# How to instruct mathematica that $\delta^*$ is the conjugate of $\delta$

I have this code:

S = (Sqrt[2]/2)*{{1 + Conjugate[δ], 0}, {0,1 - Conjugate[δ]}}(**   Suppose a+b=1 and δ=((a-b)/(a+b))\[Conjugate]   **)
k = (1/Sqrt[2])*{{S[[1, 1]] + S[[2, 2]]}, {S[[1, 1]] - S[[2, 2]]}, {2 S[[1, 2]]}} // Simplify
Subscript[T, 0] = Dot[k, ConjugateTranspose[k]]
Subscript[T, 0] // MatrixForm


$$\left( \begin{array}{ccc} 1 & \delta & 0 \\ \delta ^* & \delta \delta ^* & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$$

As you see at the end the product of $\delta$ and $\delta^*$ is not printed as $|\delta|^2$ but as $\delta\delta^*$
Someone told me in one of my questions that this is because:

It seems that you did not instruct Mma that δ∗ is a conjugated value of δ. Using simply a conjugate symbol is not enough. You should use Conjugate[δ] instead and then apply ComplexExpand

so far I have tried several ways like
Using the UpsetDelayed operator in the begining of code as:

δ\[Conjugate] ^:= Conjugate[δ]


or using:

ComplexExpand[Subscript[T,0], δ, TargetFunctions -> {Abs, Conjugate}]


But I couldn't change any thing?!

Following the the first answer posted to the question I wrote:

FullSimplify[Subscript[T, 0]] // TraditionalForm


$$\left( \begin{array}{ccc} 1 & \delta & 0 \\ \delta ^* & \left| \delta \right| ^2 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$$
But when I continue the code and apply the same trick on another matrix, the trick doesn't work!

R[ψ_] := {{1, 0, 0}, {0, Cos[2 ψ], Sin[2 ψ]}, {0, -Sin[2 ψ], Cos[2 ψ]}}
T[ψ_] := Dot[R[ψ], Subscript[T, 0], Transpose[R[ψ]]]


$$\left( \begin{array}{ccc} 1 & \delta (\cos (2 \psi )) & -\delta (\sin (2 \psi )) \\ \delta ^* (\cos (2 \psi )) & \delta \delta ^* \left(\cos ^2 (2 \psi )\right) & -\frac{1}{2} \delta \delta ^* (\sin (4 \psi )) \\ -\delta ^* (\sin (2 \psi )) & -\frac{1}{2} \delta \delta ^* (\sin (4 \psi )) & \delta \delta ^* \left(\sin ^2 (2 \psi )\right) \\ \end{array} \right)$$

• I am genuinely surprised, that MMA, treating all symbols as complex quantities by default, does not automatically convert d d* to Abs[d^2] or Abs[d]^2, however there is a sort of explanation to this. IIRC, MMA judges the complexity of an expression by its LeafCount. Both d d* and Abs[d]^2 (as well as Abs[d^2]) have a LeafCount of 4, so I guess, it feels no need to "simplify". I completely agree, that it is frustrating. – LLlAMnYP Sep 18 '15 at 7:01
• It may actually have its uses in matrix multiplication or the application of operators, where multiplication/order of application is non-commutative. – LLlAMnYP Sep 18 '15 at 7:02

If everything else fails, you always can write an explicit substitution rule

subc = {x_ Conjugate[x_] -> Abs[x]^2};


And apply it to your expression

FullSimplify[T[\[Psi]]] /. subc // TraditionalForm


FullSimplify@Subscript[T, 0] // TraditionalForm


does the trick.

• thanks it solved so there's no need to instruct anything just FullSimplify instead of Simplify. Mathetica itself knows that $\delta^*$ is the conjugate of $\delta$ and so can simplify! – Sepideh Abadpour Sep 17 '15 at 22:17