# [qm]find the angular opertor given total angular momentum wavefunction

1. Sep 5, 2009

### JayKo

1. The problem statement, all variables and given/known data

consider a system with total angular momentum, l=1 in the state

|$$\psi$$>=$$\frac{1}{\sqrt{2}}|1>-\frac{1}{2}|0>+\frac{1}{2}|-1>$$
find |$$^{^}L_{\psi}>$$

2. Relevant equations
$$^{^}L_{z}|\psi>=\hbar m|\psi>$$

3. The attempt at a solution

the basis in the wavefunction given are|1> , |0>, |-1> and there are orthogonal. but i'm not sure what the question is really asking. anyone care to shed some light on this question.thanks

it shoud be L subsript y not psi. just couldnt read the handwriting

Last edited: Sep 5, 2009
2. Sep 5, 2009

### kuruman

I cannot interpret what | Ly> might mean. The symbol for an operator is not normally placed inside a ket. Since there is handwriting involved, is it possible that you are asked to find the expectation value < Ly>? That makes more sense.

3. Sep 5, 2009

### latentcorpse

i've done no work in like four months for summer but am going back soon so can somebody tell me if ive done this correctly please.

say we wanted to find $<L_y=1>$, we would do:

$<L_y=1>=\int_{-\infty}^{\infty} \psi^{\star} \cdot 1 \cdot \psi dx=\int_{-1}^{1} \frac{1}{2}+\frac{1}{4}+\frac{1}{4} dx$ where i have used the orthogonality of the kets.

and so $<L_y=1>=\int_{-1}^{1} dx = \frac{x^2}{2} |_{x=-1}^{x=1}=0$

well that's definitely wrong. im pretty sure i can't just change the limits from infinity to 1 etc in this case. jeez i need to get some work done in the next couple of weeks to get back up to speed lol.

4. Sep 5, 2009

### kuruman

I do not understand what you mean by

$$\left\langle L_{y} } = 1 \right\rangle$$

An expectation value is usually written as

$$\left\langle \psi | L_{y} | \psi \right\rangle$$

an abbreviated form of which is

$$\left\langle L_{y} \right\rangle$$

You have to "sandwich" Ly between the bra and the ket of the wavefunction |ψ> that you have then distribute it among all nine possibilities of bra-kets.

5. Sep 5, 2009

### latentcorpse

bleh...it's going to be a long road back.

ok try this:

$<\psi^{\star}|L_y|\psi>=\int_{-\infty}^{\infty} \psi^{\star} L_y \psi dx$

when we dot product $\psi^{\star}$ and $\psi$ we get 1 though due to ket orthogonality.
what value do I use for $L_y$?
do the limits change due to the allowed values for $L_y$?

6. Sep 5, 2009

### JayKo

sorry for the confusion, what i mean is L$$_{y}$$| 〉 i am going to try it out myself and post here my answer when finish.

7. Sep 5, 2009

### kuruman

Now that makes sense.

8. Sep 5, 2009

### javierR

If you mean <L_y>, then you compute the matrix element
$$\left(\begin{array}{ccc} 1/\sqrt{2} & -1/2 & 1/2 \end{array}\right) \left(\begin{array}{ccc} ... & ... & ...\\ ... & ... & ...\\ ... & ... & ...\end{array}\right)\left(\begin{array}{c} 1/\sqrt{2}\\ -1/2\\ 1/2 \end{array}\right)$$ where the matrix is that of L_y in the 3x3 representation (spin 1). Otherwise, don't know what you meant.

9. Sep 6, 2009

### JayKo

alright here is my solution, please comment.

$$L_{\pm}=L_{x}+iL_{y}$$
$$L_{\pm}|l,m>=\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}|l,m_{\pm}1>$$

$$L_{y}=(L_{+}-L{-})/2i$$

$$L_{y}=<\psi|L_{y}|\psi>=\frac{1}{2i}[<\psi|L_{+}|\psi>-<\psi|L_{-}|\psi>]$$
$$=\frac{1}{2i}[\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{2}}[<1|L_{+}|1>-\frac{1}{2}<0|L_{+}|0>+\frac{1}{2}<-1|L_{+}|-1> -\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{2}]<1|L_{-}|1>- \frac{1}{2}<0|L_{-}|0>+\frac{1}{2}<-1|L_{-}|-1>]]$$
$$=\frac{1}{2i}[\frac{1}{\sqrt{2}}<1|2>-\frac{1}{2}<0|1>+\frac{1}{2}<-1|0>]$$
$$-\frac{1}{\sqrt{2}}[<1|0>+\frac{1}{2}<0|-1>-\frac{1}{2}<-1|-2>]=0$$

Last edited: Sep 6, 2009
10. Sep 6, 2009

### JayKo

interesting this is another solution, beside my way of doing it

11. Sep 6, 2009

### gabbagabbahey

Is there a reason why you've written the $\pm$ as subscripts on the RHS?....Surely you mean

$$L_{\pm}|l,m\rangle=\hbar\sqrt{(l\mp m)(l \pm m+1)}|l,m \pm 1\rangle$$

Right?

First, $L_y$ is just an operator, $\langle L_y\rangle=\langle\psi\vert\L_y\vert\psi\rangle$ is its expectation value.

Secondly, why are you calculating the expectation value? I thought you said the problem asked you to calculate $L_y\vert\psi\rangle[/tex].... You need to be careful here, the states [itex]|0\rangle$, $|1\rangle$ and $|-1\rangle$ correspond to the different values of $m$, so the coefficient $\sqrt{(l\mp m)(l \pm m+1)}$ will have different values when $L_y$ operates on each state. For example,

$$L_+\vert0\rangle=\hbar\sqrt{(1-0)(1+ 0+1)}|1,0+1\rangle=\sqrt{2}\hbar|1,1\rangle$$

while

$$L_+\vert1\rangle=\hbar\sqrt{(1-1)(1+ 1+1)}|1,1+1\rangle=0$$.

You also need to operate on the state $|\psi\rangle$ with $L_y$, before you multiply by $\langle\psi|=\frac{1}{\sqrt{2}}\langle1|-\frac{1}{2}\langle0|+\frac{1}{2}\langle-1|$ and distribute the different inner products. For example,

$$\left(\langle0|+\langle1|\right)L_{-}\left(|0\rangle+|1\rangle\right)=\langle0|L_{-}|0\rangle+\langle0|L_{-}|1\rangle+\langle1|L_{-}|0\rangle+\langle1|L_{-}|1\rangle\neq\langle0|L_{-}|0\rangle+\langle1|L_{-}|1\rangle$$

Last edited: Sep 6, 2009
12. Sep 6, 2009

### gabbagabbahey

First, the integral on the RHS is what you get when you expand $\psi$, $\psi^*$ and $L_y$ in the x-basis....if you don't know what those expanded versions are, there is not much point in using this method.

Second, $L_y$ operates on $\psi(x)$ before you take the product with $\psi^*$, so unless the effect of the operator is to simply multiply by a scalar (say,$\alpha$ ), you can't
say that $\oint\psi^*(x)L_y\psi(x)dx=\alpha\oint \psi^*(x)\psi(x)dx$.

Last edited: Sep 6, 2009
13. Sep 6, 2009

### JayKo

thank for the thorough explanation,
to answer part 1, the +/- at RHS is not a subscript, the latex formatted it that, it doesn't meant to be. after reading your code, i understand how to format it already.

part II, i am calculating the expectation value, not the
$L_y\vert\psi\rangle$. as i read the question wrongly (it was a handwritten one)

part III, the operation is distributive.Noted. thanks for the good effort ;)

14. Sep 6, 2009

### latentcorpse

thanks a lot. had a look over some notes from last year as well as your reply - helped a lot!