Recent content by Addez123

To say that the beam "split up" yet demand that it results in a single beam. What exactly is happening is what Im asking. Is it actually two separate rays of light? Could an electron split in two? If not then shouldn't it always be able to be defined by a single wave function and not have to be...

A wave is described as a sinus shaped wave, the √ of the probability function. It's not described as two separate wave functions interacting with eachother. Hypothetically, what happens if after the slit I separate the two areas with a sheet of paper. Now these "two waves" are forever separated...

When you do the double slit experiment with photons or electrons you get a wave pattern. At certain points no electrons are detected. This is said to be caused by destructive interference. Destructive interference of what? If we shoot single electrons, one at a time, from where is this...

Ahh yes yes! You get 1/2 from derivating the sqrt(E)! Now it all makes sense. Unbelivably grateful, thanks alot :) :)

The 1/8th is because we calculate all states as if it were in a cartesian coordinate system and since n cant be negative we only cover the first octant. Its the surface of a sphere in the 1st octant * dn as they explain it.

I literally can not explain how thankful I am for this response. I dont know how I couldnt read dn/dE as ##\frac{d}{dE}(n)## but it just never clicked. One final problem though. Their result has ##\sqrt{2}## in the denominator. I've done it twice but I get it in the numurator, isnt that correct?

$$n = \sqrt{n_x^2 + n_y^2 +n_z^2}$$ $$E = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$$ $$n = \sqrt{ \frac{2mL^2E}{\pi^2 \hbar^2} }$$ This is all given by the textbook. It's even as friendly as to say $$\text{differential number of states in dE} = \frac{1}{8}4 \pi n^2 dn$$ $$D(E) = \frac{...

If charge q = e, moves through 200V, the energy given the electron is: $$\Delta U = qV = 200 V * e = 3.2 \cdot 10^{-17} J$$ I have no clue how to convert that into eV. I havnt done this kind of physics since middleschool, there was no introduction to any of this and next week we have to do...

I have the equations to calculate transmission probability, my problem is that the barrier is given in Volts not electron volts. $$200V = e \cdot 200 eV = 3.2 \cdot 10^{-17} eV$$ I am not even sure if that's a correct conversion. But if it is then this "barrier" is extremly small and 99.999%...

Good tip! I solved the ##\Psi^* \Psi## and got correct value though. Solved it again with your solution and it gave correct result too!

Great! But I still get the wrong answer. $$4a^3 \int_0^{inf} x^3 \cdot e^{-2ax} = 4a^3 \cdot \frac {3}{8a^4} = \frac {3}{2a}$$ not 1/a as the answer suggests.

That is true! So since it has no imaginary part the conjugate is simply itself? No changes?

Isn't the conjugate simply reversing the minus sign on the exponential? Conjugate of ##xe^{-iax}## is ##xe^{iax}## according to wolfram alpha too.

To get expected value I use $$E = \int \Psi^* Q \Psi dx$$ where Q = x $$4a^3 \int xe^{ax} \cdot x \cdot xe^{-ax} dx = 4a^3 \int_0^{\inf} x^3 dx$$ which is undefined. But the answer is suppose to be 1/a.

1. Go to: https://www.physicsforums.com/help/latexhelp/ WITHOUT being logged in. 2. Find a symbol you wish to use, such as \Psi 3. Then copy it and paste it into your equation 4. Voila! You ruined your entire document with this monstrocity