Question

Question: (a) Write the wave function $\mathbf{\psi }\left(x\right)$displayed in Eq.38-27 in

the form $\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{+}\mathbf{ib}$, where $\mathbf{a}$and $\mathbf{b}$are real quantities. (Assume

that ${\mathbf{\psi }}_{\mathbf{0}}$is real.) (b) Write the time-dependent wave function $\mathbf{\psi }\left(x,t\right)$that corresponds to $\mathbf{\psi }\left(x\right)$ written in this form.

Short answer

(a) $\psi \left(x\right)={\psi }_0\cos kx+i{\psi }_0\sin kx$is $\psi \left(x\right)=a+ib$form of wave function where

$a={\psi }_0\cos kx$and $b={\psi }_0\sin kx$.

(b) The time-dependent function $\psi \left(x,t\right)$is

$\psi \left(x\right)={\psi }_0\cos \left(kx-\omega t\right)+i{\psi }_0\sin \left(kx-\omega t\right)$

Step-by-step solution

Identifying the data given in the question.

The wave function given in Eq.38-27 is

$\psi \left(x\right)=A{e}^{ikx}$

${\psi }_0$ is real

Concept used to solve the question.

A matter wave can be described by a wave function $\mathbf{\psi }\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}$,

which can be separated into a space-dependent part $\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and a time-dependent part ${\mathbf{e}}^{\mathbf{i\omega t}}$, where $\mathbf{\omega }$ is the angular frequency associated with the wave.

We can convert wave function into$\mathbf{a}\mathbf{+}\mathbf{ib}$ using Euler’s formula

Euler’s formula

${\mathbf{e}}^{\mathbf{i\varphi }}\mathbf{=}\mathbf{cos\varphi }\mathbf{+}\mathbf{isin\varphi }$

(a) Writing ψ(x) into ψ(x)=a+ib form.

Given wave function is,

$\psi \left(x\right)=A{e}^{ikx}$

It is given$x=0,\psi \left(x\right)={\psi }_0$

Therefore,

$$\begin{gathered} \psi \left(0\right)=A{e}^{ik\times 0} \\ {\psi }_0=A \end{gathered}$$

So, we can write the wave function as

$\psi \left(x\right)={\psi }_0{e}^{ikx}$

Now applying Euler’s formula,

$$\begin{gathered} \psi \left(x\right)={\psi }_0\left(\cos kx+i\sin kx\right) \\ \psi \left(x\right)={\psi }_0\cos kx+{\psi }_{0}i\sin kx \end{gathered}$$

$\psi \left(x\right)={\psi }_0\cos kx+i{\psi }_0\sin kx$is $\psi \left(x\right)=a+ib$form of wave function where

$a={\psi }_0\cos kxanda={\psi }_0\sin kx$

(b) Writing the time-dependent wave function ψ(x,t)   

The wave function is

$\psi \left(x\right)={\psi }_0{e}^{ikx}$

The time-dependent part can be given as

$\psi \left(x,t\right)=\psi \left(x\right)e^{i\omega t}$

Now substituting wave function

$\psi \left(x,t\right)={\psi }_0{e}^{ikx}{e}^{-i\omega t}={\psi }_0{e}^{i\left(kx-wt\right)}$

Now applying Euler’s formula

$$\begin{gathered} \psi \left(x,t\right)={\psi }_0\left(\cos \left(kx-\omega t\right)+i\sin \left(kx-\omega t\right)\right) \\ \psi \left(x,t\right)={\psi }_0\cos \left(kx-\omega t\right)+{\psi }_{0}i\sin \left(kx-\omega t\right) \end{gathered}$$

Hence, the time-dependent function $\psi \left(x,t\right)$is

$\psi \left(x,t\right)={\psi }_0\cos \left(kx-\omega t\right)+{\psi }_{0}i\sin \left(kx-\omega t\right)$