[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 &lt;L_y=1&gt;, we would do:

&lt;L_y=1&gt;=\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 &lt;L_y=1&gt;=\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:

&lt;\psi^{\star}|L_y|\psi&gt;=\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&gt;-\frac{1}{2}|0&gt;+\frac{1}{2}|-1&gt;
find |^{^}L_{\psi}&gt;
...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}<br /> 1/\sqrt{2} &amp; -1/2 &amp; 1/2 \end{array}\right)<br /> \left(\begin{array}{ccc}<br /> ... &amp; ... &amp; ...\\<br /> ... &amp; ... &amp; ...\\<br /> ... &amp; ... &amp; ...\end{array}\right)\left(\begin{array}{c} <br /> 1/\sqrt{2}\\<br /> -1/2\\<br /> 1/2<br /> \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&gt;=\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}|l,m_{\pm}1&gt;

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

L_{y}=&lt;\psi|L_{y}|\psi&gt;=\frac{1}{2i}[&lt;\psi|L_{+}|\psi&gt;-&lt;\psi|L_{-}|\psi&gt;]
=\frac{1}{2i}[\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{2}}[&lt;1|L_{+}|1&gt;-\frac{1}{2}&lt;0|L_{+}|0&gt;+\frac{1}{2}&lt;-1|L_{+}|-1&gt;<br /> -\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{2}]&lt;1|L_{-}|1&gt;-<br /> \frac{1}{2}&lt;0|L_{-}|0&gt;+\frac{1}{2}&lt;-1|L_{-}|-1&gt;]]
=\frac{1}{2i}[\frac{1}{\sqrt{2}}&lt;1|2&gt;-\frac{1}{2}&lt;0|1&gt;+\frac{1}{2}&lt;-1|0&gt;]
-\frac{1}{\sqrt{2}}[&lt;1|0&gt;+\frac{1}{2}&lt;0|-1&gt;-\frac{1}{2}&lt;-1|-2&gt;]=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 &lt;L_y=1&gt;, we would do:

&lt;L_y=1&gt;=\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 &lt;L_y=1&gt;=\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&gt;=\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}|l,m_{\pm}1&gt;

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}=&lt;\psi|L_{y}|\psi&gt;=\frac{1}{2i}[&lt;\psi|L_{+}|\psi&gt;-&lt;\psi|L_{-}|\psi&gt;]

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]...<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> =\frac{1}{2i}[\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{ 2}}[&amp;lt;1|L_{+}|1&amp;gt;-\frac{1}{2}&amp;lt;0|L_{+}|0&amp;gt;+\frac{1}{2}&amp;lt;-1|L_{+}|-1&amp;gt;&lt;br /&gt; -\hbar\sqrt{(l_{\mp}m)(l_{\pm}m+1)}(\frac{1}{\sqrt{ 2}}]&amp;lt;1|L_{-}|1&amp;gt;-&lt;br /&gt; \frac{1}{2}&amp;lt;0|L_{-}|0&amp;gt;+\frac{1}{2}&amp;lt;-1|L_{-}|-1&amp;gt;]] </div> </div> </blockquote><br /> 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, <br /> <br /> L_+\vert0\rangle=\hbar\sqrt{(1-0)(1+ 0+1)}|1,0+1\rangle=\sqrt{2}\hbar|1,1\rangle<br /> <br /> while<br /> <br /> L_+\vert1\rangle=\hbar\sqrt{(1-1)(1+ 1+1)}|1,1+1\rangle=0.<br /> <br /> You also need to operate on the state |\psi\rangle with L_y, <b>before</b> 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,<br /> <br /> \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:

&lt;\psi^{\star}|L_y|\psi&gt;=\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 L_y\vert\psi\rangle[/tex]...<br /> 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{\sqrt{(l\mp m)(l \pm m+1)}} will have different values when L_y operates on each state. For example, &lt;br /&gt; &lt;br /&gt; L_+\vert0\rangle=\hbar\sqrt{(1-0)(1+ 0+1)}|1,0+1\rangle=\sqrt{2}\hbar|1,1\rangle&lt;br /&gt; &lt;br /&gt; while&lt;br /&gt; &lt;br /&gt; L_+\vert1\rangle=\hbar\sqrt{(1-1)(1+ 1+1)}|1,1+1\rangle=0.&lt;br /&gt; &lt;br /&gt; You also need to operate on the state |\psi\rangle with L_y, &lt;b&gt;before&lt;/b&gt; 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,&lt;br /&gt; &lt;br /&gt; \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

&lt;br /&gt; &lt;br /&gt; thank for the thorough explanation,&lt;br /&gt; to answer part 1, the +/- at RHS is not a subscript, the latex formatted it that, it doesn&amp;#039;t meant to be. after reading your code, i understand how to format it already.&lt;br /&gt; &lt;br /&gt; part II, i am calculating the expectation value, not the &lt;br /&gt; L_y\vert\psi\rangle. as i read the question wrongly (it was a handwritten one)&lt;br /&gt; &lt;br /&gt; 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!