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Welcome To Simply Lecture

Simply lecture is a platform that assists students on difficult lectures and help them in any area. This blog serves as a personal tutor for students in any part of the world..
All students, have to do is to complete the contact form with their most difficult topic and subscribe to the blog, to get the most understandable form of the topic requested.
One good thing, about this innovation is that students are not allowed to pay any cash! Sounds good right?  Now all you need to do is:
1. Complete the contact form
2. Subscribe to get notified when your simplified lectures are posted.
3. Done
Also, Simply lecture is also ready to surprise you with great features soon, such like:
1. Online step by step calculators.
2. Free Video lectures.
3. Slideshare review of power point lectures.
4. Download simplified lecture notes.
5. Help you with assignments for a little pay.
6. And a whole lot more.
All you have to do is to stay tuned on this blog and enjoy a whole lot of lecture fun..  Schooling don't have to be boring, right??  Now get on the track, subscribe and also post this to friends and family and let them benefit the joy of e-education.

Warm Regards, 
Simply lecture.

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Schrodinger equation as a law in physics

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Maxwell first equation

The Maxwell first equation in electrostatics is called the Gauss law in electrostatics. Statement:  It states that the total electric flux \(\psi_E\) passing through a closed hypothetical surface is equal to \(\frac{1}{\epsilon_0}\) enclosed by the surface. Integral Form: $$\phi_E = \int E.ds = \frac{q}{\epsilon_0}$$ $$\int D.ds = q$$ where, $$D = \epsilon_0 E = displacement-vector$$ Let the change be distributed over a volume v and \(\rho\) be the volume charge density. Hence, $$q = \int \rho dv$$ Therefore; $$\int D.ds = \int_{v} \rho dv$$ .........(1) Equ(1) is the integral form of Maxwell first law Differential form: Apply Gauss divergence theorem to the L.H.S of equ(1) from surface integral to volume integral. $$\int D.ds = \int (\nabla.D)dv$$ Substituting this equation to equ(1) $$\int(\nabla.D)dv = \int_{v} \rho dv$$ As two volume integrals are equal only if their integrands are equal. Thus; $$\nabla.D = \rho v$$ ............(2) Equ(2) is the differe
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Auxiliary time dependent Schrodinger equation

Recall the relationship: $$\phi(t) = C_+ exp[-i(\frac{E}{\hbar})t] + C_- exp[i(\frac{E}{\hbar})t]$$ .............(1) Recall, from the present special case: $$C_+ = 0$$ ........(2) Substitute equ (2) into equ (1), we have that: $$\phi(t) = C_- exp[i(\frac{E}{\hbar})t]$$ .......(3) Also,  recall the relationship: $$\Psi(r,t) = \psi(r) \phi(t)$$ ..........(4) Substitute equ (3) into equ (4): $$\Psi(r,t) = C_- exp \psi (r)  [i(\frac{E}{\hbar})t]$$ ..............(5) Equ (5) can be differentiated with respect to t, and by rearranging we have: $$\Psi(r,t) = -i(\frac{\hbar}{E}) \frac{d \Psi (r, t)}{dt}$$ ..............(6) Recall the relationship: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = E \Psi(r,t)$$ ...........(7) Substituting equ (6) into the right hand side of equ (7) we have: $$-\frac{\hbar^{2}}{2m} \nabla^{2} \Psi(r,t) + V(r) \Psi(r,t) = i\hbar \frac{d \Psi(r,t)}{dt}$$ .........(8) Equ (8) is the auxiliary time dependent Schrodinger equation
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