i often use Physics package to finish my work,i appreciate it given a 

effective work for quantum computaion.it still maybe get promoted for  the mixture of  sum and integral,the mixture of c-number and q-number.i believe it will be more powerful in next version.

this is quite motivating ,i like Physics package,significant enhancements of physics will bring me more pleasant surprise.

in caculations,we meet mixed calculation of integation and sum,here is a question as

sup:= Int(conjugate(f[m](r))*f[n](r), r) = delta[m, n]   # heremention orthogonality

leq := Int(I*conjugate(f[m](r))*`ℏ`*(sum((diff(c[n](t), t))*f[n](r)*exp(-I*omega[n]*t), n = l .. k)), r)

req := Int((1/2)*conjugate(f[m](r))*E[0](e_.r_)*e*(sum(c[n](t)*f[n](r)*omega[n]*(exp(I*t*(-omega[n]+Omega))+exp(-I*t*(omega[n]+Omega))), n = l .. k)), r)

just i want to do like

  applyrule(sup,Int(conjugate(f[m](r))*leq,r)=Int(conjugate(f[m](r))*req,r))

yield

I*`ℏ`*(diff(c[m](t), t))*exp(-I*omega[m]*t) = sum(int(Typesetting[delayDotProduct](conjugate(f[m](r)), e_.r_, true)*f[n](r)*dr*omega[n](exp(I*t*(-omega[n]+Omega))+exp(-I*t*(omega[n]+Omega))), r), n = l .. k)

thanks a lot

in caculations,we meet mixed calculation of integation and sum,here is a question as

sup:= Int(conjugate(f[m](r))*f[n](r), r) = delta[m, n]   # heremention orthogonality

leq := Int(I*conjugate(f[m](r))*`ℏ`*(sum((diff(c[n](t), t))*f[n](r)*exp(-I*omega[n]*t), n = l .. k)), r)

req := Int((1/2)*conjugate(f[m](r))*E[0](e_.r_)*e*(sum(c[n](t)*f[n](r)*omega[n]*(exp(I*t*(-omega[n]+Omega))+exp(-I*t*(omega[n]+Omega))), n = l .. k)), r)

just i want to do like

  applyrule(sup,Int(conjugate(f[m](r))*leq,r)=Int(conjugate(f[m](r))*req,r))

yield

I*`ℏ`*(diff(c[m](t), t))*exp(-I*omega[m]*t) = sum(int(Typesetting[delayDotProduct](conjugate(f[m](r)), e_.r_, true)*f[n](r)*dr*omega[n](exp(I*t*(-omega[n]+Omega))+exp(-I*t*(omega[n]+Omega))), r), n = l .. k)

thanks a lot

hi friends,if we konw the structions of the expression,we can make it simplest,if not,we have no  common way to get the result effectively

hi friends,if we konw the structions of the expression,we can make it simplest,if not,we have no  common way to get the result effectively

pagan 2269

    yes ,you solve this problem in realdomain.but ,the complex solves may be lost.for example b[2]=1/2,b[3]=I/2,

b[4]=-1/2,b[5]=-I/2.how can i do?

   another example:eq:=abs(x)^2=1;ans:=solve(eq,Explicit);ans=1,-1

the sovles I and -I lost.

pagan 2269

    yes ,you solve this problem in realdomain.but ,the complex solves may be lost.for example b[2]=1/2,b[3]=I/2,

b[4]=-1/2,b[5]=-I/2.how can i do?

   another example:eq:=abs(x)^2=1;ans:=solve(eq,Explicit);ans=1,-1

the sovles I and -I lost.