Question

Suppose we put $\mathit{A}\mathbf{=}\mathbf{0}$in Eq. 38-24 and relabeled $\mathit{B}$as localid="1664290358337" ${\mathit{\psi }}_{\mathbf{0}}$.

(a) What would the resulting wave function then describe?

(b) How, if at all, would Fig. 38-13 be altered?

Short answer

(a) The new wave function is $\psi (x,t)={\psi }_0{e}^{-i(kx+\omega t)}$.

(b) The figure is not altered.

Step-by-step solution

Concept and Formula used:

The wave function of a quantum mechanical system is controlled by the Schrödinger equation, which is a linear partial differential equation.

The general solution of Schrodinger’s equation is given into the book as Eq. 38-24.

$\frac{d^2\psi }{d{x}^2}+k^2\psi =0$ ….. (1)

$\psi \left(x\right)=A{e}^{ikx}+B{e}^{-ikx}$ ….. (2)

Here, $\psi$is the wave function, $x$is the displacement, $k$is the wave number, $A$and $B$are the constants.

(a) Find a new wave function:

Putting $0$for $A$and ${\psi }_0$for $B$into equation (1).

The new wave function becomes (by adding time dependence ),

$\psi (x,t)={\psi }_0{e}^{-i(kx+\omega t)}$

Hence, the new wave function is $\psi (x,t)={\psi }_0{e}^{-i(kx+\omega t)}$.

(b) Explain alterations:

The new function describes a plane matter wave traveling in the negative x-direction.

An example of the actual particles that fit this description is a free electron with linear momentum

$\overrightarrow{p}=-\frac{hk}{2\pi }\widehat{i}$

And the kinetic energy is,

localid="1664290375009" $\begin{array}{rcl}K& =& \frac{p^2}{2m_e}\\ & =& \frac{h^2{k}^2}{8{\pi }^2{m}_e}\end{array}$

Hence, the figure is not altered since ${\left|\psi \right|}^2$is constant for all values of $x$.