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My apologies, I thought I attached a screenshot of the problem but I might have accidentally deleted it. I figured out the problem though. Thank you!

The eigenvalue for this ##H_0## is given by ##\hbar \omega(n+1) ; (n_x+n_y = n)## At the ground state, ##nx = ny = 0## so the eigenvalue is simply ##\hbar\omega## Next we turn the perturbation potential on and I know that the first order shift in the energy is the expectation value of the...

I am to consider a basis function ##\phi_i(x)##, where ##\phi_0 = 1 ,\phi_1 = cosx , \phi_2 = cos(2x) ## and where the scalar product in this vector space is defined by ##\braket{f|g} = \int_{0}^{\pi}f^*(x)g(x)dx## The functions are defined by ##f(x) = sin^2(x)+cos(x)+1## and ##g(x) =...

This is for a Quantum Mechanics class but part b of this question seemed like it relied more on math than physics so I think it appropriate to post here. If not, Mods please move to appropriate place. For the ##\Pi xf(\vec r)+x\Pi f(\vec r)=0## I have my answer circled in red on the first...

I figured it out. I had a big misunderstanding on the equations I was using but took the time to read through my book and was able to come to the correct answer. Thanks all for the help!

You're right. I should have ##\vec E(t) =Acos{(-kz-\Omega t+\delta)} \hat x## ##\delta## is the phase So I need to make ##\hat x## be ##\hat - x## as well as account for it is a wave traveling left by the ##-kz## in the ##cos##? I assumed the ##-kz## in the ##cos## accounted for the negative...

I am unsure of my solutions and am looking for some guidance. a.)The real part of the wave in complex notation can be written as ##\widetilde{A} = A^{i\delta}##. Writing the Complex Wave equation, we have ##\vec E(t) = \widetilde{A}e^{(-kz-\Omega t)} \hat x##. Therefore the real part is...

I understand what you are saying. I worked it out with a friend via your method and we got the right answer. Thank you. I will post my work when I get the chance.

Ah I think your distance from the wiire ##s=L/2+y.## is a lot easier to visualize forme. So if I integrate wrt y, then: ##\vec B=\frac{μI}{2piy}##. If I try to evaluate the lower corner my integral should look like this (ignoring constants for now): ##\int_0^\frac{L}{\sqrt(2)}...

Setting up the integral to find the flux that is giving me trouble. I know that I will have to break up this integral into 2 parts, the first part account for when the start of the loop is increasing in area, and another right when I pass the halfway mark of the loop and the area begins to...

##K = \frac{N}{m} = \frac{3eV}{bondlength^2} = \frac{4.806*10^-19 J}{(2*10^-10)^2} = 12.015## Then I know that ##H = \frac{1}{2}mωx^2 ## where ## k = mω^2 ## and also ##H=ħω(n+\frac{1}{2}) ## Therefore, ## \frac{1}{2}kx^2 = ħω(n+\frac{1}{2})## Solving for n, ## n = \frac{1/2kx^2}{ħω} -...

I am trying to convert the attached picture into dirac notation. I find the LHS simple, as it is just <ψ,aφ>=<ψIaIφ> The RHS gives me trouble as I am interpreting it as <a†ψ,φ>=<ψIa†Iφ> but if I conjugate that I get <φIaIψ>* which is not equiv to the LHS. *Was going to type in LaTex but I...

Thanks to all, I have seen the trivial mistake I made. I was able to get the correct answer now.

I am going through my class notes and trying to prove the middle commutator relation, I am ending up with a negative sign in my work. It comes from [a†,a] being invoked during the commutation. I obviously need [a,a†] to appear instead. Why am I getting [a†,a] instead of [a,a†]?

Below is my attempt at the problem. I used the distributive law and applied to what was given, (A∪B)∪Z which equates to (A∪Z)∩(B∪Z). I then applied the 2 set I/E formula to each union. Since there is an intersection between these two sets I added them. But I end up with 2 P(Z) which doesn't...

I did something similar to this in a 2-D laplace equation but now that it is 3-D the exponential is raised to a half power, π/a√(n2+m2)(l-x) So then I would have Cosh(π/a√(n2+m2)(l-x)) Is that correct?

Ohh that is write I can write the exponential as for condition V as 2cosh(x). so then I would get 2cosh(l-x)?

I believe what I have to do to solve this problem is find the potential at each end face and then use the super position principle to find the net potential. So my boundary condition v and iv will give the potential at each respective position. Im just a bit confused about my boundary V...

That makes a lot of sense. I see what you were saying now, I was being really stubborn about going about it the other way. And so dA should be, ##dA = R^2sin\theta d\theta d\phi## since its a surface integral?

Ok I can accept that. So from post #10, are you saying I should use: as ##\vec r##.?

It may have come off that i was trying to use the fact that ##\sigma(\theta) =\sigma_0 \cos\!\theta## in another problem but that wasn't my intention. I was using it as a reference to show how my integral should look. Why can't I make the argument that this sphere is surrounded by a uniform...

I did a similar problem, where it was two insulators with bound charges stuck to each surfaces. When finding the dipole moment for one of the surfaces, I followed these general steps And what happens with these problems is the equatorial plane has no charge leaving just the z component. I...

Sorry, mistake. ##\vec r## = ##x\hat x + y\hat y +z\hat z## This is what you mean correct?

It should look like this? ##\vec r## = ##x\vec x + y\vec y +z\vec z##

ahh. I wrote one thing but was thinking something different. so the dA should be, $$\mathbf{p}=\int \sigma(\theta)\mathbf{r}'r^2sin(\theta)d(\theta)d(\phi).$$ , right? And with the electric field point in the z direction, the equatorial plane would have 0 free charges, so $$\mathbf{r}'$$...

So is it pretty much this? The electric field that i got for the dielectric is confirmed in griffiths. I know that E inside the conductor is 0. So the dipole moment would be equal to: p = ∫rεE(r)*rdrdθ ?

I worked this problem out in griffiths and my work checks out for for the potentials, b.c. and the coefficients. I will post the solutions just because my work is a little harder to read. What I am having trouble finding is the dipole moment of the conductor. I know the formula for dipole...

I have found the total dipole moment of for this problem but am having trouble finding the electric field. I believe my electric field when r>2R ( I mistakenly wrote it as r<2R on my work, but it is the E with a coefficient of 2/3) is correct as it fits the equation: . I don't believe this...

I see what you were saying now in regards to this and can see how my thinking is wrong. I understand the problem now and thank you for your help with this problem I took a lot away from it.

So this proves the answer is zero because for an odd function, the integral over a symmetric interval equals zero?

Im kind of lost now. So we have symmetric bounds over an odd function i get that. I am stuck here

ohhhh I think I see it. Since I have that sinθ (and that's its odd) in my dV, and we have spherical symmetry or even bounds this whole integral should just be zero.

0 because we are integrating symmetric bounds over an odd function.

So the integrand then is negative, and the integration volume would just be r2dr?

Its points in the opposite direction right?

Ok I agree with your first two statements, that makes sense to me. If they are spherical they are symmetric. Now what about ##\vec r##. Well I know cos is an even function and sin in an odd function, so would I neglect ##\hat x## and ##\hat y## and would only care about the ##\hat z##...

l=0 states mean the orbitals are the shape of a sphere. Ok that's illuminating, so I should be integrating in spherical coordinates, r2sinθdrdθφ. So r should be from 0→∞. I feel like I am still missing something. If I multiply this out its going to be a mess.

I am a little lost on how to approach this problem. What I know is the following: The r vector is in terms of x y and z hat. I know my two l=0 states can be the 1s and 2s normalized wave function for Hydrogen. Should I be integrating over dxdydz?

thats fair enough, I didnt think about that.

So the probability is nonzero. It is small but still not zero. So the next question I have to ask is, is there any location where it is NOT possible to find an electron? My friend is saying it can't be between energy levels. It is quantized because the only energy values are energy eigen...

Not sure if you read my edit as our post are very close but I did get a value of 3.72*10-32 after doing it by hand. Does this seem correct? But to answer your question, I would think maybe just the first term? There isn't much change in the values if its a magnitude of 10-5

EDIT: Now I did it by hand using the tabular method and a0 = 5.3*10-11 and, taking the absolute value, 3.72*10-32

I used a different program and I got 1x10-11, but that was with using the value 5.3*10-11 for a0 that i found on google.

I thought since I was using the radial wave function I just need to integrate wrt to r^2dr?

I used wolframalpha and got this Which I guess makes sense. The probability of this happening is very unlikely, but not zero.

It should be 0 to R. Sorry I misunderstood you at first but waited for you to respond. So given that the radius is typically 1x10-15m, I guess you could say the limit of integration should be 0 to 1x10-15m?

So does that mean instead of integrating ∫ψ2r2dr, I should be integrating∫ψ2R2dR?

It should just be plugging in. Do you know where I can find the value of a? I wasnt given that information

So I used the radial wave functions from this website, http://plato.mercyhurst.edu/chemistry/kjircitano/InorgStudysheets/InorgWaveFunction.pdf and basically squared each wave function. I am given the radius of a nucelus which is r=1x10-15. So I plug that in for r in the function and also need...

For this problem, Is it as simple as using the probability density function, P = Ψ2 and plugging in the radius value given to me? So essentially I am just squaring the wave function and plugging in?