Question

CALC Normalization of the Wave Function. Consider a particle moving in one dimension, which we shall call the x-axis. (a) What does it mean for the wave function of this particle to be normalized? (b) Is the wave function $\mathit{\psi }\left(x\right)\mathbf{=}{\mathit{e}}^{\mathbf{a}\mathbf{x}}$where is a positive real number, normalized? Could this be a valid wave function? (c) If the particle described by the wave function $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}\mathbf{b}\mathbf{x}}$where and are positive real numbers, is confined to the range$\mathit{x}\mathbf{\ge }\mathbf{0}$ determine A (including its units) so that the wave function is normalized.

Short answer

1. From the expression ${\int }_{-\infty }^{\infty }\left|\psi \right(x){|}^{2}dx=1$, the wave functions are normalized means that the total probability of finding the particle somewhere is unity i.e., $\int |\psi {|}^{2}dx=1$.
2. The function$\psi \left(x\right)=e^{2ax}$ is not normalized and not a valid wave function.

c. The wave-function $A=\sqrt{2b}$ is normalized the A of is $m^{-1/2}$.

Step-by-step solution

Define the wave function and normalization.

The wave function $\psi \left[x\right]$and its derivative$\psi \left[x\right]/dx$must be continuous everywhere except where the potential-energyfunction has an infinite discontinuity U(x). Wave functions are normalized so that the total probability of finding the particle somewhere is unity.

The normalization condition is:

Explain the meaning of the wave function of the particle to be normalized and find whether the wave-function ψ(x)=eax is normalized or noy.

From the expression ${\int }_{\infty }^{\infty }\left|\psi \right(x){|}^{2}dx=1$, the wave functions are normalized means that the total probability of finding the particle somewhere is unity i.e., $\int |\psi {|}^{2}dx=1$.

The given function is $|\psi {|}^2=e^{2ax}$. Integrate the given function over the range of x.

$$\begin{gathered} {\int }_{-\infty }^{\infty }{e}^{2ax}dx={\left[\frac{e^{2ax}}{2a}\right]}_{-\infty }^{\infty } \\ =\infty \\ \ne 1 \end{gathered}$$

The function$\psi \left(x\right)=e^{2}ax$is not normalized.

Hence, for the function to be a valid wave-function it need be a normalized function.

Describe the wave function ψ(x)=Ae-bx.

The given function is$|\psi {|}^2=A^2{e}^{-2bx}$. Integrate the given function over the range of x.

$$\begin{gathered} {\int }_0^{\infty }{A}^2{e}^{-2bx}dx=\frac{A^2}{-2b}{\left[e^{-2bx}\right]}_0^{\infty } \\ =\frac{A^2}{-2b} \end{gathered}$$

The function$|\psi {|}^2=A^2{e}^{-2bx}$is normalized. So,

$$\begin{gathered} \frac{A^2}{2b}=1 \\ \therefore A=\sqrt{2b} \end{gathered}$$

Hence, for $A=\sqrt{2b}$the unit of b be $m^{-1}$and unit of A is $m^{-1/2}$.