They are, as a matter of fact, indistinguishible. Of course their dynamical properties differ, but indistinguishible means you cannot a priori know which one has that particular property. That is to say, if you somehow know that one electron has spin up, and other has spoin down, the...
I'm wondering how it really is useful.
The input for the, say 2-qubit, quantum computer that is running Grover's algoritm is
|\Psi \rangle = (|1 \rangle + |2 \rangle + |3 \rangle + |4 \rangle) / \sqrt{4}
And let us say we're looking the 3rd element in the so-called database.
Now, Grover...
Homework Statement
(from Goldstein, problem 3.12)
Suppose that there are long-range interactions between atoms in a gas in the form of central forces derivable from potential
U(r) = \frac{k}{r^m},
where r is the distance between any pair of atoms and m is a positive integer. Assume further...
Electron may be somewhere near the nucleus or in Hawaii. I'd also like to add that electron is, as far as we know, a structureless point particle. What is smeared out is the wave-function associated with it. When you go measure it, you'll find that it's a point particle.
The abrupt "quantum...
The star means complex conjugate: replaces all is with -is. In your case, \psi^* = A^* sin(\pi x / L), but you can takeA to be real, making \psi = \psi^*.
Homework Statement
(Sakurai 1.27)
[...] evaluate
\langle \mathbf{p''} | F(r) | \mathbf{p'} \rangle
Simplify your expression as far as you can. Note that r = \sqrt{x^2 + y^2 + z^2}, where x, y and z are operators.
Homework Equations
\langle \mathbf{x'} | \mathbf{p'} \rangle = \frac{1}{ {(2 \pi...
Solved it.That equation I wrote will have an integration factor f(V).
Using \frac{\partial S}{\partial V}_T = \frac{\partial P}{\partial T}_V, we have the solution for S, this time with an integration factor of T. By compraison of these two statements of S, it's perfectly defined.
Heat capacity of a liquid is C=T^4 and the state function is V(T,P) = Aexp(aT-bP)
Derive an equation for entropy. Use the relevant Maxwell relations.
dU = T dS - PdV
\frac{\partial U}{\partial T}_V = C = T^4 \Rightarrow U = \frac{T^5}{5} + f(V)
Since it's a liquid, and there're no...
Hello,
As a path-integral newbie, I've been trying to calculate the amplitude for an electron which enters a box (potential within the box is given) at a point to emerge the other edge of the box (it doesn't matter when it exits). For simplicity, I first tried to work out the problem in one...
Griffiths develops an intelgral equation for Scrödinger equation in his QM book. As doing so, he requires Green's function for Helmholtz equation
(k^2 + \nabla^2) G( \mathbf r) = \delta^3(\mathbf r)
A rigourious series of steps, including Fourier transforms and residue integrals follow...
I've recently started Feynman & Gibbs. I was sure exercises will be fun, but i can't enjoy myself when i fail solving the first one! Exercise 1-1 says: show that free particle action is
\frac{m}{2} \frac{x_b^2 - x_a^2}{t_b-t_a}
I tried finding anti-derivative of \dot x^2, ended up with...
Cylindrical solution gave a better insight though. Since the system is symmetric around z-axis, the electric field should be independent of \theta, therefore, the solution is \mathbf E = -k \left(\frac{s}{2} \mathbf e_\theta \right).
Wish that I had a more rigorous way to show it, though.
Thanks for the replies!
Meir Achuz, I'd ask "why?"
In cylindirical coordinates, z-component of curl is
\frac{1}{s} \left( \frac{\partial (s A_{\theta}) }{\partial s} -\frac{\partial A_s}{\partial \theta} \right) = -k
which has again 3 mathematically possible solutions
-k/2(\frac{s}{2}...
Imagine a solenoid with n turns per length. Now, for an instant, in which everything looks static, the magnetic field inside the solenoid will be n \mu_0 I \mathbf e_z (choosing solenoid alinged with z-axis), and zero field outside. Now, what would happen if we change the current in time?
To...
Thanks for the replies!
I think i've developed a clear answer to the "paradox":
The point particle configuration assumes no friction, the object freely slides under the gravity. However, in the rolling ball configuration, the is a friction, and it's responsible for the torque which causes...