Angular momentum polar coordinates

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SUMMARY

The discussion focuses on deriving the z component of angular momentum, L_z, in polar coordinates from its Cartesian definition. The operator is expressed as L_z = -ih(d/dθ) after transforming the Cartesian coordinates (x, y) into polar coordinates (r, θ) using the relationships x = rcosθ and y = rsinθ. Participants highlight the challenge of this derivation, noting that most references utilize spherical coordinates instead. A key identity for transforming derivatives is provided, emphasizing the need to rewrite d/dx and d/dy appropriately during the conversion.

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  • Understanding of angular momentum in quantum mechanics
  • Familiarity with polar coordinate transformations
  • Knowledge of partial derivatives and their applications
  • Basic concepts of quantum operators
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  • Learn about the transformation of derivatives in polar coordinates
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Students and professionals in physics, particularly those focusing on quantum mechanics and mathematical physics, will benefit from this discussion. It is especially relevant for those tackling angular momentum in different coordinate systems.

johnnyies
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Homework Statement



from the cartesian definition of angular momentum, derive the operator for the z component in polar coordinates

L_z = -ih[x(d/dy) - y(d/dx)]

to

L_z = -ih(d/dθ)

Homework Equations


x = rcosθ
y = rsinθ

r^2 = x^2 + y^2

r = (x^2 + y^2)^1/2

The Attempt at a Solution



first of all I'm not sure how this is even possible. Every derivation of the angular momentum operator I've seen requires spherical coordinates, not polar.

I tried taking the derivative of r with respect to x to get cosθ and with respect to y to get sinθ

and dx/dθ = -rsinθ dy/dθ = rcosθ but it's not getting me anywhere.

is there something i should be rewriting d/dx and d/dy as?
 
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Since you are transforming from one basis to another, you can use the identity (written for your particular case):

\frac{∂}{∂θ} = \frac{∂x}{∂θ}\frac{∂}{∂x} + \frac{∂y}{∂θ}\frac{∂}{∂y}.

Does that help?
 

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