# Why does my plot look different depending on PlotRange values I choose [closed]

I plotted the following function twice using different PlotRange values:

Φ[p_, r_, ϕ_, α_] :=
1/Sqrt[Cosh[r]] (1/(π h m ω))^(1/4) *
Exp[
-(1/(2 h m ω)) p^2 -
Conjugate[α]/2 (α + Conjugate[α] E^(I ϕ) Tanh[r]) +
1/(h m ω) E^(I ϕ) Tanh[r] p^2 -
I(α + Conjugate[α] E^(I ϕ) Tanh[r]) Sqrt[2 /(m ω h)] p +
1/2 (α + Conjugate[α] E^(I ϕ) Tanh[r] +
I Sqrt[2/(m ω h)] E^(I ϕ) Tanh[r] p)^2 -
1/2 Log[1 +E^(I ϕ)Tanh[r]] -
((E^(I ϕ) Tanh[r])/(1 +E^(I ϕ) Tanh[r])) *
1/2 (α + Conjugate[α] E^(I ϕ) Tanh[r] +
I Sqrt[2/(h m ω)] E^(I ϕ)Tanh[r] p)^2]


where I set m = 1, h = 1, ω = 1.

Here's a picture of the same function

I used

Plot[Re[Φ[p, 3, 0, 1 (1 + I)]], {p, -100, 100},
PlotRange -> All, PlotStyle -> {Brown, Thick},
Axes -> {False, False}, Frame -> True]


and I obtained the following plot

Notice there is a straight line between 40 and 50. However, if I now plot

Plot[Re[Φ[p, 3, 0, 1 (1 + I)]], {p, 35, 55},
PlotRange -> All, PlotStyle -> {Brown, Thick},
Axes -> {False, False}, Frame -> True]


(I only modified PlotRange) I get the following graph:

It is the same function but now the straight line disappears. This is the behavior that I expect. I do not want that straight line. Is there a way to plot the same function using PlotRange -> {-100,100} but without that annoying straight line?

• I cannot copy your definition of the function. A lot of brackets seem to be unbalanced. Commented Oct 13, 2020 at 18:46

This is a plotting artefact. Sometimes you need to tell Plot to use more initial points to sample the function when the automatic refinement algorithm misses parts. PlotPoints -> 100 seems to work:
Plot[Re[Φ[p, 3, 0, 1 (1 + I)]], {p, -100, 100},