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Topic: Operators and Eigenfunctions/values  (Read 10036 times)

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camariela

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Operators and Eigenfunctions/values
« on: April 10, 2006, 09:25:10 PM »
Hey all,
I'm having trouble with this problem; has to do with operators and eigenvalues and functions.
I will note the caps of the operators by putting it next to the letter.
I have to find the result operating the operators C^ = id/d(phi) on f(phi) = 3exp(i phi) where i is sqrt(-1). Is this an eigenfunction? What is the eigenvalue?

I really really appreciate this.
Thanks and best regards,
camariela

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Re: Operators and Eigenfunctions/values
« Reply #1 on: April 10, 2006, 09:50:55 PM »
Hey all,
I'm having trouble with this problem; has to do with operators and eigenvalues and functions.
I will note the caps of the operators by putting it next to the letter.
I have to find the result operating the operators C^ = id/d(phi) on f(phi) = 3exp(i phi) where i is sqrt(-1). Is this an eigenfunction? What is the eigenvalue?

I really really appreciate this.
Thanks and best regards,
camariela

Apply C to your f.  You get i * d/dphi[ 3 * exp(i*phi) ] = i * 3 * i * exp(i*phi) = -3 * exp(i*phi).
(The rules for derivatives of exp() are the same with complex numbers.)

Is this result the same as the original f, 3 * exp(i*phi), times some constant?
If so, then it is an eigenfunction, and that constant is the eigenvalue.
Otherwise, it isn't an eigenfunction and there is no eigenvalue.

camariela

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Re: Operators and Eigenfunctions/values
« Reply #2 on: April 13, 2006, 12:47:55 AM »
Thanks for the response! It was helpful.
How would you go about doing linear operators such as:
C = (1/r^2)d/dr(r^2d/dr) + 2/r on the function Y = A.exp(-br)
I need to find out what values must the constant b have in order for this to be an eigenfunction.

How do I go about applying the operator C to the function? Do i take the derivative of 1/r^2 wrt r, then   multiply by the derivative of r^2 wrt r + 2/r. Then all of this multiplied by the function Y?

Offline pantone159

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Re: Operators and Eigenfunctions/values
« Reply #3 on: April 13, 2006, 01:36:09 AM »
How do I go about applying the operator C to the function? Do i take the derivative of 1/r^2 wrt r, then   multiply by the derivative of r^2 wrt r + 2/r. Then all of this multiplied by the function Y?

Yea, this notation is confusing.  First, about the +2/r part.  The result of C*Y (C operating on Y) is (2/r)*Y, and is added to the result of the other part.

Now for the (1/r2) (d/dr) (r2 * (d/dr)) part...  You put Y in at the end, so this means (1/r2) (d/dr) (r2 * (d/dr)Y), in other words, take (d/dr) of Y, then multiply by r2, then take (d/dr) of this result, then multiply by 1/r2.

Or, you can expand out the whole operator.  You get:
(1/r2) * (2r * (d/dr) + r2 * (d2/dr2)) + (2/r) =
d2/dr2 + (2/r) * (d/dr) + (2/r)

To make this halfway readable, write Y' for dY/dr, etc. 
Then, C*Y = Y'' + (2/r) Y' + (2/r) Y
which is now halfway sane to read.

camariela

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Re: Operators and Eigenfunctions/values
« Reply #4 on: April 13, 2006, 01:58:48 AM »
Thanks so much Mark for replying!
I attempted it and I got the following when applying the C operator to the Y:
C*Y = exp(-br)bA - 2bA/r + (2A/r)e^-br

so what values must b have in order for this to be an eigenfunction?

camariela

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Re: Operators and Eigenfunctions/values
« Reply #5 on: April 13, 2006, 02:07:12 AM »
C*Y = A exp(-br) - (2bAexp(-br))/r + (2Aexp(-br))/r

so when b = 1, the last two terms cancel out leaving only the first term. so its an eigenfunction when b = 1?
thanks!
camariela

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Re: Operators and Eigenfunctions/values
« Reply #6 on: April 13, 2006, 03:28:30 AM »
(1/r2) * (2r * (d/dr) + r2 * (d2/dr2)) + (2/r) =
d2/dr2 + (2/r) * (d/dr) + (2/r)

To make this halfway readable, write Y' for dY/dr, etc. 
Then, C*Y = Y'' + (2/r) Y' + (2/r) Y
which is now halfway sane to read.

We should ask Mitch for LaTeX extensions.
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Offline pantone159

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Re: Operators and Eigenfunctions/values
« Reply #7 on: April 13, 2006, 02:41:37 PM »
so when b = 1, the last two terms cancel out leaving only the first term. so its an eigenfunction when b = 1?

Yep!  Make those non-constant (1/r) terms go away.

Offline pantone159

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Re: Operators and Eigenfunctions/values
« Reply #8 on: April 13, 2006, 02:43:06 PM »
We should ask Mitch for LaTeX extensions.

That could make the equations stuff a lot more readable.  I'd have to re-learn how to use LaTex though, it has been awhile.

camariela

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Re: Operators and Eigenfunctions/values
« Reply #9 on: April 27, 2006, 08:22:03 PM »
Thanks Mark for your *delete me*

I'm trying to practice more and more of these eigenfunction/value problems to improve but I'm stuck in this one:

Show that the Legendre polynomials P1 = cos? and P2 = 3cos2?-1 are eigenfunctions of:

C = (1/sin?)d/d? (sin?d/d(?)) and find the eigenvalues.

[just in case some characters don't appear, ? = theta]

thanks again!!
camariela

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Re: Operators and Eigenfunctions/values
« Reply #10 on: May 01, 2006, 03:40:19 PM »
Show that the Legendre polynomials P1 = cos? and P2 = 3cos2?-1 are eigenfunctions of:

C = (1/sin?)d/d? (sin?d/d(?)) and find the eigenvalues.

First, you can expand the operator to write C*Y as C*Y = Y'' + (cos(theta)/sin(theta)) Y'

P1 is pretty easy to check.  P2 is more confusing.  BTW - The way you wrote it implied cos (2 * theta) when you really meant (cos(theta))2, usually written cos2(theta).  If you stick P2 into the expanded operator and 'plug-and-chug', you'll find some sin2 terms that don't seem to fit.  Remember that cos2(theta) + sin2(theta) = 1.
You can add any multiple of zero to your equation, so you can add any multiple of cos2(theta) + sin2(theta) - 1.
There is such a multiple that reveals that P2 is indeed an eigenfunction.

camariela

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Re: Operators and Eigenfunctions/values
« Reply #11 on: May 02, 2006, 08:39:14 PM »
Thanks a lot.
I got for P1 = -2, P2 = -6.

Thanks again!!
camariela

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