It's better to use sympy.I rather than 1j as it is exact and does not
use floating point.
You say that the result is bulky but do you have a reason to think
that it can be any simpler?
This kind of explosion is typical if you have matrices where every
element is an arbitrary symbol. That's why it's useful to write
equations where each symbol is a whole matrix.
On Thu, 19 Mar 2020 at 15:10, Ilshat Garipov <igj...@gmail.com> wrote:
>
> I recently faced with the need to do small calculations with matrices opf
> complex number in any symbolic math environment, so I decided to try with
> simpy. And here is an example of my code:
>
> import sympy as sym
> from sympy import *
> from sympy.physics.quantum.dagger import Dagger
>
> a1, a2, a3, a4 = sym.symbols('a1, a2, a3, a4', real=True)
> b1, b2, b3, b4 = sym.symbols('b1, b2, b3, b4', real=True)
> g1, g2, g3 = sym.symbols('g1, g2, g3', real=True)
> h1, h2, h3 = sym.symbols('h1, h2, h3', real=True)
>
> U = sym.Matrix([[a1 + b1*1j, a2 + b2*1j], [a3 + b3*1j, a4 + b4*1j]])
> G = sym.Matrix([[g1, g2 + g3*1j], [g2 - g3*1j, -g1]])
> H = sym.Matrix([[h1, h2 + h3*1j], [h2 - h3*1j, -h1]])
> U1 = Dagger(U)
>
> Rg= U*G*U1
> Rh = U*H*U1
> tr = (Rg*Rh).trace()
>
> The problem is that result looks awesomely bulky:
>
> ((a1 - 1.0*I*b1)*(g1*(a1 + 1.0*I*b1) + (a2 + 1.0*I*b2)*(g2 - 1.0*I*g3)) + (a2
> - 1.0*I*b2)*(-g1*(a2 + 1.0*I*b2) + (a1 + 1.0*I*b1)*(g2 + 1.0*I*g3)))*((a1 -
> 1.0*I*b1)*(h1*(a1 + 1.0*I*b1) + (a2 + 1.0*I*b2)*(h2 - 1.0*I*h3)) + (a2 -
> 1.0*I*b2)*(-h1*(a2 + 1.0*I*b2) + (a1 + 1.0*I*b1)*(h2 + 1.0*I*h3))) + ((a1 -
> 1.0*I*b1)*(g1*(a3 + 1.0*I*b3) + (a4 + 1.0*I*b4)*(g2 - 1.0*I*g3)) + (a2 -
> 1.0*I*b2)*(-g1*(a4 + 1.0*I*b4) + (a3 + 1.0*I*b3)*(g2 + 1.0*I*g3)))*((a3 -
> 1.0*I*b3)*(h1*(a1 + 1.0*I*b1) + (a2 + 1.0*I*b2)*(h2 - 1.0*I*h3)) + (a4 -
> 1.0*I*b4)*(-h1*(a2 + 1.0*I*b2) + (a1 + 1.0*I*b1)*(h2 + 1.0*I*h3))) + ((a1 -
> 1.0*I*b1)*(h1*(a3 + 1.0*I*b3) + (a4 + 1.0*I*b4)*(h2 - 1.0*I*h3)) + (a2 -
> 1.0*I*b2)*(-h1*(a4 + 1.0*I*b4) + (a3 + 1.0*I*b3)*(h2 + 1.0*I*h3)))*((a3 -
> 1.0*I*b3)*(g1*(a1 + 1.0*I*b1) + (a2 + 1.0*I*b2)*(g2 - 1.0*I*g3)) + (a4 -
> 1.0*I*b4)*(-g1*(a2 + 1.0*I*b2) + (a1 + 1.0*I*b1)*(g2 + 1.0*I*g3))) + ((a3 -
> 1.0*I*b3)*(g1*(a3 + 1.0*I*b3) + (a4 + 1.0*I*b4)*(g2 - 1.0*I*g3)) + (a4 -
> 1.0*I*b4)*(-g1*(a4 + 1.0*I*b4) + (a3 + 1.0*I*b3)*(g2 + 1.0*I*g3)))*((a3 -
> 1.0*I*b3)*(h1*(a3 + 1.0*I*b3) + (a4 + 1.0*I*b4)*(h2 - 1.0*I*h3)) + (a4 -
> 1.0*I*b4)*(-h1*(a4 + 1.0*I*b4) + (a3 + 1.0*I*b3)*(h2 + 1.0*I*h3)))
>
> And i am wondering if there are any built-in methods to shorten this
> expression? Am I doing any thing wrong in my code? I am new in python so
> there might be things I don't yet know.. Thanks in advance for helping.
>
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