Quantum Angular Momentum [SOLVED]

[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 $$<\vec{L}^{2}>$$ I would apply operator of L^2 to angular part of wave function, just like I have done for Lz.

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

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.