Recent content by thecourtholio

I was just using the indecies as good practice so i remember to use all indecies of Q or P later if they need it. But for [Q,P] I find $$[Q,P] = \bigg[\frac{\partial Q}{\partial x}\frac{\partial P}{\partial p_x} - \frac{\partial Q}{\partial p_x}\frac{\partial P}{\partial x}\bigg] + \bigg[...

Homework Statement Consider a charge ##q##, with mass ##m##, moving in the ##x-y## plane under the influence of a uniform magnetic field ##\vec{B}=B\hat{z}##. Show that the Hamiltonian $$ H = \frac{(\vec{p}-q\vec{A})^2}{2m}$$ with $$\vec{A} = \frac{1}{2}(\vec{B}\times\vec{r})$$ reduces to $$...

Ok I think I understand it all now. Thank you so much for y'all's help!

Ok I got $$ \int e^{2\xi y-y^2}e^{2\xi z-z^2}e^{-\xi^2}d\xi = \sqrt{\pi}e^{2yz} = \sqrt{\pi} \sum \frac{y^lz^l}{l!} $$ Is this the correct series? Cause in comparison to $$ \int \frac{y^m}{m!}H_m \frac{z^n}{n!}H_n e^{-\xi^2} d\xi $$ the series and gaussian would cancel for m=n=l leaving just...

Ok I think I see what you are saying. So I evaluate ## \int e^{2\xi y-y^2}e^{2\xi z-z^2} e^{-\xi^2} d\xi ## and compare it to ##\int \bigg[\sum \frac{y^m}{m!}H_m\bigg]\bigg[\sum \frac{z^n}{n!}H_n\bigg] e^{-\xi^2}d\xi ## something like that?

Oh yeah I see. I should've noticed that after seeing everyones comments about mod squared. Yeah I was questioning that as I was typing it since ##e^{2\xi-z^2} = \sum \frac{z^n}{n!}H_n## and not just ##H_n## but I wasn't sure if it was possible to rearrange the summation. So combining your...

Oh thank you I hadn't caught that mistake! So it should be ##\int_{-\infty}^{\infty} A^2 e^{4\xi z-z^2}e^{-\xi^2}d\xi## ? And should it be an integral of ##d\xi## or ##dx##? But I'm still not sure how ## e^{4\xi z-2z^2} = \sum \frac{z^{2n}}{(n!)^2}H_n^2## gives me cross terms that vanish? What...

Homework Statement Prove that ##\psi_n## in Eq. 2.85 is properly normalized by substituting generating functions in place of the Hermite polynomials that appear in the normalization integral, then equating the resulting Taylor series that you obtain on the two sides of your equation. As a...

Oh okay. So if the particle's energy is lower than the maximums of ##U(x)## it will oscillate indefinitely but if its energy is greater than that potential maximum then it can escape right? In this case that means that if ##E<U(\pm\sqrt{\frac{a}{b}})## it will oscillate but if...

What would quantify "how deep" the valley is? And that was another thing that was confusing me was the constant energy. I've seen a lot of the plots of ##U(x)## vs. ##x## in the literature include a horizontal line indicating the total energy but none of them really explained why they chose...

Homework Statement A particle of mass m moves along the x–axis under the influence of force ##F_x=-ax+bx^3## , where a and b are known positive constants. (a) Find, and sketch, the particle's potential energy, taking U(0) = 0 (b) Identify and classify all equilibrium points (c) Find the...

Omg I think I got it. So ##\vec{r}## would be the vector describing the distance from the point charge to the reference point right? So then at some reference point P, the vector from ##q_1## would be ##\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}## and from ##q_2## it would be...

Well my thinking was that if ##q_1## is at the origin then ##\vec{r_1}=(0,0,0)## and if ##q_2## is on the z axis then ##\vec{r_2}=(0,0,a)##. So then ##\vec{r_1}\cdot\vec{r_2}=r_{x1}r_{x2}+r_{y1}r_{y2}+r_{z1}r_{z2}=(0)(0)+(0)(0)+(0)(a)=0## right?

Homework Statement Find the interaction energy ( ##\epsilon_0 \int \vec{E_1}\cdot\vec{E_2}d\tau##) for two point charges, ##q_1## and ##q_2##, a distance ##a## apart. [Hint: put ##q_1## at the origin and ##q_2## on the z axis; use spherical coordinates, and do the ##r## integral first.]Homework...

Homework Statement 1) Calculate the angular diameter distance to the last scattering surface in the following cosmological models: i) Open universe, ΩΛ= 0.65, Ωm = 0.30 ii) Closed universe, ΩΛ = 0.75, Ωm = 0.30 ii) Flat universe, ΩΛ = 0.75, Ωm = 0.25 Describe how the CMB power spectrum...