Question

does the wave function have a well-defined $\mathit{\psi }\left(x\right)\mathbf{=}\mathit{A}\left(e^{ikx}+e^{-ikx}\right)$momentum? Explain.

Short answer

The constant in the front multiplies the initial function and two opposing moving plane waves are added, so the momentum is not clearly defined.

Step-by-step solution

Identification of given data

• The wave function is$\psi \left(x,t\right)=A\left(e^{ikx}+e^{-ikx}\right).$

Concept/significance of Eigenvector

The eigenvectors of a linear transformation are those vectors where the transformation only modifies the magnitude of the vectors and not the angle. The "eigenvalue" of vectors refers to the ratio by which they differ from the original, and the vectors it works with are referred to as the "eigenvectors" corresponding to the eigenvalue.

Determination ofthe wave function has well-defined momentum: 

The momentum operator is:

$\hat{p}\psi \left(x,t\right)=-i\hslash \frac{\partial }{\partial x}\psi \left(x,t\right).$

Replace the value of the wave function in the above equation.

$\begin{array}{rcl}\hat{p}\psi \left(x,t\right)& =& -i\hslash \frac{\partial }{\partial x}A\left(e^{ikx}+e^{-ikx}\right)\\ & =& -i\hslash A\left(ik{e}^{ikx}-ik{e}^{-ikx}\right)\\ & =& -i^2\hslash kA\left(e^{ikx}-e^{-ikx}\right)\\ & =& \hslash kA\left(e^{ikx}-e^{-ikx}\right).\\ & & \end{array}$

Hence, the constant in the front multiplies the initial functionand two opposing moving plane waves are added, so the momentum is not clearly defined.