What Are the Expectation Values of Quantum Angular Momentum Components?

[natugnaro]

[SOLVED] Qunatum Angular Momentum

Homework Statement

Particle is in state

\psi=A(x+y+2z)e^{-\alpha r}

r=\sqrt{x^{2}+y^{2}+z^{2}
A and alpha are real constants.

a) Normalize angular part of wave function.
b) Find <\vec{L}^{2}> , <L_{z}>
c) Find probability of finding L_{z}=+\hbar.

Homework Equations

{L}^{2}= \hbar^{2}l(l+1)|lm>
L_{z}=m \hbar|lm>

The Attempt at a Solution

I have found a) part

T(\theta,\phi)=\frac{1}{2\sqrt{3}}(1+i)Y^{-1}_{1}-\frac{1}{2\sqrt{3}}(1-i)Y^{1}_{1}+\frac{2}{\sqrt{6}}Y^{0}_{1}

b)

Since Y^{m}_{l}=|lm>

Using L_{z}=m \hbar|lm>

<Lz>= 1/(4*3)*2<1-1|Lz|1-1> + 1/(4*3)*2<11|Lz|11> + 4/6<10|Lz|10> = 0

To find &lt;\vec{L}^{2}&gt; I would apply operator of L^2 to angular part of wave function, just like I have done for Lz.

{L}^{2}= \hbar^{2}l(l+1)|lm&gt;

Is this is the way to find expectation values ?

c)

P(\hbar)=|-\frac{1}{2\sqrt{3}}(1-i)|^{2}=\frac{2}{12}

 

[Avodyne]

All correct!

You do realize that the value \langle\vec{L}^{2}\rangle is immediately obvious, right?

 

[natugnaro]

You do realize that the value \langle\vec{L}^{2}\rangle is immediately obvious, right?

No, can you explain ?
I can see that l=1 , and m={-hbar, 0 , hbar}, but why is \langle\vec{L}^{2}\rangle obvious ?

 

[malawi_glenn]

No, can you explain ?
I can see that l=1 , and m={-hbar, 0 , hbar}, but why is \langle\vec{L}^{2}\rangle obvious ?

You only have states with l = 1 , then what can you say about the expactation value?

 

[natugnaro]

Possible values for measurment of L are L=hbar*sqrt(l(l+1)) so L^2=hbar^2*(l(l+1)) .
Because I have l=1
L^2=2*(hbar)^2 , but I only have states with l=1 so it must be also
<L^2>=L^2=2*(hbar)^2 right ?

 

[Avodyne]

Right. Your state is an eigenstate of \vec{L}^{2} with eigenvalue 2\hbar^2. So the expectation value (for a normalized state) is the same as the eigenvalue.

 

[natugnaro]

ok, thanks for hints and replies.