Question

Find the transmission coefficient for the potential in problem 2.27

Short answer

The transmission coefficient for the potential is$\mathrm{T}={\left|\frac{\mathrm{F}}{\mathrm{A}}\right|}^2=\frac{8{\mathrm{g}}^4}{(8{\mathrm{g}}^4+4{\mathrm{g}}^2+1)+(4{\mathrm{g}}^2-1)\mathrm{cos\varphi }-4\mathrm{gsin\varphi }}$

Step-by-step solution

 Define the Schrodinger equation

A differential equation that 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.

Define the transmission coefficient

The boundary conditions

$\mathrm{\psi }\left(\mathrm{x}\right)=\left\{\begin{array}{ll}{\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}& (\mathrm{x}<-\mathrm{a})\\ {\mathrm{Ce}}^{\mathrm{ikx}}+{\mathrm{De}}^{-\mathrm{ikx}}& (-\mathrm{a}<\mathrm{x}<\mathrm{a})\\ {\mathrm{Fe}}^{\mathrm{ikx}}& (\mathrm{a}>\mathrm{x})\end{array}\right.$

Using the continuity at$-a$

$A{e}^{ika}+B{e}^{-ika}=C{e}^{-ika}+D{e}^{ika}$

Let $\beta =e^{-2ika}$

So $\beta A+B=\beta C+D$ …(i)

Using the continuity at$+a$

$C{e}^{ika}+D{e}^{-ika}=F{e}^{ika}$

So $F=C+\beta D$ …(ii)

Using the discontinuity of ${\psi }^{\text{'}}$at$-a$

$ik(C{e}^{-ika}-D{e}^{ika})-ik(A{e}^{-ika}-B{e}^{ika})=\frac{-2m\alpha }{{\hslash }^2}(A{e}^{-ika}+B{e}^{ika})$

Let $\gamma =i2m\alpha /{\hslash }^{2}k$

So $\beta C-D=\beta (\gamma +1)A+B(\gamma -1)$ … (iii)

Using the discontinuity of ${\psi }^{\text{'}}$at$+a$

$ikF{e}^{ika}-ik(C{e}^{ika}-D{e}^{-ika})=\frac{-2m\alpha }{{\hslash }^2}\left(F{e}^{ika}\right)$

So $C-\beta D=(1-\gamma )F$ …(iv)

Adding (2) and (4)

$2C=F+(1-\gamma )F$

So $2C=(2-\gamma )F$

Subtract (ii) and (iv)

$2\beta D=F-(1-\gamma )F$

so , $2D=(\gamma /\beta )F$

Add (i) and (iii)

$2\beta C=\beta A+B+\beta (\gamma +1)A+B(\gamma -1)$

So$2C=(\gamma +2)A+(\gamma /\beta )B.$

Determine the transmission coefficient

Equation 2C in the equations

$(2-\gamma )F=(\gamma +2)A+(\gamma /\beta )B$ …(v)

Equation 2D in the equations

$(\gamma /\beta )F=-\gamma \beta A+(2-\gamma )B$ …(vi)

$\beta {(2-\gamma )}^{2}F=\beta (4-{\gamma }^2)A+\gamma (2-\gamma )B$

$({\gamma }^2/\beta )F=-\beta {\gamma }^{2}A+\gamma (2-\gamma )B$

$[\beta {(2-\gamma )}^2-{\gamma }^2/\beta ]F=\beta [4-{\gamma }^2+{\gamma }^2]A=4\beta A$

So $\frac{\mathrm{F}}{\mathrm{A}}=\frac{4}{(2-{\mathrm{\gamma }}^2)-{\mathrm{\gamma }}^2/{\mathrm{\beta }}^2}$

Let $g=i/\gamma =\frac{{\hslash }^{2}k}{2m\alpha }$ and $\varphi =4kA$

So$\gamma =\frac{i}{g},{\beta }^2=e^{-i\varphi }$

$\frac{F}{A}=\frac{4g^2}{{(2g-i)}^2+e^{i\varphi }}$

The Denominator: $4g^2-4ig-1+\cos \varphi +i\sin \varphi =(4g^2-1+\cos \varphi )+i(\sin \varphi -4g)$

${\left[\text{TheDenominator}\right]}^2={(4g^2-1+\cos \varphi )}^2-{(\sin \varphi -4g)}^2$

$=16g^4+1+{\cos }^2\varphi -8g^2-2\cos \varphi +8g^2\cos \varphi +\sin 2\varphi -8g\sin \varphi +16g^2$

$=16g^4+8g^2+2+(8g^2-2)\cos \varphi -8g\sin \varphi .$

Therefore, the transmission coefficient for the potential is

$T={\left|\frac{F}{A}\right|}^2=\frac{8g^4}{(8g^4+4g^2+1)+(4g^2-1)\cos \varphi -4g\sin \varphi }$.