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Math Help - Differentiation - a quantum mechanics equation

  1. #1
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    Differentiation - a quantum mechanics equation

    P(r) = [rR(r)]^2,

    R(r) = , a0 = 0.529 * 10^-10


    Find P'(r).

    This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
    Last edited by Sasuke12; March 29th 2007 at 03:11 AM.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Sasuke12 View Post
    P(r) = [rR(r)]^2,

    R(r) = , a0 = 0.529 * 10^-10


    Find P'(r).

    This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
    Is it R(r), or P(r)? let's say P(r), and i'll call a0 a

    P = (2 - r/a)[(e^-r/2a)/2a^(3/2)] .............let's try the product rule

    => P' = (-1/a)[(e^{-r/2a})/2a^(3/2)] + (2 - r/a)[(-1/2a)(1/2a)^(3/2) * e^{-r/2a}]

    => P' = -e^{-r/2a}/[a(2a)^(3/2)] - (2 - r/a)(1/2a)^(5/2)*e^{-r/2a}

    and this can be simplified further

    now let's do it by the way you suggested, by logarithmic differentiation (which i think would be more complicated than it has to be)

    P = (2 - r/a)[(e^-r/2a)/2a^(3/2)]
    => lnP = ln(2 - r/a)[(e^-r/2a)/2a^(3/2)]
    => lnP = ln(2 - r/a) + ln[(e^-r/2a) - ln[2a^(3/2)]
    => lnP = ln(2 - r/a) - r/2a - ln[2a^(3/2)] ..............now differentiate (implicitly)
    => P'/P = 1/(2 - r/a)*(-1/a) - 1/2a
    => P'/P = -1/(2a - r) - 1/2a
    => P' = P(-1/(2a - r) - 1/2a)
    => P' = {(2 - r/a)[(e^-r/2a)/2a^(3/2)]}(-1/(2a - r) - 1/2a)
    hey that's not so bad after all, at least in comparison to the first answer we got. of course, both methods can be simplified further, which you might want to do if you're going to do a lot more work with P'. Good luck
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Sasuke12 View Post
    P(r) = [rR(r)]^2,

    R(r) = , a0 = 0.529 * 10^-10


    Find P'(r).

    This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
    Nope. He got it right. This is an n = 2, l = 0 Hydrogen electron wave function. (Or at least the radial part of an n = 2 wavefunction.)

    For simplicity, let's just call a0 = a for now.

    So:
    P(r) = r^2*R(r)^2 = r^2*[(2 - r/a)^2 * e^{-r/a}] / (2a)^3

    Now use the product and chain rules:
    P'(r) = 2r*[(2 - r/a)^2 * e^{-r/a}] / (2a)^3 + r^2*[2(2 - r/a) * (-1/a) * e^{-r/a}] / (2a)^3 + r^2*[(2 - r/a)^2 * e^{-r/a} * (-1/a)] /(2a)^3

    Now we need to simplify a bit.

    P'(r) = 2r*[(4 - 4r/a + r^2/a^2) * e^{-r/a}] / (2a)^3 + r^2*[(-4/a + 2r/a^2) * e^{-r/a}] / (2a)^3 + r^2*[(-4/a + 4r/a^2 - r^2/a^3) * e^{-r/a}] /(2a)^3

    Now factor a e^{-r/a}/(2a)^3 and do some further multiplying:

    P'(r) = {8r - 8r^2/a + 2r^3/a^2 + -4r^2/a + 2r^3/a^2 - 4r^2/a + 4r^3/a^2 - r^4/a^3} * e^{-r/a}/(2a)^3

    Now collect powers of r:

    P'(r) = {-(1/a^3)r^4 + (8/a^2)r^3 - (16/a)r^2 + 8r} * e^{-r/a}/(2a)^3

    Now make things nice with that (2a)^3 at the end:

    P'(r) = {-(1/(8a^6))r^4 + (1/a^5)r^3 - (2/a^4)r^2 + (1/a^3)r} * e^{-r/a}

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Nope. He got it right. This is an n = 2, l = 0 Hydrogen electron wave function. (Or at least the radial part of an n = 2 wavefunction.)

    For simplicity, let's just call a0 = a for now.

    So:
    P(r) = r^2*R(r)^2 = r^2*[(2 - r/a)^2 * e^{-r/a}] / (2a)^3

    Now use the product and chain rules:
    P'(r) = 2r*[(2 - r/a)^2 * e^{-r/a}] / (2a)^3 + r^2*[2(2 - r/a) * (-1/a) * e^{-r/a}] / (2a)^3 + r^2*[(2 - r/a)^2 * e^{-r/a} * (-1/a)] /(2a)^3

    Now we need to simplify a bit.

    P'(r) = 2r*[(4 - 4r/a + r^2/a^2) * e^{-r/a}] / (2a)^3 + r^2*[(-4/a + 2r/a^2) * e^{-r/a}] / (2a)^3 + r^2*[(-4/a + 4r/a^2 - r^2/a^3) * e^{-r/a}] /(2a)^3

    Now factor a e^{-r/a}/(2a)^3 and do some further multiplying:

    P'(r) = {8r - 8r^2/a + 2r^3/a^2 + -4r^2/a + 2r^3/a^2 - 4r^2/a + 4r^3/a^2 - r^4/a^3} * e^{-r/a}/(2a)^3

    Now collect powers of r:

    P'(r) = {-(1/a^3)r^4 + (8/a^2)r^3 - (16/a)r^2 + 8r} * e^{-r/a}/(2a)^3

    Now make things nice with that (2a)^3 at the end:

    P'(r) = {-(1/(8a^6))r^4 + (1/a^5)r^3 - (2/a^4)r^2 + (1/a^3)r} * e^{-r/a}

    -Dan
    Oh, ok, i guess i had to have some knowledge of quantum mechanics to answer this question. i thought R(t) was a typo or something, i never knew R(t) was different from P(t) and you had to relate them by another function. Luckily there's a physics maestro on this site!
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Sasuke12 View Post
    P(r) = [rR(r)]^2,

    R(r) = , a0 = 0.529 * 10^-10


    Find P'(r).

    This is an equation from quantum mechanics, I have tried putting log on both sides and pull the powers down, but I'm not sure whether I'm doing the correct thing, please help me. Thanks!!
    I almost didn't see it at first either, but P(r) = [rR(r)]^2

    If you had noticed that, you would likely have gotten the correct derivative.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    I almost didn't see it at first either, but P(r) = [rR(r)]^2

    If you had noticed that, you would likely have gotten the correct derivative.
    O wow, you know i still didn't see that until just now! I thought topsquark pulled that formula out of his head.

    i seriously think i need to relearn how to read

    but still, the font size for that thing was like 8 while for the other function its like 500, you won't notice that if your itching to differentiate something like i usually am
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