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

[JayKo]

Homework Statement

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 |$$^{^}L_{\psi}>$$

Homework Equations

$$^{^}L_{z}|\psi>=\hbar m|\psi>$$

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 couldn't read the handwriting

 

[kuruman]

I cannot interpret what | L_y> 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 < L_y>? That makes more sense.

 

[latentcorpse]

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

 

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

 

[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 $L_y$?

thanks for your help.

 

[JayKo]

I cannot interpret what | L_y> 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 < L_y>? That makes more sense.

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.

 

[kuruman]

Now that makes sense.

 

[javierR]

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 |$$^{^}L_{\psi}>$$
...it shoud be L subsript y not psi. just couldn't read the handwriting

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.

 

[JayKo]

alright here is my solution, please comment.

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

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

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

 

[JayKo]

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

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

 

[gabbagabbahey]

alright here is my solution, please comment.
$$L_{\pm}|l,m>=\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}|l,m_{\pm}1>$$

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?

$$L_{y}=<\psi|L_{y}|\psi>=\frac{1}{2i}[<\psi|L_{+}|\psi>-<\psi|L_{-}|\psi>]$$

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 [itex]L_y\vert\psi\rangle[/tex]...

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

You need to be careful here, the states $|0\rangle$, $|1\rangle$ 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,

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

 

[gabbagabbahey]

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 $L_y$?

thanks for your help.

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 $\oint\psi^*(x)L_y\psi(x)dx=\alpha\oint \psi^*(x)\psi(x)dx$.

 

[JayKo]

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 [itex]L_y\vert\psi\rangle[/tex]...
You need to be careful here, the states $|0\rangle$, $|1\rangle$ and $|-1\rangle$ correspond to the different values of $m$, so the coefficient [itex]\sqrt{\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,

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

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 ;)

 

[latentcorpse]

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