H atom electron in combined spin/position state

[outhsakotad]

Homework Statement

An electron in a H atom occupies the combined spin and position state: R21{(sqrt(1/3)Y10χ+) + (sqrt(2/3)Y11χ-)} If you measured both the z component of spin and the distance from the origin, what is the probability density for finding the particle with spin up and at radius r?

Homework Equations

The Attempt at a Solution

The answer should just be |R21|^2*(1/3)*|Y10|^2*|χ+|^2 = (r^2)/(96πa^5) * exp(-r/a) * (cosθ)^2, right? Or do I need to do an integral? The theta dependence of my answer is bugging me, but I'm not entirely sure if I need to integrate over theta and phi to just get an r dependent answer? Could somebody please help me think through this? Thanks very much.

 

[Matterwave]

If the question just asks for "radius r", then, yes, you do have to integrate out the theta and phi dependence.

 

[outhsakotad]

Sorta makes sense, but if the probability density is not isotropic in theta, it still seems wrong to me to integrate that dependence away.

 

[vela]

If you don't integrate over the angles, the expression you have doesn't give the probability density that the electron is at a distance r; it's the probability density that the electron is at a distance r, angle θ, and angle φ.

 

[paranormal]

I would first clarify that the given state represents an electron in a hydrogen atom in the second energy level (n=2), with the first subscript indicating the principal quantum number (n), and the second subscript indicating the orbital angular momentum (l). The state is a combination of the spin state (χ+ or χ-) and the position state (Y10 or Y11), with the coefficients indicating the relative probabilities of the two states.

To calculate the probability density for finding the particle with spin up and at radius r, we need to consider the square of the amplitude of the combined state. This would involve taking the square of the coefficients and the square of the wavefunction for each state, and then multiplying them together. The result would be an expression that is dependent on both r and θ (the angle between the z-axis and the position vector).

To obtain the probability density at a particular radius r, we would need to integrate this expression over all possible values of θ. This would involve integrating from θ=0 to θ=π, as the electron can be found at any angle around the nucleus. This would give us the overall probability density at that radius r.

Therefore, the correct expression for the probability density at radius r would be: |R21|^2*(1/3)*|Y10|^2*|χ+|^2 * (1/2π) * ∫(0 to π) (cosθ)^2 dθ. This integral can be simplified to give a final expression that is dependent on r only.

I hope this helps clarify the solution for you. Keep in mind that in quantum mechanics, we usually have to integrate over all possible values of the variables to obtain the probability density at a particular point.