# Derivatives in spherical coordinates

• maria clara
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.

#### maria clara

In quantum mechanics the momentum operator is a constant multiplied by the 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?

That doesn't look like d/dx, it looks like the radial part of the laplacian. It's a second derivative. In general to convert derivatives between coordinate systems you use the chain rule for partial derivatives.

the momentum operator is often written as -ihd/dx... but in that case x is the position vector. More generally, the momentum op is -ihDel (Del = the gradient operator, more easily generalized to different coordinate systems). (note also that i have written 'h' but mean "h-bar").
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).

sorry, my mistake.
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
?

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]

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BTW there are much better ways to derive this using differential geometry, but it gives you an idea of WHY the answer isn't as simple as you think. No need to follow all of the details.

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

Dick said:
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]
Just want to thank you, this website is basically my homework assignment this week. I have been getting my *** worked until i found this.
THANK YOU

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## 1. What are spherical coordinates?

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.

## 2. What is the purpose of using spherical coordinates?

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.

## 3. How do you convert between spherical and Cartesian coordinates?

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)

## 4. What are the derivatives in spherical coordinates?

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.

## 5. How do you find the gradient of a function in spherical 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.