Recent content by B3NR4Y

Homework Statement An intensity current I descends down the z-axis from z = \infty to z = 0, where it spreads out in an isotropic way on the plane z = 0. Compute the magnetic field. Homework Equations The only relevant equation I can think of is Ampere's Law, \oint_\gamma \vec{B} \cdot...

Okay I have that |1> can be written: |1\rangle = \frac{1}{\sqrt{2}i} \left(|a\rangle-|b\rangle\right) With this I apply H, which I can use the schrodinger equation to find. The only thing I am uneasy on now is that I think the coefficients will be imaginary, and I am not sure how to reconcile...

I see, that is shockingly easy. So once I have that I should compute <1|\hat{H}|1> and the coefficient of each term is the probability of that energy?

Yeah my assumption about the basis was stupid. I did some extra reading to refresh my memory from my first quantum course and realized that. Since |1> and |2> are any two linearly independent vectors, then an arbitrary state |ψ> can be written c1 |1> + c2 |2>. In the case that part c says, c2...

Homework Statement |1> and |2> form an orthonormal basis for a two-level system. The Hamiltonian of this system is given by: \hat{H} = \epsilon \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} a.) Is this Hamiltonian hermitian? What is the significance of a hermitian operator? b.) Find the...

Homework Statement An aqueous solution of silver nitrate, AgNO3, reacts with an aqueous solution ammonium carbonate, (NH4)2CO3. What are the spectator ions for the reaction? AgNO3 (aq) + (NH4)2CO3 -> A 17.5 mL sample of hydrochloric acid HCl solution required 29.6 mL of 0.250M Ba(OH)2 for...

This isn't true, sleep isn't just 8 hours. You can stay up later for two hours, and sleep in for two hours and still feel tired. The best is to have a natural sleep/wake time. I go to bed at 20, wake up at 4:30. Get to school at 5, and get work done before my classes and after classes. I go home...

Homework Statement Let {b k } be a sequence of positive numbers. Assume that there exists a sequence {a k}, such that a k is greater than or equal to 0 for all k, a_k is decreasing, the limit of a_k is 0 and b_k = a_k - a _(k+1). Show that the sum from k=1 to infinity of b k exists and equals...

Ok so applying the mean value theorem, there is an x in (am,an) so that ## \frac{f(a_m) - f(a_n)}{a_m - a_n} = f'(x) ## taking the absolute value of both sides, ## |f(a_m) - f(a_n)| < |a_m - a_n| ##. Since an is convergent, |am - an| is always less than some epsilon greater than zero (because it...

Homework Statement Assume f:(a,b)→ℝ is differentiable on (a,b) and that |f'(x)| < 1 for all x in (a,b). Let an be a sequence in (a,b) so that an→a. Show that the limit as n goes to infinity of f(an) exists. Homework Equations We've learned about the mean value theorem, and all of that fun...

Homework Statement I and J are open subsets of the real line. The function f takes I to J, and the function g take J to R. The functions are in C1. Use the mean value theorem to prove the chain rule. Homework Equations (g o f)' (x) = g'(f (x)) f'(x) MVT The Attempt at a Solution [/B] I know...

Okay I think I am able to prove it using what SteamKing said, but I see what axmls means by some of the lower power terms may overcome the upper power terms. That's why I preferred writing it as $$p(x) = x^{2n} (a_{2n} + ... + \frac{a_1}{x^{2n-1}} + \frac{a_0}{x^2n} ) $$ it seems to show that...

Unfortunately that's not an assumption I can make, I have to prove everything.

Homework Statement Let $$p(x) = a_{2n} x^{2n} + ... + a_{1} x + a_{0} $$ be any polynomial of even degree. If $$ a_{2n} > 0 $$ then p has a minimum value on R. Homework Equations We say f has a minimum value "m" on D, provided there exists an $$x_m \in D$$ such that $$ f(x) \geq f(x_m) = m $$...

Okay, so for the next parts I'm thinking I should follow in the same way, but it seems to be redundant to continue in the same direction for the two directions orthogonal to the ##\hat z##. I was thinking maybe multiplying by the rotation matrix but that seems silly to do, but also not silly to...