Expectation values for angular momentum

In summary: Once you have that, I'd try applying ##\hat{L}_z## to that state. What result do you get?You get the eigenvalue h^{2}l(l+1), and similarly for L_{z} with eigenvalue hm.
  • #1
Albereo
16
0
Consider a quantum system with angular momentum 1, in a state represented by the vector
[itex]\Psi=\frac{1}{\sqrt{26}}[1, 4, -3][/itex]​
Find the expectation values <L[itex]_{z}[/itex]> and <L[itex]_{x}[/itex]>



I'm reviewing my quantum mechanics; I had a pretty horrible course on it during undergrad. I feel like this should be fairly simple, but I'm just not sure how to start. I think I can say that the state above is an eigenfunction of L[itex]^{2}[/itex] with eigenvalue h[itex]^{2}[/itex]l(l+1), and similarly for L[itex]_{z}[/itex] with eigenvalue hm, but is this on the right track? How do I proceed from there?
 
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  • #2
Albereo said:
Consider a quantum system with angular momentum 1, in a state represented by the vector
[itex]\Psi=\frac{1}{\sqrt{26}}[1, 4, -3][/itex]​
Find the expectation values <L[itex]_{z}[/itex]> and <L[itex]_{x}[/itex]>


I'm reviewing my quantum mechanics; I had a pretty horrible course on it during undergrad. I feel like this should be fairly simple, but I'm just not sure how to start. I think I can say that the state above is an eigenfunction of L[itex]^{2}[/itex] with eigenvalue h[itex]^{2}[/itex]l(l+1), and similarly for L[itex]_{z}[/itex] with eigenvalue hm, but is this on the right track? How do I proceed from there?
You need to get the basics down first before you can solve the problem. A convenient basis to work with when dealing with angular momentum consist of those states which are simultaneous eigenstates of ##\hat{L}^2## and ##\hat{L}_z##. You can label these states using the quantum numbers ##l## and ##m_l##. For a given ##l##, ##m_l## can assume the values ##l, l-1, \ldots, -(l-1), -l##. So the first couple of eigenstates correspond to:

\begin{align*}
l &= 0, m_l = 0 \\
\\
l &= 1 \ \
\begin{cases}
m_l = 1 \\
m_l = 0 \\
m_l = -1
\end{cases}
\\\\
l &= 2 \ \
\begin{cases}
m_l = 2 \\
m_l = 1 \\
m_l = 0 \\
m_l = -1 \\
m_l = -2
\end{cases}
\end{align*}
In light of this information, what linear combination of eigenstates is given in your problem?
 
  • #3
vela said:
You need to get the basics down first before you can solve the problem. A convenient basis to work with when dealing with angular momentum consist of those states which are simultaneous eigenstates of ##\hat{L}^2## and ##\hat{L}_z##. You can label these states using the quantum numbers ##l## and ##m_l##. For a given ##l##, ##m_l## can assume the values ##l, l-1, \ldots, -(l-1), -l##. So the first couple of eigenstates correspond to:

\begin{align*}
l &= 0, m_l = 0 \\
\\
l &= 1 \ \
\begin{cases}
m_l = 1 \\
m_l = 0 \\
m_l = -1
\end{cases}
\\\\
l &= 2 \ \
\begin{cases}
m_l = 2 \\
m_l = 1 \\
m_l = 0 \\
m_l = -1 \\
m_l = -2
\end{cases}
\end{align*}
In light of this information, what linear combination of eigenstates is given in your problem?

I've got a linear combination of the states |##l## , ##m##[itex]_{l}[/itex]> = |1, -1>, |1, 0>, and |0, 1>. Do I now need to determine the coefficients, or do I proceed some other way?
 
  • #4
|0, 1>? That's not an allowed combination.
 
  • #5
Oops, I meant |1, 1>.
 
  • #6
OK, good. You're given the coefficients. How would you express the state ##\Psi## in Dirac notation?

Once you have that, I'd try applying ##\hat{L}_z## to that state. What result do you get?
 
  • #7
Okay, the Dirac notation is definitely one spot where I'm a bit fuzzy. But I think it would be something like:

[itex]\Psi[/itex]=[itex]\frac{1}{\sqrt{26}}[/itex]|1, -1> + [itex]\frac{4}{\sqrt{26}}[/itex]|1, 0> - [itex]\frac{3}{\sqrt{26}}[/itex]|1, 1>​

But is the ordering of states important here? (If this is correct)?
 
  • #8
Yeah, you have the right idea. I'd write ##\vert \Psi \rangle## on the lefthand side, though. The order is indeed important. You should look up what convention your class is using to get it right.
 
  • #9
Thanks for all your help so far. This isn't coursework, so perhaps I'll look up the conventions later and just pick my ordering for the moment.

So with that written for the state, to find the expectation values, I'd have to do
<##L##[itex]_{z}[/itex]> = <[itex]\Psi|\frac{h}{i}\frac{\partial}{\partial \varphi}|\Psi>[/itex]​
and so on, where the |##l##, ##m##> is shorthand for the spherical harmonics?

I can calculate those with the aid of a spherical harmonics table, but I wonder if I'm missing a quicker way to do it? I only ask because this comes from an old qualifying exam, and with time and memory constraints this would seem a bit unreasonable.
 
  • #10
No, you're making it more complicated than needed. ##\lvert l \ m \rangle## is an eigenstate of ##\hat{L}_z##, which means that ##\hat{L}_z \lvert l \ m \rangle = m\hbar \lvert l \ m \rangle##.

If you were working in the ##\theta##, ##\phi## basis, then you'd have
$$\langle \theta\ \phi\ \vert\ l\ m \rangle = Y_{lm}(\theta,\phi)$$ and
$$\hat{L}_z \rightarrow \frac{\hbar}{i}\frac{\partial}{\partial \phi}.$$ In this representation, the eigenvalue equation would be
$$ \frac{\hbar}{i}\frac{\partial}{\partial \phi} Y_{lm} = m\hbar Y_{lm},$$ which you should easily be able to verify, but going to this basis doesn't buy you anything in this problem. It's just needless work. Stick with the abstract notation for now.
 
  • #11
Ah, got it!

So then the expectation value for the z-component is straightforward, as well as the one for ##L##[itex]^{2}[/itex].

But I can't do the same thing for ##L##[itex]_{x}[/itex] because the state isn't an eigenfunction of that operator. Do I then turn to the commutation relations?
 
  • #12
Try looking up the matrix representation of ##\hat{L}_x## and see if you can figure it out from there.
 

Related to Expectation values for angular momentum

What is an expectation value for angular momentum?

An expectation value for angular momentum is a mathematical concept used in quantum mechanics to predict the average value of a quantity, such as angular momentum, when measured on a large number of particles. It is calculated using the wave function of a system and takes into account all possible outcomes of a measurement.

How is the expectation value for angular momentum calculated?

The expectation value for angular momentum is calculated by taking the integral of the product of the wave function and the operator for angular momentum. This integral is then divided by the integral of the squared magnitude of the wave function, known as the normalization constant.

What is the significance of the expectation value for angular momentum?

The expectation value for angular momentum provides a way to predict the average outcome of a measurement of angular momentum for a large number of particles. It also helps to determine the most probable outcome of a measurement.

Can the expectation value for angular momentum be negative?

Yes, the expectation value for angular momentum can be negative. This indicates that there is a higher probability for the angular momentum to be measured in the opposite direction of the positive value. However, the magnitude of the expectation value is still the same as the positive value.

How does the expectation value for angular momentum relate to uncertainty?

The expectation value for angular momentum does not directly determine the uncertainty of a measurement. However, it can be used to calculate the standard deviation, which is a measure of the spread or uncertainty in the measurements of angular momentum.

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