- #1

maria clara

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*partial*derivative d/dx. In spherical coordinates it's turning into something like that:

constant*(1/r)(d^2/dr^2)r

can anyone explain please how this result is obtained?

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- Thread starter maria clara
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- #1

maria clara

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constant*(1/r)(d^2/dr^2)r

can anyone explain please how this result is obtained?

- #2

Dick

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- #3

lzkelley

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As dick said, the operator you've shown doesn't look quite right... if something is only changing radially, then the grad in spherical is just d/dr (i believe).

- #4

maria clara

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I meant the radial part of the (momentum)^2 in spherical coordinates.

If the Hamiltonian is:

(Pr)^2/2m + L^2/2mr^2 +V(r)

what's wrong in writing (Pr)^2 as

-(hbar)^2*(d/dr)^2

?

why is it written like that:

(-(hbar)^2/r)*(d/dr)^2*r

?

- #5

Dick

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Because the grad^2 operator is only the sum of coordinate derivatives squared in cartesian coordinates. Spherical coordinates are curved. Here's a derivation.

http://planetmath.org/encyclopedia/%3Chttp://planetmath.org/?method=l2h&from=collab&id=76&op=getobj [Broken]

http://planetmath.org/encyclopedia/%3Chttp://planetmath.org/?method=l2h&from=collab&id=76&op=getobj [Broken]

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- #6

Dick

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- #7

maria clara

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wow... I'll take a look at that, thanks!

- #8

l_ron_hubbard

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Just want to thank you, this website is basically my homework assignment this week. I have been getting my *** worked until i found this.Because the grad^2 operator is only the sum of coordinate derivatives squared in cartesian coordinates. Spherical coordinates are curved. Here's a derivation.

http://planetmath.org/encyclopedia/%3Chttp://planetmath.org/?method=l2h&from=collab&id=76&op=getobj [Broken]

THANK YOU

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