Acemanhattan, you're not making sense. Try proofreading your posts, or learn to be less sloppy with the terminology.

BTW the zero polynomial is usually said to have degree minus infinity, which is consistent with the formulae for the degrees of sums and products.

You're confusing subsets, subspaces, and polynomials. A polynomial can't "contain" anything. If you mean the set of all polynomials of the form P(t) = a + t^2 for all real a, then this set does not contain the zero polynomial, and so it does not form a vector space. If P_n is a vector space of all polynomials of degree at most n, then the set of polynomials P(t) = a + t^2 is not a subspace of P_n for any n because it is not a vector space. It is however a subset of P_n for n >= 2.

I just realized I was using subset instead of subspace. As you've pointed out. Sorry about that.

QM question.

Ive been going back over my QM textbook since I did badly in the class.

The part that I am reading now is about tunneling. The book says that a quantum particle can tunnel through barriers with a higher potential energy than it posesses, I get that.

It gives an example where it says, and I paraphrase, "imagine a particle in one dimensional space that hits a finite potential energy barrier with potential V where V > E (E is translational energy of the particle). The book splits space up into three regions I, II, III with potentials 0, V and 0 respectively where regions I and III are the spaces before and after the barrier.

Then it gives the solutions for stationary Schrodinger equation in the three regions as the wave function.

region I:
Ae^(ikx)+Be^(-ikx) where the first term is for the particles with momentum kh and the second for those with momentum -kh

region II:
it solves the schrodinger eq and gets Ce^(Rx)+De^(-Rx)

region III:
is just Ee^(ikx) because if the particle penetrates it will continue on.

I just don't understand how this problem is time independant. It seems that the wave function should evolve in time. It should start as the free particle solution in region I and as time goes to infinity it should go to some A'^(-ikx) in region I and some B'^(ikx) in region III because eventually it has to hit the barrier.

hope I made myself clear

When the potential is not a function of time, the wavefunction solution to the time-DEPENDENT Schrodinger equation can be written

PSI(x,t) = psi(x)*phi(t)

where psi(x) is the solution to the time-INDEPENDENT Schrodinger equation, and phi(t) will just be a complex exponential e^(-jEt/h_bar) which is a solution an ordinary differential equation that depends on only on t. The time-independent portion psi(x) is all you need to compute the probabilities P(x,t) for the particle location since

P(x,t) = |PSI(x,t)|^2 = PSI^*(x,t)PSI(x,t) = psi^*(x)psi(x).

That is, the time dependent portion e^(-jEt/h_bar) does does not change the magntitude, so it does not change the probabilities. If you know the constants, then you can use the probabilities of being in the 3 regions to compute the transmission coefficient for the tunneling.

The time-independent solution psi(x) is called the eigenfunction for the potential, and its eigenvalues are the discrete energy levels for the particle. Separating the space dependence from the time dependence is not unlike using phasor notation to "freeze time" when analyzing AC circuits. You could think of phi(x) as information which is modulated onto a carrier e^(-jEt/h_bar).

Hi Bruce, thanks for the response its definitely helping me understand this.

I still dont completely understand though. It just seems so counterintuitive that if I have a particle on one side of a potential energy barrier that the probability distribution should not be a function of time. I mean, it makes sense that in the limit there should be like a "equilibrium distribution" but definitely I dont understand how, in the limit, the wave function in region I still has particles with positive momentum. It seems that given enough time they should pass through.

They pass through every instant, it's just that they pass through with a certain probability. They also don't pass though. They also hang out in the middle. Each electron does all these things at once. That's what the wavefunction means. Each electron is in a superposition of states. It is in many places at the same time. When we measure the current, we conclude that a certain percentage of electrons tunneled, but actually before we make a measurement, an individual electron is still in a superposition of states. When we make the measurement, each electron chooses one of the states in accordance with the probability function. Yeah, quantum mechanics is weird. There is no classical analogue to this.

noname6520 as Bruce said when the Hamiltonian is time independent you can separate the time and space dependency in the wavefuction and result in 2 differential equations.

One of them the time independent one is what you solve in your problem ie here;

http://en.wikipedia.org/wiki/Rectang...ential_barrier

Maybe in earlier problems you have seen discrete energy eigenvalue cases as a result of the application of boundary conditions and time independent Schrödinger equation. In collision problems however (such as this here) the energy eigenvalues are continuously distributed. In a collision problem the energy is specified in advance and the behavior of the wavefuction at great distances (asymptotic solutions) is found in terms of that energy and the potential (which is used to model the interaction/collision). So you start with a particular well defined value of energy or equivalently the momentum of the wave (particle). You deal with what is called free particles or plane wave solutions of given momentum. Obviously due to uncertainty principle when you have perfect knowledge of the momentum of a particle the uncertainty in its position is infinite. Basically its not a localized wave packet. ie see here http://en.wikipedia.org/wiki/Wave_packet and notice what it says;

"In physics, a wave packet (or wave train) is a short "burst" or "envelope" of localized wave action that travels as a unit. A wave packet can be analyzed into, or can be synthesized from, an infinite set of component sinusoidal waves of different wavenumbers, with phases and amplitudes such that they interfere constructively only over a small region of space, and destructively elsewhere.[1] Depending on the evolution equation, the wave packet's envelope may remain constant (no dispersion, see figure) or it may change (dispersion) while propagating."

Essentially here you are imagining having a particle as a wave packet that is somewhat localized (ie you have uncertainties in both its momentum and position) and traveling from say left to right and you are asking yourself how can it be that the entire thing is not time dependent? I mean you can imagine that when the "particle" wave packet is very far to the left just starting to move, near the potential nothing is happening so the wave function there is 0 but as it eventually gets there you will start seeing interesting things and then after it has interacted it will leave and go through or reflected as calculated leaving again nothing happening at the potential neighborhood after enough time has passed. So you ask yourself this is pretty much time dependent to me why do we treat it as time independent? Of course the answer is that when you fix the energy(ie given momentum) you have stationary solutions, plane waves, and the the 3 different regions your problem discusses. You do not have a localizable wave packet in that sense. But of course as you saw above you can describe it in terms of composition of plane waves (eg read the example of Gaussian wavepackets in quantum mechanics in the above wavepacket link).

To help you visualize what you were already imagining as time development of the collision see the numerical solution of the time development of a gaussian wavepacket as it interacts with the potential barrier of your problem in the next simulation .

(Also notice one of the first such computer-generated images/simulation was discussed by A. Goldberg, H. M. Schey, and J. L. Schwartz, "Computer-generated Motion Pictures of One-dimensional Quantum-mechanical Transmission and Reflection Phenomena," Am. J. Phys. 35, 177-186 (1967))




See also the pictures and descriptions in this link for further understanding;

http://en.wikipedia.org/wiki/Quantum_tunneling

eg and



You can also read chapter 5 of the classic QM book by Leonard Schiff.

Trying to figure out what part of this mortgage payment formula I am misunderstanding.



I= 4.3% => .043
i= .043/12 => .00358
T = 30
n = 12 * 30 => 360
L = 500,000

500,000(.043)/(1-1/(1+.00358)^360)) = $29,705

It looks like the answer is, in practical terms, prob about 10x too high, but I can't figure out what computational error I am making to cause this.

more precisely it is exactly 12 times too high. You have the wrong i in the numerator.

That's a bummer. I even asked myself "what are you assuming that could be wrong" about 10 times.

What is the probability of rolling two 4's when rolling two dice?

I got 1/12 and I know that is way wrong.

Answer is 1/36. Can someone show the work to get that? (I know its really basic lol im trying to learn how to actually do a calculation instead of just relying on shortcuts.)

Thanks!

Do you want exactly two fours, or at least two fours?

Edit: Either way I wouldn't call it really basic.

1/36 is definitely wrong.

For exactly two fours we can count as follows: First we need to decide which two dice show the fours, and then we need to decide what numbers the other two dice show (only five possibilities each, since fours are forbidden). Thus we get

I think exactly two 4's at the same time, in one roll. Like in a game of craps or something I guess?

Assuming each is 6 sided standard dice and fair.

See above.

You need to know what is the probability that 2 independent events both happen.

Basically P(A and B)= P(A)*P(B)

Now here you have A=dice one = 4 call it d1=4
B= dice 2= 4 d2=4

So P(A and B)=P(d1=4 and d2=4)=P(44)=P(d1=4)*P(d2=4)

P(d1=4)=P(d2=4)=1/6

So P(44)=1/6*1/6=1/36

Try now Probability of 3 dices all coming 2 say for practice. Try to answer also what is the probability you get with 2 dices 64 as result.

read more here;

http://en.wikipedia.org/wiki/Conditional_probability

http://en.wikipedia.org/wiki/Probability_axioms


here is an introductory probability set of lectures that may help if you are motivated to read it slowly and think about its examples.

http://www.stat.berkeley.edu/~aldous/134/gravner.pdf


Ps: Re Cangurino, thats ok he will have to read yours anyway when he gets to 4 dices and he is now looking for all kinds of things that will keep him busy for days lol, this is why i linked a set of lectures to motivate him and your version will illuminate for him further the need to get systematic about it.

Sorry I misread the question

Thanks for the help, Im gonna be trying to figure this out for the next couple days. Thanks for the links too, should help when I inevitably ask "why?". More practice problems seems like a great idea too, ill keep doing them till I get it down.




Np, appreciate the help though. And I had no idea what that answer you posted meant either lol hopefully with some work I can figure out what you meant.

I thought you were throwing four dice and wanted exactly two fours. Much to complicated

First off let me say, thanks for all the QM replies. I have read them all. Unfortunately I have had to take a little break on my QM review due to being busy with other things. I plan on getting back on it as soon as I can and when I do you can be sure I will go back to those responses and read them again until I understand the issues well.

I have two other questions,

question 1:
suppose you have a dehydration reaction characterized by:

A -> 3/2 H2O + B

And suppose you have a sample of a known amount of A which undergoes the reaction mentioned above and produces a mixture of A and B in unknown amounts. You stick it into a machine and it gives you a number X, where:

X=m_b/(m_a + m_b)

where m_a and m_b are the masses of A and B in the sample respectively.

The objective is to find the amount of water it would take to rehydrate the B in the sample.

I would like criticism on my solution:

I have one equation above namely,

X=m_b/(m_a + m_b)

and another,

n_a + n_b = m_ai/MW_a

where n_a and n_b are the moles of A and B, m_ai is the initial mass of A and MW_a is the molecular weight of A.

Ok, now here is the step that I am really unsure about. I define:

MW_avg = X*MW_b + (1-X)*MW_a

and then say that, m_a + m_b = MW_avg*(n_a + n_b)

after this the system is closed and it is a simple matter of algebra to solve m_b as a function of knowns.

then m_h20 = 3/2*m_b


question 2:
suppose I put an object with a known amount of moisture into a drying oven whose temperature varies with time and I would like to calculate the drying efficiency of the oven.

what I was thinking was,

first, drying efficiency = E_ideal/E_actual

second, since ideally I would go into the object and input enough energy into all the water in there so that its temperature would rise to 100C (this is at atmospheric pressure) and then vaporize.

which leads to E_ideal = m_h2o * (integral Cp_h2o dT + H_v)

where m_h2o is the mass of the water in the object
the integral Cp dT is evaluated from the initial temperature of the object to 100C
and H_v is the latent heat of water at 1 atm

and E_actual =m_air' * integral ( integral Cp_air dT ) dt

where m_air' is the rate of air flowing into the oven
the inner integral is evaluated from the initial temperature of the air to T(t)
the outer integral is evaluated from t=0 to t=final.

for a final expression of

dryer efficiency = [m_h2o * (integral Cp_h2o dT + H_v)]/[m_air' * integral ( integral Cp_air dT ) dt]

criticism please, thanks in advance.

When calculating Fold Equity from the button, I assume we multiply folding range by folding range i.e if both blinds call 10% of hands it is .90*.90. the percentage of fold equity here is 81%. How do I add overcall to this calculation? i.e if villain one calls my equation is no longer .90x.90 it is more like .90*.95.

Thanks.

If you are only interested in fold equity then the product already takes care of overcalls.

Hey guys. I've never done any complex analysis, but just found some lecture notes that I'll try to work through. So... I might bother you with some trivial questions, the first of which comes here:

I want to find



where C is the circle of radius 2 centered at zero orientated counterclockwise.

I parametrize the curve C by: f(t) = 2cost + 2isint = 2exp(it). Now f'(t) = if(t) and the integral becomes



Is this correct?

looks good to me, and it passes my sanity check, which is called the residue theorem (which you'll learn later)