When the hydrogen atomic 2p_x is acted upon by the operator for z-component of t

Question

When the hydrogen atomic 2p_x is acted upon by the operator for z-component of the angular momentum L_z we find that....

Attached image.

Since the answer is already provided, what I need is actually the LOGIC/EXPLANATION between each step of the solution process, obviously I am not understanding how one led to the next. Such as where the sqrt 2 came from, and how the h-bar comes into play and the overall conclusion that Lz is not a eigenfunction of Lz.

When the hydrogen atomic 2p, is acted upon by the operator for the z-component of the angular HC-9 we find that momentum, ( (A)2 (B)2p. 2p, is not an eigenfunction of L, , thus there is no definite value 2p, is an eigenfunction of L1 , with eigenvalue 0 . 2p, is an eigenfunction of L,, with eigenvalue l /i 2Pa is an eigenfunction of L‘ , with eigenvalue-l (D) . Knowledge Required: An eigenvalue o, of an operator for a function/ satisfies (s, a; th proerties of operators. Hydrogen orbitals designated px and py are linear combinations of eigenfunctions of L a. ) ; the properties of Thinking it Through: Applying the L, operator to the 2p, function yields 2 p. L Va +LV21-1 V2 (Van-va-1) constant × 2(v211 + 21-1 Since the result of L, acting on 2px is not a constant multiplied by the 2px function, there is no definite eigenfunction. The correct response is (A)

Explanation / Answer

The Lz operator is given as -ih(x/y – y/x) where we read h as (h cross) (sorry I couldn’t write that symbol in MS word.

Also, 2px = 1/2(2 1 1 + 2 1 -1).

When we use the Lz operator on this function, we obtain this result. Point to remember in this case is since Lz is the angular momentum operator in Cartesian co-ordinate system, the wavefunction must be chosen in the same system.

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