Given that \$\alpha\$ and \$\beta\$ are related by \$\alpha = \frac{\beta}{1+\beta}\$ as stated in the wiki article, obviously you can do your sums with either.
However, which is going to be easier to use? I personally always use \$\beta\$, regardless of the transistor configuration.
In common emitter \$I_c = \beta\times I_b\$, so I can say 'I need to control \$I_c\$ collector current, I need at least \$\frac{I_c}{\beta}\$ of base current to do it'.
But as \$\beta >> 1\$ (for most transistors), \$\alpha \approx 1\$, and \$I_c \approx I_e\$. You may object to the approximation, but given the way that \$\beta\$ varies with temperature, \$I_c\$, and between transistors of the same type, that is a far far better approximation than insisting that \$\beta\$ is constant. Any good transistor design will allow for operation with a range of \$\beta\$, at least \$2:1\$, preferably more.
Once you have made the approximation \$I_c \approx I_e\$, then common collector operation is given by 'I need to allow for a base current of \$\frac{I_c}{\beta}\$ to flow in the base circuit, without upsetting operation'.
With a common base stage, you say much the same thing, allowing an amount of base current, however you also say that the emitter to collector gain is slightly less than \$1\$, a fraction of \$\frac{1}{\beta}\$ less than one. The error of the gain from \$1\$ will usually be a smaller error than resistor tolerances and other sources of gain error.
Given that you can write an equation for \$\alpha\$, does that mean that you need to? For most practical engineering designs, the answer is no. If you are in college, and the tutor really likes to use \$\alpha\$, then the answer is yes.
Both quantities (\${h_{fe}}\$ and \${h_{ie}}\$) are linked to one another via the transconductance \${g_m}\$.
Derivation:
\${h_{fe}} = \frac{{{I_C}}}{{{I_B}}} = \frac{{d\left( {{I_C}} \right)}}{{d\left( {{I_B}} \right)}}\$,
with
\$d\left( {{I_B}} \right) = \frac{{d({V_{BE}})}}{{{h_{ie}}}}\$.
Therefore:
\$\frac{{{h_{fe}}}}{{{h_{ie}}}} = \frac{{d\left( {{I_C}} \right)}}{{d\left( {{V_{BE}}} \right)}} = {g_m}\$,
with
\${g_m} = \frac{{{I_C}}}{{{V_T}}}\$ (\${V_T}\$: temperature voltage)
(The transconductance \${g_m}\$ is identical to the slope of the \${I_C} = f\left(V_{BE}\right)\$ characteristics.)
Best Answer
\$Hybrid~ Model\$
more complex
\$h_{ie}=\dfrac{\delta V_{BE}}{\delta I_B} @Q_{point}\$ This is the small signal input impedance. This is a product of the dynamic current gain and many variables. \$h_{ie}\$ is normally in kΩ range for low linear currents. The transistor has both a DC current gain of \$\beta =H_{fe}\$ and a smaller AC current gain of \$h_{fe}\$ .
Simpler \$r_e~Model\$
-simpler model for DC to mid-frequency use with only 2 variables \$h_{fe}\$ and temperature.
Vt = γ = kT/q as the thermal voltage with temp. T [°K] and k=Boltzman's constant and charge, q
Vt = 26mV @ 30°C , 25mV @ 20°C
\$r_e = V_t/I_e~~~~\$ [Ω= mV / mA ] while \$r_e\$ is typically in the 10~70 ohm range.
since \$I_e=I_c+I_b ~,~~ ~ I_c=βI_b~,~~ I_e/I_b=(β+1)\$
\$h_{ie}=(β+1)r_e\$
ref's
Berkeley
Xi'an University of Electronic Science and Technology
Other stuff
Now if we were to add an emitter resistor Re=100Ω;
\$r_e=50,~~ Z_{in(f)} = β * [r_e+Ze(f)]=100 * (50+100)=15kΩ\$
But if Vce is saturated as a switch, β drops to 10-20 and Zin(f) drops to 10% of the above value.
When Vbe saturates this become more linear and depends on the PN junction size inverse to Pd of the diode junction. But we know that Ic doubles for every 17mV rise in Vbe if at constant temp.
But consider when Vbe is saturated and consider just diodes with no current gain.
I generalize this /$r_e~/$ as ESR~1/Pd so a 20mW junction is 50 Ohms, a 65mW LED as 16 Ohms, a 1W LED as 0.5 to 1 Ohm at 85'C but due to tolerances this is generally +/-25% typ and +/-50% worst case over temp.
Thermal design affects this as does temp rise, so generally SMD parts for ESR=k/Pd(max) k is drops from 1 to 0.5 and the very best thermal bulk power diodes are k=0.25 .
You won't find this in datasheets or textbooks, but it is my observation of doing above over thousands of devices.
The same can be done for the Rce value when a transistor is used as a switch.
Vce /Ice is saturated Vce(sat) @ I spec, then you can expect an Rce value similar to a power diode Rce = x/Pd but depends on design of chip and thermal capacity. so k has a wider tolerance and an offset that depends on doping differences of BE-CB, but x = 0.25 to 1 is a good fit from SMD power transistor to small signal plastic transistors.