How to plot the transmission coefficient as a function of #E//V_0#?

Here is the Excel sheet I made while doing this.

Here we follow what is readily found in McQuarrie & Simon (pp. 142-143). We found that:

#T = {([1 + [sinh^2[v_0^(1//2)(1 - r)^(1//2)]]/(4r(1 - r))]^(-1),0 < r < 1),([1 + [sin^2[v_0^(1//2)(r - 1)^(1//2)]]/(4r(r - 1))]^(-1),r >= 1):}#

where:

• #T# is the transmission coefficient (i.e. the fraction of particles that tunnel).
• #r = E/V_0 = epsilon/v_0# is the ratio of the particle energy to the barrier height.
• #v_0 = (2ma^2V_0)/ℏ^2# is a unitless variable for the height of the barrier.
• #epsilon = (2ma^2E)/ℏ^2# is a unitless variable for the energy of the particle.

and because I thought it would be interesting, besides #v_0 = 10#, I also graphed #v_0 = 10, 20, 40#:

It is rather interesting that the higher the barrier gets, the more values of #E# there are that allow for #100%# transmission. Here we have:

#ul(V_0" "" "" "" "" "E//V_0" "" "" "%T" ")#
#10ℏ^2//2ma^2" "color(white)(.)1.99" "" "" "" "100%#
#" "" "" "" "" "" "4.95#

#20ℏ^2//2ma^2" "color(white)(.)1.49" "" "" "" "#
#" "" "" "" "" "" "2.97#

#40ℏ^2//2ma^2" "color(white)(.)1.25" "" "" "" "#
#" "" "" "" "" "" "1.99#
#" "" "" "" "" "" "3.22#
#" "" "" "" "" "" "4.95#

It just so happens that at the classical limit, we have

#lim_(V_0 -> oo) T = 100%# for #E > V_0#,

with a rigid cutoff at #E = V_0#. That's just what we expect. Classically, particles can't tunnel unless they have enough energy. Here is the same graph at #v_0 = 10^6#:

DISCLAIMER: LONG AND RIGOROUS DERIVATION!

The potential is set up as:

#V(x) = {(0, x < 0),(V_0, 0 < x < a),(0, x > a):}#

We of course set #E < V_0# to allow tunnelling, and the eigenfunctions are:

#psi_I(x) = Ae^(ikx) + Be^(-ikx), " "x < 0#
#psi_(II)(x) = Ce^(k'x) + De^(-k'x), " "0 < x < a#
#psi_(III)(x) = Fe^(ikx) + Ge^(-ikx), " "x > a#

where #k = sqrt((2mE)/ℏ^2)# and #k' = sqrt((2m(V_0 - E))/ℏ^2)#.

If we restrict the particle to only come in from the left, #G = 0#. Then, the transmission coefficient is given by

#T = (|F|^2)/(|A|^2)#

The continuity conditions are:

#ul(psi_I(0) = psi_(II)(0))#:
#A + B = C + D# #" "" "" "" "" "" "bb((1))#

#ul((dpsi_I(0))/(dx) = (dpsi_(II)(0))/(dx))#:
#ik(A - B) = k'(C - D)# #" "" "" "bb((2))#

#ul(psi_(II)(a) = psi_(III)(a))#:
#Ce^(k'a) + De^(-k'a) = Fe^(ika)# #" "" "" "bb((3))#

#ul((dpsi_(II)(a))/(dx) = (dpsi_(III)(a))/(dx))#:
#k'Ce^(k'a) - k'De^(-k'a) = ikFe^(ika)##" "bb((4))#

Now, we take #ik(1) + (2)# to get:

#color(green)(2ikA = (ik + k')C + (ik - k')D)#

Then, take #k'(3) + (4)# to get #C# in terms of #F#:

#2k'Ce^(k'a) = (k' + ik)Fe^(ika)#

#color(green)(C = ((k' + ik)/(2k'))Fe^(ika)e^(-k'a))#

and take #k'(3) - (4)# to get #D# in terms of #F#:

#2k'De^(-k'a) = (k' - ik)Fe^(ika)#

#color(green)(D = ((k' - ik)/(2k'))Fe^(ika)e^(k'a))#

Plug it back into the first result to get:

#2ikA = (ik + k')((k' + ik)/(2k'))Fe^(ika)e^(-k'a) + (ik - k')((k' - ik)/(2k'))Fe^(ika)e^(k'a)#

#= [(ik - k')(k' - ik)e^(k'a) + (ik + k')(k' + ik)e^(-k'a)] (Fe^(ika))/(2k')#

#= [(k^2 - k'^2 + 2ikk')e^(k'a) + (k'^2 - k^2 + 2ikk')e^(-k'a)] (Fe^(ika))/(2k')#

#= [(k^2 - k'^2)(e^(k'a) - e^(-k'a)) + 2ikk'(e^(k'a) + e^(-k'a))] (Fe^(ika))/(2k')#

Now, we use the identities #2sinh(x) = e^x - e^(-x)# and #2cosh(x) = e^x + e^(-x)# to get:

#2ikA = [2(k^2 - k'^2)sinh(k'a) + 4ikk'cosh(k'a)] (Fe^(ika))/(2k')#

Solve for #F/A# to get:

#F/A = (2ik)/{[2(k^2 - k'^2)sinh(k'a) + 4ikk'cosh(k'a)]e^(ika)/(2k')}#

#= (4ikk'e^(-ika))/[2(k^2 - k'^2)sinh(k'a) + 4ikk'cosh(k'a)]#

Using the definition of #T#,

#T -= (F^"*"F)/(A^"*"A) = [(4ikk'e^(-ika))/[2(k^2 - k'^2)sinh(k'a) + 4ikk'cosh(k'a)]]^"*" [(4ikk'e^(-ika))/[2(k^2 - k'^2)sinh(k'a) + 4ikk'cosh(k'a)]]#

# = (4k^2k'^2)/((k^2 - k'^2)^2sinh^2(k'a) + 4k^2k'^2cosh^2(k'a))#

Next, use the identity that #cosh^2(x) = 1 + sinh^2(x)# to get:

#T = (4k^2k'^2)/((k^2 - k'^2)^2sinh^2(k'a) + 4k^2k'^2 + 4k^2k'^2sinh^2(k'a))#

#= (4)/(4 + (k^2 - k'^2)^2/(k^2k'^2)sinh^2(k'a) + 4sinh^2(k'a))#

#= (4)/(4 + (k^4 - 2k^2k'^2 + k'^4)/(k^2k'^2)sinh^2(k'a) + (4k^2k'^2)/(k^2k'^2) sinh^2(k'a))#

#= (4)/(4 + (k^4 + 2k^2k'^2 + k'^4)/(k^2k'^2)sinh^2(k'a))#

#= (4)/(4 + (k^2 + k'^2)^2/(k^2k'^2)sinh^2(k'a))#

Then, insert #k# and #k'# in terms of #E# and #V_0# to get:

#T = (4)/(4 + ((2mV_0)/ℏ^2)^2/(((2m)/ℏ^2)^2 E(V_0 - E))sinh^2(sqrt((2ma^2(V_0 - E))/ℏ^2)))#

#= 4/(4 + (V_0^2)/(E(V_0 - E))sinh^2(sqrt((2ma^2(V_0 - E))/ℏ^2)))#

Now, let the unitless quantities #v_0 = (2ma^2V_0)/ℏ^2#, and #epsilon = (2ma^2E)/ℏ^2#. This means that:

#V_0 = (ℏ^2v_0)/(2ma^2)#, #" "E = (ℏ^2epsilon)/(2ma^2)#

and we then rewrite #T# as:

#T = 4/(4 + (((ℏ^2)/(2ma^2))^2v_0^2)/((ℏ^2epsilon)/(2ma^2)((ℏ^2v_0)/(2ma^2) - (ℏ^2epsilon)/(2ma^2)))sinh^2(sqrt(v_0 - epsilon)))#

#= 4/(4 + (cancel(((ℏ^2)/(2ma^2))^2)v_0^2)/(cancel(((ℏ^2)/(2ma^2))^2) epsilon(v_0 - epsilon))sinh^2(sqrt(v_0 - epsilon)))#

#= 1/(1 + (v_0^2)/(4epsilon(v_0 - epsilon))sinh^2(sqrt(v_0 - epsilon)))#

Finally, let #r = E/V_0 = epsilon/v_0# so that we can graph #T# vs. #E//V_0 -= r#:

#T = 1/(1 + cancel(v_0^2)/((4epsilon)/v_0 cancel(v_0^2)(1 - r))sinh^2(sqrt(v_0(1 - r))))#

#= barul|" "1/(1 + [sinh^2[v_0^(1//2)(1 - r)^(1//2)]]/(4r(1 - r)))" "|#

Now, when we plot this, if we want #r > 1#, i.e. #E > V_0#, then we would need to use the identity #sinh(ix) = isinx#. So, we separate our result into two segments:

#color(blue)(T = {([1 + [sinh^2[v_0^(1//2)(1 - r)^(1//2)]]/(4r(1 - r))]^(-1),0 < r < 1),([1 + [sin^2[v_0^(1//2)(r - 1)^(1//2)]]/(4r(r - 1))]^(-1),r >= 1):})#

Let's choose #v_0 = 10#. Then this produces the following graph: