\text{complete}}{\psi }^{*}\left(x\right)\hat{p}\psi \left(x\right)dx$ ...............( 1 )

Here,${\psi }^{*}\left(x\right)$is the complex conjugate of the wave function and$\psi \left(x\right)$is wave function.

The wave function is real in this case,meaning the complex number's complex conjugate will not change.

Substitute $-i\hslash \frac{\partial }{dx}$for $\hat{p}$and $\psi \left(x\right)$for ${\psi }^{*}\left(x\right)$in equation (1), and we get,

localid="1660113671504" $$\begin{gathered} \overline{p}={\int }_{-\infty }^{\infty }\psi \left(x\right)\left(-i\hslash \frac{\partial }{\partial x}\right)\psi \left(x\right)dx \\ \overline{p}={\left.(-i\hslash ){\left(\psi \right(x\left)\right)}^2\right|}_{-\infty }^{+\infty }-{\int }_{-\infty }^{\infty }\left(-i\hslash \frac{\partial }{\partial x}\psi \left(x\right)\right)\psi \left(x\right)dx \\ p={\left.\left(-i\hslash \right){\left(\psi \left(x\right)\right)}^2\right|}_{-\infty }^{+\infty }-p \\ 2\overline{p}={\left.(-i\hslash ){\left(\psi \right(x\left)\right)}^2\right|}_{-\infty }^{+\infty } \end{gathered}$$

Substitute 0 for$\psi \left(x\right)$as the state is bound for limits from $-\infty$to$+\infty$ in the above equation, and we get,

$$\begin{gathered} 2\overline{p}=0 \\ \overline{p}=0 \end{gathered}$$

Thus, the expectation value of momentum is 0.