Is this right? Re: Finding expectation value of L_z

[Geezer]

Okay, so I'm now reviewing ladder operators (no, not homework).

While reviewing a quantum problem involving the L_z operator at this website (http://quantummechanics.ucsd.edu/ph130a/130_notes/node219.html#example:expectLz"), I found myself confused.

Okay, here's my question: don't we need to take the complex conjugate of the wavefunction for the bra? They have -i for both the bra and the ket...

Am I wrong? It's been a while since I took quantum, so it's possible I'm not recalling some stuff properly...

Thanks,
Geez

PS...Sorry, I don't know how to TeX (but feel free to teach me or show me some good resources to learn), so I don't know how else to show you what I'm talking about.

 

[kuruman]

Yes, they do and you are right. However, the final answer is correct but inconsistent with the work shown. If you multiply things out (i.e. do the angular integrals) according to the second line, you should get 2/3+1/3 not 2/3-1/3 because (-i)*(+i) = +1 for the second term. It's an honest mistake that does not affect the bottom line.

 

[Geezer]

But shouldn't the final answer, then, be 2/3 h-bar + 1/3 h-bar, which is just one unit of h-bar?

 

[Redbelly98]

No, the final answer is (2/3 - 1/3) h-bar = (1/3)h-bar, as stated. They should have +i in the ket. Then this should become -i*hbar when operated on by Lz.

When terms are multiplied out, the second term becomes

(-i)(-i)(1/3)hbar = -(1/3)hbar​

And the +(1/3)hbar final answer makes sense, since in the original wavefunction there is probability (2/3) to measure Lz=+hbar, and probability (1/3) to measure -hbar. (So the expectation value is +(1/3)hbar.)

 

[Geezer]

Okay. I see my error now. I had multiplied the "i's" correctly, but had neglected to apply the m_l, which is -1.

Thanks, everyone!