Problem with the plots of eigenvalues of the Matrix

Question

I am trying to plot the eigenvalues of the following matrix

hamil[kx_,ky_,kz_]={{-10.6`, -0.25` E^(
    I (-0.625` kx - 0.21650635094610965` ky - 
       0.43666666666666665` kz)) - 
   0.7` E^(I (0.375` kx - 0.21650635094610965` ky - 
       0.43666666666666665` kz)) - 
   0.25` E^(
    I (-0.125` kx + 0.649519052838329` ky - 0.43666666666666665` kz)),
   E^((0.` + 
      0.43666666666666676` I) kz) (-0.25` E^((0.` - 
          0.125` I) kx - (0.` + 0.649519052838329` I) ky) + 
     E^((0.` - 0.625` I) kx + (0.` + 
          0.21650635094610965` I) ky) (-0.25` - 
        0.7` E^((0.` + 1.` I) kx)))}, {E^((0.` - 
       0.375` I) kx - (0.` + 0.649519052838329` I) ky + (0.` + 
       0.43666666666666665` I) kz) (-0.25` E^((0.` + 0.5` I) kx) + 
     E^((0.` + 0.8660254037844386` I) ky) (-0.7` - 
        0.25` E^((0.` + 1.` I) kx))), -10.6`, 
  E^((0.` - 0.5` I) kx - (0.` + 0.43301270189221935` I) ky - (0.` + 
       0.43666666666666665` I) kz) (-0.25` - 
     0.25` E^((0.` + 1.` I) kx) - 
     0.7` E^((0.` + 0.5` I) kx + (0.` + 
          0.8660254037844387` I) ky))}, {E^((0.` - 
       0.375` I) kx - (0.` + 0.21650635094610965` I) ky - (0.` + 
       0.43666666666666676` I) kz) (-0.7` - 
     0.25` E^((0.` + 1.` I) kx) - 
     0.25` E^((0.` + 0.5` I) kx + (0.` + 0.8660254037844386` I) ky)), 
  E^((0.` - 0.43301270189221935` I) ky + (0.` + 
       0.43666666666666665` I) kz) (-0.7` + 
     E^((0.` - 0.5` I) kx + (0.` + 
          0.8660254037844387` I) ky) (-0.25` - 
        0.25` E^((0.` + 1.` I) kx))), -10.6`}}

Now the eigenvalues as function of kx, ky and kz are

{es1[kx_, ky_, kz_], es2[kx_, ky_, kz_], es3[kx_, ky_, kz_]} = 
  Eigenvalues[hamil[kx, ky, kz]] // FullSimplify;

Now plotting all the three eigenvalues

Plot[{Chop[es1[0, 0, z]], Chop[es2[0, 0, z]], Chop[es3[0, 0, z]]}, {z,
   0, \[Pi]/1.31}]

I don't understand why is there a sudden jump in blue plot and green plot? Is there a way to rectify this? Is this problem also occur when solving eigenfunctions?

To further stating the problem, as Bob's answer suggests that changing the precision of the equation helps somewhat. However, changing the plotting variable results in same issue that is not rectified by the Bob's solution.

Plot[{Chop[es1[x, 0, 0]], Chop[es2[x, 0, 0]], Chop[es3[x, 0, 0]]}, {x,
0, \[Pi]/1.31}]

Answer 1

Reevaluating with v12.3.1, I am able to reproduce the crossover.

$Version

(* "12.3.1 for Mac OS X x86 (64-bit) (June 19, 2021)" *)

Clear["Global`*"]

hamil[kx_, ky_, kz_] = 
   {{-10.6`, -0.25` E^(I (-0.625` kx - 0.21650635094610965` ky - 
           0.43666666666666665` kz)) - 
     0.7` E^(I (0.375` kx - 0.21650635094610965` ky - 
           0.43666666666666665` kz)) - 
     0.25` E^(I (-0.125` kx + 0.649519052838329` ky - 
           0.43666666666666665` kz)), 
    E^((0.` + 
          0.43666666666666676` I) kz) (-0.25` E^((0.` - 0.125` I) kx - (0.` + 
              0.649519052838329` I) ky) + 
       E^((0.` - 0.625` I) kx + (0.` + 0.21650635094610965` I) ky) (-0.25` - 
          0.7` E^((0.` + 1.` I) kx)))}, {E^((0.` - 0.375` I) kx - (0.` + 
           0.649519052838329` I) ky + (0.` + 
           0.43666666666666665` I) kz) (-0.25` E^((0.` + 0.5` I) kx) + 
       E^((0.` + 0.8660254037844386` I) ky) (-0.7` - 
          0.25` E^((0.` + 1.` I) kx))), -10.6`, 
    E^((0.` - 0.5` I) kx - (0.` + 0.43301270189221935` I) ky - (0.` + 
           0.43666666666666665` I) kz) (-0.25` - 0.25` E^((0.` + 1.` I) kx) - 
       0.7` E^((0.` + 0.5` I) kx + (0.` + 
              0.8660254037844387` I) ky))}, {E^((0.` - 0.375` I) kx - (0.` + 
           0.21650635094610965` I) ky - (0.` + 
           0.43666666666666676` I) kz) (-0.7` - 0.25` E^((0.` + 1.` I) kx) - 
       0.25` E^((0.` + 0.5` I) kx + (0.` + 0.8660254037844386` I) ky)), 
    E^((0.` - 0.43301270189221935` I) ky + (0.` + 
           0.43666666666666665` I) kz) (-0.7` + 
       E^((0.` - 0.5` I) kx + (0.` + 0.8660254037844387` I) ky) (-0.25` - 
          0.25` E^((0.` + 1.` I) kx))), -10.6`}};

The Eigenvalues are

{es1[kx_, ky_, kz_], es2[kx_, ky_, kz_], es3[kx_, ky_, kz_]} =
  Eigenvalues[hamil[kx, ky, kz]];

Plotting shows the crossover:

Plot[{Chop[es1[0, 0, z]], Chop[es2[0, 0, z]], Chop[es3[0, 0, z]]},
 {z, 0, π/1.31},
 PlotLegends -> Placed[{es1, es2, es3}, {.2, .3}]]

However, using arbitrary-precision instead of machine precision:

{es1p[kx_, ky_, kz_], es2p[kx_, ky_, kz_], es3p[kx_, ky_, kz_]} = 
  Eigenvalues[SetPrecision[hamil[kx, ky, kz], 15]];

Plot[{Chop[es1p[0, 0, z]], Chop[es2p[0, 0, z]], Chop[es3p[0, 0, z]]}, {z, 
  0, π/1.31},
 PlotLegends -> Placed[{es1, es2, es3}, {.2, .3}]]

EDIT: Explicitly Sort the Eigenvalues and use ListLinePlot

Manipulate[
 Module[{args},
   args = ReplacePart[{0, 0, 0}, var -> t];
   es[t_] = Eigenvalues[
     SetPrecision[hamil @@ args, prec]];
   data = Transpose[Table[Sort@Chop@es[t], {t, 0, 100 Pi/131, Pi/131}]];
   ListLinePlot[data,
    DataRange -> {0, 100 Pi/131},
    Frame -> True,
    FrameLabel -> (Style[#, 14, Bold] & /@
       {{"kx", "ky", "kz"}[[var]], 
        "es"}),
    PlotRange -> All,
    PlotLegends ->
     Placed[{es1, es2, es3}, {.9, .55}]]] // Quiet,
 {{var, 3, "Plot variable"},
  {1 -> "kx", 2 -> "ky", 3 -> "kz"}},
 {{prec, MachinePrecision, "Precision"}, {MachinePrecision, 15}}]