Determine the transmission coefficient for a rectangular barrier (same as Equation 2.148, only with V(x) = + V0 0 in the region -a

Question

Determine the transmission coefficient for a rectangular barrier (same as Equation 2.148, only with $V(x) = + V_0 > 0$ in the region -a < x < a). Treat separately the three cases $E < V_0 , E = V_0$ and $E >V_0$ . (note that the wave function inside the barrier is different in the three cases). Partial answer: for $E < V_0$ ,

$V(x)= \begin{cases}-V_{0}, & -a \leq x \leq a \\ 0, & |x|>a\end{cases}$ (2.148).

$T^{-1}=1+\frac{V_{0}^{2}}{4 E\left(V_{0}-E\right)} \sinh ^{2}\left(\frac{2 a}{\hbar} \sqrt{2 m\left(V_{0}-E\right)}\right)$ .

[latex] V(x)= \begin{cases}-V_{0}, & -a \leq x \leq a \ 0, & |x|>a\end{cases} [/latex] (2.148).

[latex] T^{-1}=1+\frac{V_{0}^{2}}{4 E\left(V_{0}-E ight)} \sinh ^{2}\left(\frac{2 a}{\hbar} \sqrt{2 m\left(V_{0}-E ight)} ight) [/latex].

Accepted Answer

[latex] E < V_0[/latex] : [latex] \psi=\left\{\begin{array}{ll} A e^{i k x}+B e^{-i k x} & (x<-a) \ C e^{\kappa x}+D e^{-\kappa x} & (-aa) \end{array} ight\} \quad k=\frac{\sqrt{2 m E}}{\hbar} ; \kappa=\frac{\sqrt{2 m\left(V_{0}-E ight)}}{\hbar} [/latex].

(1) Continuity of ψ at -a : [latex] A e^{-i k a}+B e^{i k a}=C e^{-\kappa a}+D e^{\kappa a} [/latex].

(2) Continuity of ψ′ at -a : [latex] i k\left(A e^{-i k a}-B e^{i k a} ight)=\kappa\left(C e^{-\kappa a}-D e^{\kappa a} ight) [/latex].

[latex] \Rightarrow 2 A e^{-i k a}=\left(1-i \frac{\kappa}{k} ight) C e^{-\kappa a}+\left(1+i \frac{\kappa}{k} ight) D e^{\kappa a} [/latex].

(3) Continuity of ψ at +a : [latex] C e^{\kappa a}+D e^{-\kappa a}=F e^{i k a} [/latex].

(4) Continuity of ψ′ at +a : [latex] \kappa\left(C e^{\kappa a}-D e^{-\kappa a} ight)=i k F e^{i k a} [/latex].

[latex] \Rightarrow 2 C e^{\kappa a}=\left(1+\frac{i k}{\kappa} ight) F e^{i k a} ; \quad 2 D e^{-\kappa a}=\left(1-\frac{i k}{\kappa} ight) F e^{i k a} [/latex].

[latex] 2 A e^{-i k a}=\left(1-\frac{i \kappa}{k} ight)\left(1+\frac{i k}{\kappa} ight) F e^{i k a} \frac{e^{-2 \kappa a}}{2}+\left(1+\frac{i \kappa}{k} ight)\left(1-\frac{i k}{\kappa} ight) F e^{i k a} \frac{e^{2 \kappa a}}{2} [/latex].

[latex] =\frac{F e^{i k a}}{2}\left\{\left[1+i\left(\frac{k}{\kappa}-\frac{\kappa}{k} ight)+1 ight] e^{-2 \kappa a}+\left[1+i\left(\frac{\kappa}{k}-\frac{k}{\kappa} ight)+1 ight] e^{2 \kappa a} ight\} [/latex].

[latex] =\frac{F e^{i k a}}{2}\left[2\left(e^{-2 \kappa a}+e^{2 \kappa a} ight)+i \frac{\left(\kappa^{2}-k^{2} ight)}{k \kappa}\left(e^{2 \kappa a}-e^{-2 \kappa a} ight) ight] [/latex].

But [latex] \sinh x \equiv \frac{e^{x}-e^{-x}}{2}, \cosh x \equiv \frac{e^{x}+e^{-x}}{2} [/latex], so

[latex] =\frac{F e^{i k a}}{2}\left[4 \cosh (2 \kappa a)+i \frac{\left(\kappa^{2}-k^{2} ight)}{k \kappa} 2 \sinh (2 \kappa a) ight] [/latex].

[latex] =2 F e^{i k a}\left[\cosh (2 \kappa a)+i \frac{\left(\kappa^{2}-k^{2} ight)}{2 k \kappa} \sinh (2 \kappa a) ight] [/latex].

[latex] T^{-1}=\left|\frac{A}{F} ight|^{2}=\cosh ^{2}(2 \kappa a)+\frac{\left(\kappa^{2}-k^{2} ight)^{2}}{(2 \kappa k)^{2}} \sinh ^{2}(2 \kappa a) [/latex].

But [latex] \cosh ^{2}=1+\sinh ^{2} [/latex], so

[latex] T^{-1}=1+[\underbrace{1+\frac{\left(\kappa^{2}-k^{2} ight)^{2}}{(2 \kappa k)^{2}}}_{\star}] \sinh ^{2}(2 \kappa a)= 1+\frac{V_{0}^{2}}{4 E\left(V_{0}-E ight)} \sinh ^{2}\left(\frac{2 a}{\hbar} \sqrt{2 m\left(V_{0}-E ight)} ight) [/latex].

where [latex] \star=\frac{4 \kappa^{2} k^{2}+k^{4}+\kappa^{4}-2 \kappa^{2} k^{2}}{(2 \kappa k)^{2}}=\frac{\left(\kappa^{2}+k^{2} ight)^{2}}{(2 \kappa k)^{2}}=\frac{\left(\frac{2 m E}{\hbar^{2}}+\frac{2 m\left(V_{0}-E ight)}{\hbar^{2}} ight)^{2}}{4 \frac{2 m E}{\hbar^{2}} \frac{2 m\left(V_{0}-E ight)}{\hbar^{2}}}=\frac{V_{0}^{2}}{4 E\left(V_{0}-E ight)} [/latex].

(You can also get this from Eq. 2.172 by switching the sign of [latex] V_0 [/latex] and using [latex] \sin (i heta)=i \sinh heta [/latex]).

[latex] T^{-1}=1+\frac{V_{0}^{2}}{4 E\left(E+V_{0} ight)} \sin ^{2}\left(\frac{2 a}{\hbar} \sqrt{2 m\left(E+V_{0} ight)} ight) [/latex] (2.172).

[latex] E = V_0[/latex] : [latex] \psi=\left\{\begin{array}{ll} A e^{i k x}+B e^{-i k x} & (x<-a) \ C+D x & (-aa) \end{array} ight\} [/latex].

(In central region [latex] -\frac{\hbar^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+V_{0} \psi=E \psi \Rightarrow \frac{d^{2} \psi}{d x^{2}}=0, ext { so } \psi=C+D x [/latex].)

(1) Continuous ψ at -a : [latex] A e^{-i k a}+B e^{i k a}=C-D a [/latex].

(2) Continuous ψ at +a : [latex] F e^{i k a}=C+D a [/latex].

[latex] \Rightarrow(2.5) 2 D a=F e^{i k a}-A e^{-i k a}-B e^{i k a} [/latex].

(3) Continuous ψ′ at -a : [latex] i k\left(A e^{-i k a}-B e^{i k a} ight)=D [/latex].

(4) Continuous ψ′ at +a : [latex] i k F e^{i k a}=D [/latex].

[latex] \Rightarrow(4.5) A e^{-2 i k a}-B=F [/latex].

Use (4) to eliminate D in (2.5): [latex] A e^{-2 i k a}+B=F-2 a i k F=(1-2 i a k) F [/latex], and add to (4.5):

[latex] 2 A e^{-2 i k a}=2 F(1-i k a), ext { so } T^{-1}=\left|\frac{A}{F} ight|^{2}=1+(k a)^{2} = 1+\frac{2 m E}{\hbar^{2}} a^{2} [/latex].

(You can also get this from Eq. 2.172 by changing the sign of [latex] V_0 [/latex] and taking the limit [latex] E → V_0 [/latex] using [latex] \sin \epsilon \cong \epsilon [/latex].

[latex] T^{-1}=1+\frac{V_{0}^{2}}{4 E\left(E+V_{0} ight)} \sin ^{2}\left(\frac{2 a}{\hbar} \sqrt{2 m\left(E+V_{0} ight)} ight) [/latex] (2.172).

[latex] E > V_0[/latex] : This case is identical to the one in the book, only with [latex] V_{0} ightarrow-V_{0} [/latex] . So

[latex] T^{-1}=1+\frac{V_{0}^{2}}{4 E\left(E-V_{0} ight)} \sin ^{2}\left(\frac{2 a}{\hbar} \sqrt{2 m\left(E-V_{0} ight)} ight) [/latex].

Question 2.33: Determine the transmission coefficient for a rectangular bar...

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