Probability of the outcomes of ##J^2## and ##J_{z}##?

[keyzan]

Homework Statement

determine the possible outcomes of a measurement of J^2 and Jz being J=L+S, the total angular momentum and the related probabilities.

Relevant Equations

-

Hi guys, I have a problem with point 2 of this exercise:

The electron of a hydrogen atom is initially found in the state:

having considered the quantum numbers n,l,m and epsilon related to the operators H, L^2, Lz and Sz.

I am asked: determine the possible outcomes of a measurement of J^2 and Jz being J=L+S, the total angular momentum and the related probabilities.



---------------------------------------



I find the theory of total angular momentum complicated to say the least and I can't quite understand how to calculate the probabilities of the outcomes. Considering exclusively the up component to simplify the calculations (for the moment), I obtain results of J^2 and Jz:


J^2 results: 3/2, 1/2

Outcomes of Jz for j=3/2: -3/2, -1/2, 1/2, 3/2

Outcomes of Jz for j=1/2: -1/2, 1/2



Now I can't figure out how to calculate the probabilities for each outcome. Maybe I should consider that the state with the old quantum numbers splits into 6 states? I really don't know how to continue please help me.

 

[kuruman]

Are you familiar with Clebsch-Gordan coefficients? If so, then write the given linear combination of the $|l,m_l,\frac{1}{2},m_s\rangle$ states as a linear combination of the $|J,m_J\rangle$ states.

For example the state (not one that is given to you)
$|1,1,\frac{1}{2},-\frac{1}{2}\rangle=\sqrt{\frac 1 3}|\frac{3}{2},\frac{1}{2}\rangle+\sqrt{\frac 2 3}|\frac{1}{2},\frac{1}{2}\rangle$

Looking at the right-hand side, the possible outcomes and probabilities are
$J^2=\frac{3}{2}\times\frac{5}{2}\hbar^2~;~~J_z=\frac{1}{2}\hbar~;~~P=\left(\sqrt{\frac 1 3}\right)^2.$
$J^2=\frac{1}{2}\times\frac{3}{2}\hbar^2~;~~J_z=\frac{1}{2}\hbar~;~~P=\left(\sqrt{\frac 2 3}\right)^2.$

 

[keyzan]

The professor more or less solved it this way, but there are things I don't understand. Let's consider the state:
$|2, 1, 0, +\rangle = |n=2, l=1, m=0, m_{s}=1/2\rangle$
Then we have to consider the matrix:

$\hspace{3cm}m=-1\hspace{1cm}m=0\hspace{1cm}m=1$

$m_{s} = \frac{1}{2} \hspace{2cm}°\hspace{2.5cm}°\hspace{2cm}°$

$m_{s} = -\frac{1}{2}\hspace{1.6cm}°\hspace{2.5cm}°\hspace{2cm}°$

Let's consider the state at the bottom left (the one with $\hspace{0.5cm}m_{s}=-\frac{1}{2} \hspace{0.5cm}and\hspace{0.5cm} m=-1$).

This is the state $|J=\frac{3}{2}, M = -\frac{3}{2}\rangle$ which corresponds to a state expressed in terms of $l,m,m_{s}$ as:

$|J=\frac{3}{2}, M = -\frac{3}{2}\rangle = |l=1, m=-1, -\rangle$

At this point we act with the ascent operator: $J_{+}$ to go up and get to the state that interests us, that is, the dot in the center and at the bottom with $m=0$ e $m_{s}=-\frac{1}{2}$.

So i consider:

$J_{+} |J=\frac{3}{2}, M = -\frac{3}{2}\rangle = (L_{+} + S_{+}) |l=1, m=-1, -\rangle$

So by doing some calculations we arrive at the same result as yours. Now I wonder, can this method always be applied? Does this seem correct to you?

 

[DrClaude]

So by doing some calculations we arrive at the same result as yours. Now I wonder, can this method always be applied? Does this seem correct to you?

Yes. You can start by a stretched state (i.e., $|M| = J$) and apply ladder operators.

Personally, I prefer the Clebsch-Gordan approach, as the use of ladder operators can become complicated when $J > 3/2$.

 

[kuruman]

I agree with @DrClaude. It's like deriving the Laplacian in spherical coordinates starting from Cartesian. You do it once the long way to see how it is put together and then look it up next time you need it.