Derive Equations 2.167 and 2.168.Use Equations 2.165 and 2.166 to solve C and D in terms of F:

$\mathit{C}\mathbf{=}\left(sin\left(la\right)+\frac{ik}{l}cos\left(la\right)\right){\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{a}}\mathit{F}$_;$\mathit{D}\mathbf{=}\left(cos\left(la\right)-\frac{ik}{l}sin\left(la\right)\right){\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{a}}\mathit{F}$

Plug these back into Equations 2.163 and 2.164. Obtain the transmission coefficient and confirm the equation 2.169

A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

$C\sin \left(la\right)+D\cos \left(la\right)=F{e}^{ika}$ …(i)

$l\left(C\cos \right(la)-D\sin (la\left)\right)=ikF{e}^{ika}$

$C\cos \left(la\right)-D\sin \left(la\right)=\frac{ik}{l}F{e}^{ika}$ …(ii)

$k=\sqrt{\frac{2mE}{h^2}}$ $l=\sqrt{\frac{2m(E=V_0)}{h^2}}$

$C{\sin }^2\left(la\right)+D\cos \left(la\right)\sin \left(la\right)=F{e}^{ika}\sin \left(la\right)$Multiply both sides $\sin \left(la\right)$

$C{\cos }^2\left(la\right)-D\cos \left(la\right)\sin \left(la\right)=\frac{ik}{l}F{e}^{ika}\cos \left(la\right)$Multiply both sides $\cos \left(la\right)$

Add the above 2 equations

$C=F\left(\sin \left(la\right)+\frac{ik}{l}\cos \left(la\right)\right)e^{ika}$ …(iii)

Now equation (i) multiply withwe get an equation (ii) multiply with $\sin \left(la\right)$

$C\sin \left(la\right)\cos \left(la\right)+D{\cos }^2\left(la\right)=F\sin \left(la\right)e^{ika}$

$C{\cos }^2\left(la\right)-D\cos \left(la\right)\sin \left(la\right)=\frac{ik}{l}F{e}^{ika}$

Now add the above equation

$D=F\left(\cos \left(la\right)-\frac{ik}{l}\sin \left(la\right)\right)e^{ika}$ …(iv)

Substitute equations (iii) and (iv)

$A{e}^{-ika}+B{e}^{ika}=-C\sin \left(la\right)+D\cos \left(la\right)$

Put values of $C$and $D$

$A{e}^{-ika}+B{e}^{ika}=-F\left(\sin \right(la)+\frac{ik}{l}\cos (la\left)\right)e^{ika}\left(\sin \right(la\left)\right)+F\left(\cos \right(la)-\frac{ik}{l}\sin (la\left)\right)e^{ika}\cos \left(la\right)$

$A{e}^{-ika}+B{e}^{ika}=F\left[\cos \left(2la\right)-\frac{ik}{l}\sin \left(2la\right)\right]e^{ika}$ …(v)

$ik[A{e}^{-ika}-B{e}^{ika}]=l\left[C\cos \left(la\right)-\frac{ik}{l}\sin \left(la\right)\right]$

$ik[A{e}^{-ika}-B{e}^{ika}]=\frac{l}{ik}F\left(\sin \right(la)+\frac{ik}{l}\cos (la\left)\right)e^{ika}\left(\cos \right(la\left)\right)+F\left(\cos \right(la)-\frac{ik}{l}\sin (la\left)\right)e^{ika}\sin \left(la\right)\left)\right]$

role="math" localid="1656056140172" $ik\left[A{e}^{-ika}-B{e}^{ika}\right]=F\left[\cos \left(2la\right)-\frac{il}{k}\sin \left(2la\right)\right]e^{ika}$ …(vi)

$2B{e}^{ika}=F\left[\cos \left(2la\right)-\frac{ik}{l}\sin \left(2la\right)\right]e^{ika}-F\left[\cos \left(2la\right)-\frac{il}{k}\sin \left(2la\right)\right]e^{ika}$

$2B{e}^{ika}=iF\left(\frac{l^2-k^2}{lk}\right)\sin \left(2la\right)e^{ika}$

$B=iF\left(\frac{l^2-k^2}{2lk}\right)\sin \left(2la\right)$

$2A{e}^{-ika}=F\left\{2\cos \left(2la\right)-l\left(\frac{k}{l}+\frac{l}{k}\right)\sin \left(2la\right)\right\}{e}^{ika}$

$F=\frac{A{e}^{-2ika}}{\cos \left(2la\right)+i\left(\frac{k^2+l^2}{2kl}\right)\sin \left(2la\right)}$

$\frac{F}{A}=\frac{e^{-2ika}}{\cos \left(2la\right)+i\left(\frac{k^2+l^2}{2kl}\right)\sin \left(2la\right)}$

$T={\left(\frac{F}{A}\right)}^2$

role="math" localid="1656056446552" $T={\left(\frac{e^{-2ika}}{\cos \left(2la\right)+i\left(\frac{k^2+l^2}{2kl}\right)\sin \left(2la\right)}\right)}^2$

$T=\left(\frac{e^{-2ika}}{\cos \left(2la\right)+i\left(\frac{k^2+l^2}{2kl}\right)\sin \left(2la\right)}\right)\left(\frac{e^{2ika}}{\cos \left(2la\right)-i\left(\frac{k^2+l^2}{2kl}\right)\sin \left(2la\right)}\right)$

By putting ${\cos }^2\left(2la\right)=1-{\sin }^2\left(2la\right)$

$T=\frac{1}{1+{\sin }^2\left(2la\right)\left(\frac{{(k^2-l^2)}^2}{4k^2{l}^2}\right)}$

$T=\frac{1}{1+{\sin }^2\left(2a\sqrt{\frac{2m(E-V_0)}{h^2}}\right)\left(\frac{{V_0}^2}{4E(E+V_0)}\right)}$

Where

$l=\sqrt{\frac{2m(E-V_0)}{h^2}},k=\sqrt{\frac{2mE}{h^2}}$