- #1

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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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In summary, the momentum operator in quantum mechanics is often written as -ihd/dx, but in spherical coordinates it becomes -(hbar)^2/r*(d/dr)^2*r. This is due to the fact that spherical coordinates are curved and require a more complex derivation. Understanding this concept is important in converting derivatives between coordinate systems.

- #1

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

can anyone explain please how this result is obtained?

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- #2

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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

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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

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

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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.Dick said:

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

THANK YOU

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Spherical coordinates are a system of coordinates used to describe points in three-dimensional space. They consist of a radial distance from a fixed point, an inclination angle from a fixed plane, and an azimuth angle from a fixed direction.

Spherical coordinates are useful for describing points in three-dimensional space, particularly in situations where the Cartesian coordinate system is not convenient. They are commonly used in physics and engineering, especially for problems involving spherical objects or systems.

To convert from spherical coordinates (r, θ, φ) to Cartesian coordinates (x, y, z), you can use the following formulas:x = r * sin(θ) * cos(φ)y = r * sin(θ) * sin(φ)z = r * cos(θ)To convert from Cartesian coordinates to spherical coordinates, you can use the inverse formulas:r = √(x^2 + y^2 + z^2)θ = arccos(z / r)φ = arctan(y / x)

The derivatives in spherical coordinates are the partial derivatives with respect to each coordinate (r, θ, φ). They are used to calculate the rate of change of a function in three-dimensional space, just like how derivatives are used in Cartesian coordinates.

To find the gradient of a function in spherical coordinates, you can use the following formula:∇f = (∂f/∂r)er + (1/r)(∂f/∂θ)eθ + (1/rsin(θ))(∂f/∂φ)eφwhere er, eθ, and eφ are unit vectors in the radial, azimuthal, and polar directions respectively. This formula is similar to the gradient formula in Cartesian coordinates, but also takes into account the conversion factors for spherical coordinates.

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