Can someone explain to me why the solution of \(\frac{d^{2}\Phi (\phi)}{d\phi^{2}} = -m_{l}^{2}\) is \(\Phi = e^{im_{l}\phi}\)?

 

You probably know that the derivative of exp(c*x) with respect to x is c*exp(c*x).

You also need to know that i=sqrt(-1). You may know this, or perhaps you usually use j=sqrt(-1). They tend to use i in physics, and j in electrical engineering.

Anyway, take the derivative of exp(i*ml*phi) with respect to phi twice and remember that i squared is minus one.

 

boks,

Can someone explain to me why the solution of is ?

Actually the solution is Phi = -(1/12)*m^4 + C1*m + C2 , where C1 and C2 are arbitary constants. How did you get a complex term for a solution?

Ratch

 

boks,



Actually the solution is Phi = -(1/12)*m^4 + C1*m + C2 , where C1 and C2 are arbitary constants. How did you get a complex term for a solution?

Ratch

Ratch, you misread the question. The derivatives are with respect to \( \phi \), not with respect to \( m_l \). You may have been thrown off by the typo in the equation.

That must have been a good vacation you took!

 

steveb,

Ratch, you misread the question. The derivatives are with respect to , not with respect to . You may have been thrown off by the typo in the equation.

So what is ml, a constant? And what is the correct equation? I still don't see a sinusoid in the solution.

That must have been a good vacation you took!

Yes, my wife and I went to Hot Springs, Arkansas.

Ratch

 

steveb,

So what is ml, a constant? And what is the correct equation? I still don't see a sinusoid in the solution.

Ratch

This equation shows up in the quantum mechanical solution for the hydrogen atom, and \( m_l \) is the magnetic quantum number. There is a typo in the way boks wrote it, but it was clear from the context what he meant. One issue is the proper amplitude for the solution. There is a customary way to normalize the solution, but I assumed that was not the sticking point.

 

Ratch, you misread the question. The derivatives are with respect to \( \phi \), not with respect to \( m_l \). You may have been thrown off by the typo in the equation.

That must have been a good vacation you took!

Based on the problem that was posted, Ratch's solution was a lot closer than the one the OP provided.
For the given problem, I get
\(\Phi = \frac{-m^2\phi ^2}{2} + C\phi + D\)

SteveB, you mentioned that boks had a typo in the equation he posted. My guess is that the correct DE was missing a factor of \(\Phi \) on the right side, like so:
\(\frac{d^2 \Phi}{d\phi ^2 } = -m^2 \Phi\)

 

Based on the problem that was posted, Ratch's solution was a lot closer than the one the OP provided.
For the given problem, I get
\(\Phi = \frac{-m^2\phi ^2}{2} + C\phi + D\)

SteveB, you mentioned that boks had a typo in the equation he posted. My guess is that the correct DE was missing a factor of \(\Phi \) on the right side, like so:
\(\frac{d^2 \Phi}{d\phi ^2 } = -m^2 \Phi\)

Well, I would say your answer is the correct answer based on the exact equation posted. It's a matter of judgement whether ratch or the OP is closer. Technically they are both wrong.

Your corrected equation with the \( \Phi \) put in on the right hand side is correct, of course. I guess it's good to point this out for the benefit of anyone that is not familiar with the context of the quantum mechanical equation. When I first answered the question I really didn't even notice. It's like one of those sentences that has has a repeated word, and your mind just doesn't even notice it.

 

Mark44,

you mentioned that boks had a typo in the equation he posted. My guess is that the correct DE was missing a factor of on the right side, like so: d2(Phi)/phi = -m^2*Phi

If d2(Phi)/phi = -m^2*Phi , then the solution is Phi = A*cos(m*phi) + (B/m)*sin(m*phi)

Ratch

 

Mark44,

If d2(Phi)/phi = -m^2*Phi , then the solution is Phi = A*cos(m*phi) + (B/m)*sin(m*phi)

Ratch

Nope, it's not. Two can play at this pedantic game.
The solution is
\(\Phi = A cos(m\phi) + B sin(m\phi)\)
To show that this function is the general solution, take the first and second aking derivatives.
\(\frac{d}{d\phi}\Phi = -Am sin(m\phi) + Bm cos(m\phi)\)
\(\frac{d^2}{d\phi ^2}\Phi = -Am^2 cos(m\phi) - Bm^2 cos(m\phi)\)
\(= -m^2(A cos(m\phi) + B sin(m\phi)) = -m^2\Phi\)

In addition, the linear combination of cosine and sine of the solution can also be written as a linear combination of exponential functions (of \(im\phi\)), similar to what boks originally posted.

 

Mark44,

Nope, it's not.

Oh yes it is. Our solutions are equivalent. Your "A" is what I designate as "B/m" and your "B" is what I designate as A. They are all constants, albeit different constants. Each solution's double derivative equates to -m^2*Phi . I will admit that your expression of the solution looks nicer.

In addition, the linear combination of cosine and sine of the solution can also be written as a linear combination of exponential functions (of ), similar to what boks originally posted.

Would you be so kind as to give me a quick example. Make A and B any nontrivial value, and show the equivalency of the exponential sums.

Ratch

 

I thought I would provide a link to some lecture notes available on-line about the solution of the Hydrogen atom which is the context of that equation. On page 4 of the notes, you'll see the solution and normalization in red ink.

http://inside.mines.edu/fs_home/dwu/classes/CH351/study/H_soln/Hydrogen.html

 

Mark44,
Oh yes it is. Our solutions are equivalent. Your "A" is what I designate as "B/m" and your "B" is what I designate as A. They are all constants, albeit different constants. Each solution's double derivative equates to -m^2*Phi . I will admit that your expression of the solution looks nicer.

Would you be so kind as to give me a quick example. Make A and B any nontrivial value, and show the equivalency of the exponential sums.

Ratch

I'll take your word for the equivalence of our solutions. For an example,
here's what I was talking about, which was that sine and cosine of the same whatever could be expressed in terms of exponentials. Acos(x) + B sin(x) would be just A times the first thing on the right + B times the second thing on the right. See the wikipedia article on Euler's formula at http://en.wikipedia.org/wiki/Euler's_formula.
\(cos x = \frac{e^{ix} + e^{-ix} }{2}\)
\(sin x = \frac{e^{ix} - e^{-ix} }{2i}\)

 

Mark44,

OK, thanks.

Ratch