Quantum Mechanics: Question on Angular Momentum

[Collisionman]

Homework Statement

Consider a system that is initially in the state:

$\psi\left(\theta,\phi\right)=\frac{1}{\sqrt{5}}Y_{1,-1}\left(\theta,\phi\right) + \frac{\sqrt{3}}{5}Y_{1,0}\left(\theta,\phi\right)+\frac{1}{\sqrt{5}}Y_{1,1}\left(\theta,\phi\right)$

Part 1: Find $<\psi|L_{+}|\psi>$
Part 2: If $L_{z}$ is measured, what values would one obtain and with what probabilities?

Homework Equations

• $L_{z}|lm>=mh|lm>$
• $L_{+}|lm>=[l(l+1)-m(m+1)]^{\frac{1}{2}}h|lm + 1>$
• $Probability = \frac{|<\varphi|\psi>|^{2}}{<\psi|\psi>}$

The Attempt at a Solution

For Part 1:

So I started off by putting the expression for $\psi\left(\theta,\phi\right)$ in Bra-Ket notation:

$|\psi> = \frac{1}{\sqrt{5}}|1,-1> + \frac{\sqrt{3}}{5}|1,0> + \frac{1}{\sqrt{5}}|1,1>$

Then I applied $L_{+}$ to each individual component:

• $L_{+}|1,-1> = \frac{1}{\sqrt{5}}[1(1+1)-(-1)(-1+1)]^{\frac{1}{2}}h|1,0> = \sqrt{\frac{2}{5}}h|1,0>$
• $L_{+}|1,0> = \frac{\sqrt{3}}{5}[1(1+1)-0(0+1)]^{\frac{1}{2}}h|1,1> = \frac{\sqrt{6}}{2}h|1,1>$
• $L_{+}|1,1> = \frac{1}{\sqrt{5}}[1(1+1)-(-1)(-1+1)]^{\frac{1}{2}}h|1,2> = 0$

So, $L_{+}|\psi> = \sqrt{\frac{2}{5}}h|1,0>$$+ \frac{\sqrt{6}}{2}h|1,1>$

And then,

$<\psi|L_{+}|\psi> = <1,0|\sqrt{\frac{2}{5}}h|1,0>$$+ <1,1|\frac{\sqrt{6}}{2}h|1,1>$
As $<1,1|1,1> = 1$ and $<1,0|1,0>=1$
$<\psi|L_{+}|\psi> = \sqrt{\frac{2}{5}}h$$+ \frac{\sqrt{6}}{2}h$

I think I'm going wrong here somewhere. I think I'm using the wrong complex conjugate. Can someone verify if I am or not?

For Part 2:

I took $\psi\left(\theta,\phi\right)$ in Bra-Ket notation as before, i.e.,

$|\psi> = \frac{1}{\sqrt{5}}|1,-1> + \frac{\sqrt{3}}{5}|1,0> + \frac{1}{\sqrt{5}}|1,1>$

And used $L_{z}|lm>=mh|lm>$ to try and obtain a value for $L_{z}$. I used this on individual components as follows;

• $L_{z}|1,-1> = \frac{-h}{\sqrt{5}}|1,-1>$
• $L_{z}|1,0> = \frac{\sqrt{3}}{5}(0)h|1, 0> = 0$
• $L_{z}|1,1> = \frac{h}{\sqrt{5}}|1,1>$

Then I multiplied by the complex conjugate, i.e.,

• $<1,-1|L_{z}|1,-1> = <1,-1|\frac{-h}{\sqrt{5}}|1,-1>$
• $<1,1|L_{z}|1,1> = <1,1|\frac{h}{\sqrt{5}}|1,1>$

So, $L_{z}=$$\frac{-h}{\sqrt{5}}$$+ \frac{h}{\sqrt{5}} = 0$

Again, I'm not too sure if I'm right or wrong here. If someone could verify if I am or not, I'd really appreciate it. If I know where I'm going with $L_{z}$ I can continue on with finding the probabilities, which I understand how to do.

Thanks again in advance. Any help appreciated!

 

[Collisionman]

Anyone ... please?

 

[TSny]

Generally, your work on part 1 looks ok, but I think there are some minor errors. You wrote the wavefunction as

$\psi\left(\theta,\phi\right)=\frac{1}{\sqrt{5}}Y_{1,-1}\left(\theta,\phi\right) + \frac{\sqrt{3}}{5}Y_{1,0}\left(\theta,\phi\right)+\frac{1}{\sqrt{5}}Y_{1,1}\left(\theta,\phi\right)$

First, I suspect that the numerical coefficient of the Y_1,0 term should have a denominator of √5 rather than 5.

Also, I think you need to check the numerical factors in the following:

$L_{+}|1,0> = \frac{\sqrt{3}}{5}[1(1+1)-0(0+1)]^{\frac{1}{2}}h|1,1> = \frac{\sqrt{6}}{2}h|1,1>$

Finally, when constructing $<\psi|$ you left out the numerical coefficients contained in $|\psi>$

For part 2, you have found the "expectation value" of $L_z$. But that won't give you much information about what values are possible for individual measurements of $L_z$. The only possible value that you can get for a measurement of an operator in QM is one of the eigenvalues of that operator. Note that your wavefunction is a superposition of three eigenstates of $L_z$. Each eigenstate corresponds to a specific eigenvalue of $L_z$.

So, what are the possible values of a measurement of $L_z$ for your wavefunction?

The numerical coefficients of each of the terms in the wavefunction have something to do with the probability of measuring a particular eigenvalue of $L_z$.

 

[Collisionman]

Thanks TSny, that helped a lot!