Hello,
Differentiability of f : \mathbb C \to \mathbb C is characterized as \frac{\partial f}{\partial z^*} = 0.
More exactly: \frac{\partial f(z,z^*)}{\partial z^*} := \frac{\partial f(z[x(z,z^*),y(z,z^*)])}{\partial z^*} = 0 where z(x,y) = x+iy and x(z,z^*) = \frac{z+z^*}{2} and...
[Linear Algebra] Finding T* adjoint of a linear operator
Homework Statement
Consider P_1{}(R), the vector space of real linear polynomials, with inner product
< p(x), q(x) > = \int_0^1 \! p(x)q(x) \, \mathrm{d} x
Let T: P_1{}(R) \rightarrow P_1{}(R) be defined by T(p(x)) = p'(x) +...
My textbook claims that the complex conjugate operator is linear. I can't see how this could be. Could someone give me an example of how it is not linear?
I have a rather fundamental question which I guess I've never noticed before:
Firstly, in QM, why do we define the expectation values of operators as integral of that operator sandwiched between the complex conjugate and normal wavefunction. Why must it be "sandwiched" like this?
From...
Homework Statement
Compute the complex conjugate of <p> using eq 1.35 (<p>=∫ψ*(h/i)∂/∂x ψ dx) and prove that <p> is real (<p>=<p>*)
Homework Equations
equation 1.35 is given above
The Attempt at a Solution
to take the c.c. don't i just add a minus to the i and switch the stars like...
Hi,
Is there a possibility of getting a complex conj zeros in pure RC ckt.
we never get complex conj poles but how about complex conj zeros.
regards,
Asif
Homework Statement
[PLAIN]http://img823.imageshack.us/img823/4500/85131172.png Homework Equations
Derivations and substitutions.
The Attempt at a Solution
Basically it seems like a very simple problem to me however I can't seem to get the right answer. First I just assumed that the c.c...
Homework Statement
i am supposed to prove that for the complex number z=cis\theta
the conjugate is \frac{1}{\overline{z}}
Homework Equations
if
z=a+bi
\overline{z}=a-bi
The Attempt at a Solution
all that i can think of is that \frac{1}{cos\theta i sin \theta}
=(cos \theta i sin...
I have seen discussion about it here but it is still not clear to me whether probability is square of probability amplitude or is it product of amplitude with it's complex conjugate.
I looked in HyperPhysics http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/qm.html#c5" and it says it's product...
Hi.
Sometimes in my quantum mechanics course we encounter derivatives such as \frac{d}{dz}z^*, i.e. the derivative of the complex conjugate of the complex variable z wrt z. We are told that this is just zero, even though I know that the complex conjugate is not an analytic function ... Has...
First, sorry for my poor English and any impolite behavior might happen.
Here's two wave function(pic1) and problem below(pic2).
and they are polar coordinate problem ψ(r,θ,Φ)
You can see, problem requires conjugate function of ψ1.
Is it possible to find one? or is there a possibility...
Im doing a bit of contour integration, and a question came up with a term in it am unsure of how to do: in its simplest form it would be
\int\bar{z}dz
where z is a complex number and \bar{z} is it's conjugate. Hmm i can't get the formatting to work out properly.. :S
Hey all, I was just curious:
Why is the conjugate of a complex number (a + bi) defined as (a - bi)? If we instead change the sign of the real part (-a + bi), we still get a real number when we multiply the two. Is there a particular significance to the current definition...
A wave function(psi) is a mathematical quantity which gives complete information about the state of a system at a particular instant of time. But what information does the complex conjugate of a wave function(psi*) give? Does it represent the same state as psi? Or does it just have a...
I have two questions
Homework Statement
Show that the function defined by complex conjugation is not differentiable.
Homework Equations
If z = x + iy then the complex conjugate of z is x - iy
The Attempt at a Solution
f'(z) = lim z -> z1 \frac{f(z) - f(z1)}{z - z1}
So I...
Homework Statement
Suppose that f(x) is a polynomial of degree n with real coefficients; that is,
f(x)=a_n x^n+ a_(n-1) x^(n-1)+ …+a_1 x+ a_0, a_n,… ,a_0∈ R(real)
Suppose that c ∈ C(complex) is a root of f(x). Prove that c conjugate is also a root of f(x)
Homework Equations...
i was thinking this over, and i'd like to know if the following statement is valid:
\oint\overline{z}dz = \oint1/zdz
over the contour \left|z\right|=1
any thoughts?
Hi,
Is there any trick to treat complex conjugate variables in polynomial equations as independent variables by adding some other constraint equation ? Say, we have polynomial equation $f(x,x^{*},y,...) = 0$. where x^{*} is the complex conjugate of variable $x$. I might think of taking $x = r...
Homework Statement
sun, I don't know why I am stuggling with a simple freakin problem. Its not even for me its for my friend who's in college algebra, but for some reaon I can't get the correct answer.
{9 - 11i \over 6i}
The Attempt at a Solution
I multiplied by the conjugate twice and...
It's very commong to use z and z* as two independent variables, differentiating with respect to one while keeping the other constant. Can you please give me some intuitive insight into this method, and why it works so well? Because every time I see this my first thought is that z and z* are NOT...
Hello All,
As I understand it, the wavefunction Psi(x) can be written as a sum of all the particle's momentum basis states (which is the Fourier transform of Psi(x)). I was woundering if the wavefunction's complex conjugate Psi*(x) can be written out in terms of momentum basis states, similar...
Homework Statement
show that det(A)=(detA)*= det(A$)
where * denotes complex conjugate and $ means transpose
Homework Equations
The Attempt at a Solution
Please help me to start the problem.I am not getting a way.
I'm doing a take home final and wanted reassurance that I'm doing the problem right. the question involves taking <Sz>of |psi>. I know it's
<psi|Sz|psi>. I've never done it for a spin 2 particle which is a 5*5 matrix.Do i just take the complex conjugate of the values without switching their...