Why Doesn't Chop[] Remove I.0 from the Output?

[Natthawin Cho]

As I found on many websites, they suggest to use Chop[]. I tried that already but it doesn't work.

This is my output.

{{2 (Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 1]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2])) Subscript[\[Lambda],
1] (Conjugate[I.0] + Conjugate[Subscript[v, 1]]/Sqrt[
2] + (I.0 + Subscript[v, 1]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) + (Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 2]/Sqrt[2])) Subscript[\[Lambda],
5] (Conjugate[I.0] + Conjugate[Subscript[v, 1]]/Sqrt[
2] + (I.0 + Subscript[v, 1]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) +
1/2 (2 Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2]) + (Conjugate[I.0] +
Conjugate[Subscript[v, 1]]/Sqrt[2]) (I.0 + Subscript[v, 2]/
Sqrt[2])) Subscript[\[Lambda],
6] (Conjugate[I.0] + Conjugate[Subscript[v, 1]]/Sqrt[
2] + (I.0 + Subscript[v, 1]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) -
1/2 I (-(Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2]) + (Conjugate[I.0] +
Conjugate[Subscript[v, 1]]/Sqrt[2]) (I.0 + Subscript[v, 2]/
Sqrt[2])) Subscript[\[Lambda],
8] (Conjugate[I.0] + Conjugate[Subscript[v, 1]]/Sqrt[
2] + (I.0 + Subscript[v, 1]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) - \!\(
\*SubsuperscriptBox[\(\[Mu]\), \(1\), \(2\)]\ \((Conjugate[I . 0] +
\*FractionBox[\(Conjugate[
\*SubscriptBox[\(v\), \(1\)]]\),
SqrtBox[\(2\)]] + \((I . 0 +
\*FractionBox[
SubscriptBox[\(v\), \(1\)],
SqrtBox[\(2\)]])\)\ \*
SuperscriptBox["Conjugate", "\[Prime]",
MultilineFunction->None][I . 0 +
\*FractionBox[
SubscriptBox[\(v\), \(1\)],
SqrtBox[\(2\)]]])\)\) -
1/2 (-(Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[2]) (I.0 +
Subscript[v, 1]/Sqrt[2]) + (Conjugate[I.0] +
Conjugate[Subscript[v, 1]]/Sqrt[2]) (I.0 + Subscript[v, 2]/
Sqrt[2])) Subscript[\[Lambda],
4] (-Conjugate[I.0] - Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) -
1/2 I (Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 1]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2])) Subscript[\[Lambda],
8] (-Conjugate[I.0] - Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) -
1/2 I (Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 2]/Sqrt[2])) Subscript[\[Lambda],
9] (-Conjugate[I.0] - Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) -
1/4 I (2 Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2]) + (Conjugate[I.0] +
Conjugate[Subscript[v, 1]]/Sqrt[2]) (I.0 + Subscript[v, 2]/
Sqrt[2])) Subscript[\[Lambda],
10] (-Conjugate[I.0] - Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) + 1/2 I
\!\(\*SubsuperscriptBox[\(\[Mu]\), \(4\), \(2\)]\) (-Conjugate[I.0] -
Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) +
1/2 (2 Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2]) + (Conjugate[I.0] +
Conjugate[Subscript[v, 1]]/Sqrt[2]) (I.0 + Subscript[v, 2]/
Sqrt[2])) Subscript[\[Lambda],
3] (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) +
1/2 (Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 1]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2])) Subscript[\[Lambda],
6] (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) +
1/2 (Conjugate[
I.0] I.0 + (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 2]/Sqrt[2])) Subscript[\[Lambda],
7] (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) -
1/4 I (-(Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2]) (I.0 + Subscript[v, 1]/Sqrt[2]) + (Conjugate[I.0] +
Conjugate[Subscript[v, 1]]/Sqrt[2]) (I.0 + Subscript[v, 2]/
Sqrt[2])) Subscript[\[Lambda],
10] (Conjugate[I.0] + Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]]) - 1/2
\!\(\*SubsuperscriptBox[\(\[Mu]\), \(3\), \(2\)]\) (Conjugate[I.0] +
Conjugate[Subscript[v, 2]]/Sqrt[
2] + (I.0 + Subscript[v, 2]/Sqrt[2]) Derivative[1][Conjugate][
I.0 + Subscript[v, 1]/Sqrt[2]])}}

I also would like to know the difference between Conjugate[] and Conjugate'[].

 

[JorisL]

Conjugate prime seems to be the derivative of the conjugate. See e.g. http://mathematica.stackexchange.com/questions/91356/derivative-of-conjugate-multivariate-function for a way to write it in a different way.

W.r.t. the huge block of codes, can you post a screenshot of the result?

Also what command yields that result? Its nigh impossible to formulate an answer when you haven't encountered the issue before.

 

[Natthawin Cho]

This is the screenshot.

 

[DrClaude]

What is the input that gave you this?

 

[Dale]

To remove I.0 from your output you should probably remove it from your input. It doesn't make sense to take the dot product of two scalars. I think it indicates an error in your code.

 

[Natthawin Cho]

I solved that problem already. The error is in my code. I used I dot with matrix. However, I have 1x2 matrix times 2x1 matrix and I get 1x1 matrix. Can I change this to scalar?

 

[Dale]

To me it doesn't make sense to have a dot product of I with anything.

If you read the documentation on Dot it says that if either of the arguments are not lists then the Dot remains unevaluated. So that is consistent with the behavior that we see.