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I know the magnitude of complex number a+ib is the sqrt(a^2+b^2).and the magnitude of the complex function for examble (f(z)=ab/cd) where a,b,c and d are complex numbers is |f(z)|=|a||b|/|c||d|.
My question is:
I received this explenation from some expert in matlab work , but I can not understand it
((The magnitude of f(x) corresponds to rotating each point in the complex plane over to the positive x axes, preserving vector magnitude. The result has no remaining phase.))
Can I get more explenation or referece to understand this please?
I will appreciate any help

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Torsten
Torsten 2022-8-20

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The easiest way to see what you are asking is to use the polar representation of a complex number.
If z = r*exp(i*phi), then r is the magnitude of z and phi is the phase angle.
Rotation of a complex number w by an angle theta in the complex plane is given by multiplication of w by exp(i*theta):
w' = exp(i*theta)*w.
Thus if you rotate z = r*exp(i*phi) by its negative phase angle (-phi) , you arrive at z' = r*exp(i*phi)*exp(-i*phi) = r, the magnitude of z.

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Aisha Mohamed
Aisha Mohamed 2022-8-20
Thank you so much for thse answers. It is really helpfull
From the second answer
Rotation of a complex number w by an angle theta in the complex plane is given by multiplication of w by exp(i*theta):
w' = exp(i*theta)*w.........(1)
We have the following analogos case,
when we study the time evolution of finite system, we use this relation to rotate the vector f
f' = exp(i*t*H)*f ...................(2)
where f is a vector. in finite system, H is Hamiltonian, and t is the time.
we replace (theta) in (1) by (t*H) in (2)
My question is what is the difference between (1) and (2)?
Bruno Luong
Bruno Luong 2022-8-20
编辑：Bruno Luong 2022-8-20
A Hamiltonian is a Hermitian operator, even in finite dimension, it's slightly more complex than theta where people explain you which is smply a single real number of argument of the complex number.
If you prefer H is a Hermitian matrix in finite dimension.
In short if you project f on the eigen space (eigen state) of H, theta is taken by the (real) eigen value (~ energy) of H.
Aisha Mohamed
Aisha Mohamed 2022-8-21
Thanks very much for all you.
I take the Hermitian matrix in finite dimension. and as the Hamiltonian is a Hermitian operator its eigen values are real numbers (energy of the system) and the eigen states can be used as a base of this finite system.
What I am struggling with here is in the last line (In short if you project f on the eigen space (eigen state) of H, theta is taken by the (real) eigen value (~ energy) of H.)
Did you mean that the eigen values of H are used as the angle theta?
But the Hamiltonian has ( for examle) 3 real eigen values when ( H_3x3) .How can I find theta from these eigen values?
I would appreciate if someone could further elaborate an explanation regarding this issue.
Bruno Luong
Bruno Luong 2022-8-21
编辑：Bruno Luong 2022-8-21
Ran in:
Ran in:
theta is the argument of the exp function without i (=sqrt(-1)), therefore theta = t*Ek, where Ek is the eigen value #k of the Hamiltonian.
This must be written in all textbook tht explains Schrodinger equation.
Under MATLAB call expm function to compoute exponential of matrix
H=rand(3)
H = 3×3
0.1663 0.2985 0.0529 0.7941 0.2471 0.0493 0.1516 0.9316 0.6585
t=rand
t = 0.2454
expm(1i*t*H)
ans =
0.9918 + 0.0404i -0.0052 + 0.0728i -0.0017 + 0.0128i -0.0101 + 0.1941i 0.9897 + 0.0600i -0.0026 + 0.0119i -0.0259 + 0.0349i -0.0266 + 0.2263i 0.9854 + 0.1606i
Torsten
Torsten 2022-8-21
If you diagonalize H with S,
D = S*H*S^(-1)
you get
f' = exp(S^(-1)*(i*t*D)*S)*f = (S^(-1)*exp(i*t*D)*S)*f
thus
S*f' = exp(i*t*D) * (S*f)
where D is a real diagonal matrix.
exp(i*t*D) is a diagonal matrix with exp(i*t*d(j)) on the diagonal.
Thus theta(j) is kind of t*d(j) with d(j) as j-th eigenvalue of H.
But I think Bruno can better comment on this from the application side.
Aisha Mohamed
Aisha Mohamed 2022-8-21
Thanks very much Bruno Luong and Torsten. this explanation is very useful and helped me a lots.

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dpb
dpb 2022-8-20

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Basically, just what it says -- albeit somewhat wordily, perhaps... :)
A vector in 2D has X,Y components; a complex variable can be represented as a vector in a 2D plane with X-->Re, Y-->Im components.
In that plane, the magnitude is the vector from the origin to the point at which the intersection of the X (Re) and Y(Im) lines intersect; the angle of that vector represents the phase. By Pythagoras, the magnitude is abs() value, but if you compute only it, then you don't know what the two components were any more; you've gained the size but lost the phase (angle). Hence, all you can do then is plot a point on the X axis.

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