Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: camariela on April 10, 2006, 09:25:10 PM
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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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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.
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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?
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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.
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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?
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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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(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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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.
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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.
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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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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.
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Thanks a lot.
I got for P1 = -2, P2 = -6.
Thanks again!!
camariela