Using the complex representation to find scattering amplitudes

Now that we are able to write down the forms (8) and (10) directly, it is then a relatively simple matter to compute scattering amplitudes such as $T_{0 \rightarrow 1}$ and $R_{0\rightarrow 1}$ directly from the boundary conditions. For notational convenience, in this section we shall drop the writing of the subscripts `` $0 \rightarrow 1$'' and refer to these amplitudes as $T$ and $R$, respectively. Eqs. (8,10) then become

$$s_0(x \le 0,t) = {\mbox{Re\,}}\left(e^{-iwt} \underline{A} \left[ e^{ik_0x}+R e^{-ik_0x} \right] \right)$$ (13)

$$s_1(x \ge 0,t) = {\mbox{Re\,}}\left(e^{-iwt} \underline{A} \left[ T e^{ik_1x} \right] \right)$$

Substituting (13) into the consistency condition, $s_0(x=0,t)=s_1(x=0,t)$, we find

$$\begin{eqnarray*} {\mbox{Re\,}}\left(e^{-iwt} \underline{A} \left[ e^{ik_0 \cdot... ...box{Re\,}}\left(e^{-iwt} \underline{A} \left[ T \right] \right), \end{eqnarray*}$$

so that

$${\mbox{Re\,}}\left(e^{-iwt} \underline{A} \left[1+R-T\right] \right) = 0.$$

Now, ${\mbox{Re\,}}\left( e^{-i\omega t} \underline{Q} \right)$ can only be zero for all times $t$ if the harmonic motion which it represents has zero amplitude, which implies $\underline{Q}=0$. We thus conclude $1+R-T=0$, so that

$$1+R = T$$ (14)

Next, substituting (13) into the force balance condition, $B_0 \partial s_0(x=0,t)/\partial x=B_1 \partial s_1(x=0,t)/\partial x$, we find

$$\begin{eqnarray*} B_0 {\mbox{Re\,}}\left(e^{-iwt} \underline{A} \left[ ik_0 e^{i... ...\left(e^{-iwt} \underline{A} B_1 \left[ ik_1 T \right] \right), \end{eqnarray*}$$

where the factors of $i k$ come down as we take the derivatives with respect to $x$ before substituting $x=0$. Combining terms we find

$${\mbox{Re\,}}\left(e^{-iwt} \underline{A} i \left[B_0 k_0(1-R)-B_1 k_1 T\right]\right)=0,$$

for all times $t$. Following the same logic which led to (14), we find

$$B_0 k_0 (1-R) = B_1 k_1 T.$$

Using the fact that $B k=B (\omega/c)=Z \omega$, this simplifies to

$$Z_0 \omega (1-R) = Z_1 \omega T$$

$$Z_0 (1-R) = Z_1 T.$$ (15)

Finally, to find the reflection and transmission amplitudes, we combine (14) and (15). To find $T$, we take $Z_0 \times$(14)$+$(15):

$$2 Z_0 = (Z_0 + Z_1) T.$$

This gives precisely our previous result,

$$T=\frac{2 Z_0}{Z_0+Z_1}.$$

And, to find $R$, we take $Z_1 \times$(14)$-$(15):

$$(Z_1-Z_0)+(Z_1+Z_0) R = 0.$$

Again, we find our previous result,

$$R=\frac{Z_0-Z_1}{Z_0+Z_1}.$$