# QM: Angular Momentum Matrices (Rotating Molecule)

• Hart
In summary, the three angular momentum components, Lx, Ly, and Lz, can be represented by 3x3 matrices using the given equations. To confirm the commutation relations, the commutator definition is used to show that the matrices fulfill the desired relations. The Hamiltonian of a rotating molecule can be calculated using the given equations, and the energy levels of the molecule can be found by further calculations using matrix multiplication.

## Homework Statement

For $$l=1$$ the angular momentum components can be represented by the matrices:

$$\hat{L_{x}} = \hbar \left[ \begin{array}{ccc} 0 & \sqrt{\frac{1}{2}} & 0 \\ \sqrt{\frac{1}{2}} & 0 & \sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & 0 \end{array} \right]$$

$$\hat{L_{y}} = \hbar \left[ \begin{array}{ccc} 0 & -i\sqrt{\frac{1}{2}} & 0 \\ i\sqrt{\frac{1}{2}} & 0 & -i\sqrt{\frac{1}{2}} \\ 0 & i\sqrt{\frac{1}{2}} & 0 \end{array} \right]$$

$$\hat{L_{z}} = \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right]$$

Q(a). Confirm that the matrices fulfill the commutation relations of angular momentum.

Q(b) Calculate the matrix which represents the Hamiltonian:

$$\hat{H} = \frac{1}{2I}\hat^{L}^{2} + \alpha \hat{L}_{z}$$

of a rotating molecule, where $I$ and $\alpha$ are constants and:

$$\hat{L}^{2} = \hat{L}_{x}^{2} + \hat{L}_{y}^{2} + \hat{L}_{z}^{2}$$

Q(c) Calculate the energy levels of the molecule.

## Homework Equations

Commutation Relations of angular momentum:

$$\hat{L_{x}},\hat{L_{y}} = i\hbar \hat{L_{z}}$$

$$\hat{L_{y}},\hat{L_{z}} = i\hbar \hat{L_{x}}$$

$$\hat{L_{z}},\hat{L_{x}} = i\hbar \hat{L_{y}}$$

Commutator Definition:

$$\hat{A},\hat{B} = \hat{A}\hat{B} - \hat{B}\hat{A}$$

Rest as relevant within the question statement (and subsequent answers)

## The Attempt at a Solution

I have seen examples using Pauli matrices that are 2x2 but I don’t know how to go about this using these 3x3 matrices, i.e. how to adapt to these matrices from the standard Pauli ones.

A bit of help and advice to get me going in the right direction would be great, then I think I should hopefully be OK.

Can you multiply two 3x3 matrices together? If yes, then take the matrix product LxLy, subtract from it the matrix product LyLx and see if result is equal to i*(hbar)*Lz.

Repeat as necessary for the rest of the commutation relations.

.. it was more help with the latter questions, I know now how to do the first part to show the commutation relations.

Hart said:
.. it was more help with the latter questions, I know now how to do the first part to show the commutation relations.

This is vague and not very helpful. Can you do part (b) and write the Hamiltonian as a 3x3 matrix? If not, exactly what can you not do in part (b)? How about part (c)? What troubles you there?

Sorry, I do see that obviously my last post wasn't very helpful or informative of problems.

Right, so:

For the first question (proving the commutation relations):

I have managed to do this for the first two cases without any problems, but I just can't get it to work for the last case, so I must be overlooking something but I can't see what even after checking calculations. This is what I have:

$$\left[\hat{L_{z}},\hat{L_{x}}\right] = -\hbar \left[ \begin{array}{ccc} 0 & -\sqrt{\frac{1}{2}} & 0 \\ \sqrt{\frac{1}{2}} & 0 & -\sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & 0 \end{array} \right]$$

So I don't see where the $i$ terms come from?

For the second question (with the hamiltonian):

I have that:

$$\left(\hat{L_{x}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 0 & {\frac{1}{2}} & 0 \\ \frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & {\frac{1}{2}} & 0 \end{array} \right]$$

$$\left(\hat{L_{y}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 0 & \frac{1}{2}} & 0 \\ -\frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & -\frac{1}{2}} & 0 \end{array} \right]$$

$$\left(\hat{L_{z}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]$$

So then:

$$\hat{L}^{2} = \hat{L}_{x}^{2} + \hat{L}_{y}^{2} + \hat{L}_{z}^{2} = 3\hbar^{2}\left(\left[ \begin{array}{ccc} 0 & {\frac{1}{2}} & 0 \\ \frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & {\frac{1}{2}} & 0 \end{array} \right] + \left[ \begin{array}{ccc} 0 & \frac{1}{2}} & 0 \\ -\frac{1}{2}} & 0 & \frac{1}{2}} \\ 0 & -\frac{1}{2}} & 0 \end{array} \right] + \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]\right) = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right]$$

Which means:

$$\hat{H} = \frac{1}{2I}\hat^{L}^{2} + \alpha \hat{L_{z}} = \frac{1}{2I}\left[ \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{array} \right] + \alpha \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right]$$

.. and then I assume I need to add the matrices together, and combine the constants, which gives:

$$\hat{H} = \left(\frac{1}{2I} + \alpha \hbar\right) \left[ \begin{array}{ccc} 2 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right]$$

.. is this correct?! Not sure if need to do anything else.

For the thirds question (energy levels):

.. I don't know how to do this.

Hart said:
$$\left[\hat{L_{z}},\hat{L_{x}}\right] = -\hbar \left[ \begin{array}{ccc} 0 & -\sqrt{\frac{1}{2}} & 0 \\ \sqrt{\frac{1}{2}} & 0 & -\sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & 0 \end{array} \right]$$

So I don't see where the $i$ terms come from?
The matrix on the right is related to Ly. How?

The rest of your work shows that you are confused about how the square of a matrix is to be calculated. For example, if you want to find Lx2, you don't just square every single matrix element and call that the square of the matrix. That works only if the matrix is diagonal. e.g. with Lz. To do it right, you have to take the 3x3 matrix product LxLx.

.. oh wait, I do actually see where the ' i ''s come from, since the two matrices are just different by a multiplication of i which has been factored outside of the matrix, if that makes sense. My question therefore should rather be, what's going on with the other negative sign that shouldn't be there?!

Ok. I've looked up how to take the matrix product but I don't really understand how to do it still.. any tips?

Hart said:
.. oh wait, I do actually see where the ' i ''s come from, since the two matrices are just different by a multiplication of i which has been factored outside of the matrix, if that makes sense. My question therefore should rather be, what's going on with the other negative sign that shouldn't be there?!
Calculate i*hbar*Ly. This is what you want on the right side of the equation. How does it compare with what you already got?

Ok. I've looked up how to take the matrix product but I don't really understand how to do it still.. any tips?
How did you calculate the matrix products such as LxLy that you needed for the commutators? Do the same thing except that you need to replace Ly with Lx to calculate LxLx and likewise for the other squares.

i.

It differs in that what I got has a factor of $\hbar$ opposed to the $\hbar^{2}$ that I need.. oh wait, no, my correct calculation should be:

$$\left[\hat{L_{z}},\hat{L_{x}}\right] = -\hbar^{2} \left[ \begin{array}{ccc} 0 & -\sqrt{\frac{1}{2}} & 0 \\ \sqrt{\frac{1}{2}} & 0 & -\sqrt{\frac{1}{2}} \\ 0 & \sqrt{\frac{1}{2}} & 0 \end{array} \right]$$

.. so I think it's all good now, just have to double check it through though.

ii.

Oh right OK I'll give that another go now then at calculating those..

$$\left(\hat{L_{x}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} {\frac{1}{2}} & 0 & {\frac{1}{2}} \\ 0 & 1 & 0 \\ {\frac{1}{2}} & 0 & {\frac{1}{2}} \end{array} \right]$$

$$\left(\hat{L_{y}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} {\frac{1}{2}} & 0 & -{\frac{1}{2}} \\ 0 & 1 & 0 \\ -{\frac{1}{2}} & 0 & {\frac{1}{2}} \end{array} \right]$$

and obviously from previously:

$$\left(\hat{L_{z}}\right)^{2} = \hbar^{2} \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]$$

So calculating the Hamiltonian expression again:

$$\hat{L}^{2} = \hat{L}_{x}^{2} + \hat{L}_{y}^{2} + \hat{L}_{z}^{2} = 3\hbar^{2}\left(\left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right]\right) = \frac{3\hbar^{2}}{2}\left(\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\right)$$

and that matrix is an identity matrix, so can denote it just as I, therefore:

$$\hat{L}^{2} = \frac{3\hbar^{2}I}{2}$$

hence:

$$\hat{H} = \frac{1}{2I} \hat^{L^{2}} + \alpha \hat{L_{z}} = \frac{1}{2I}\left(\frac{3 \hbar^{2}I}{2}\right) + \alpha \hat{L_{z}} = \frac{3\hbar^{2}}{4} + \alpha \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right]$$

.. better?!

Hart said:
So calculating the Hamiltonian expression again:

$$\hat{L}^{2} = \hat{L}_{x}^{2} + \hat{L}_{y}^{2} + \hat{L}_{z}^{2} = 3\hbar^{2}\left(\left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right]\right) = \frac{3\hbar^{2}}{2}\left(\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\right)$$
Better, but in need of improvement.

Where did that factor of 3 come from? If you first put the (hbar)2 inside each matrix for Lx2, Ly2 and Lx2, then add the three matrices, what do you get?

Also, when you finally construct the Hamiltonian, it should be a 3x3 matrix because it is the sum of two 3x3 matrices. What you have written down is a constant added to a 3x3 matrix. L2 that is part of the Hamiltonian is a 3x3 matrix the product of a constant times the 3x3 identity matrix.

When the times comes for you to assemble the Hamiltonian, I suggest that you bring all constants inside the matrices, then add everything together.

$$\left(\hat{L_{x}}\right)^{2} = \left[ \begin{array}{ccc} {\frac{\hbar^{2}}{2}} & 0 & {\frac{\hbar^{2}}{2}} \\ 0 & \hbar^{2} & 0 \\ {\frac{\hbar^{2}}{2}} & 0 & {\frac{\hbar^{2}}{2}} \end{array} \right]$$

$$\left(\hat{L_{y}}\right)^{2} = \left[ \begin{array}{ccc} {\frac{\hbar^{2}}{2}} & 0 & -{\frac{\hbar^{2}}{2}} \\ 0 & \hbar^{2} & 0 \\ -{\frac{\hbar^{2}}{2}} & 0 & {\frac{\hbar^{2}}{2}} \end{array} \right]$$

$$\left(\hat{L_{z}}\right)^{2} = \left[ \begin{array}{ccc} \hbar^{2} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \hbar^{2} \end{array} \right]$$

Hence:

$$\left(\hat{L}\right)^{2} = \left[ \begin{array}{ccc} 2\hbar^{2} & 0 & 0\\ 0 & 2\hbar^{2} & 0 \\ 0 & 0 & 2\hbar^{2} \end{array} \right]$$

Therefore:

$$\frac{1}{2I}\left(\hat{L}\right)^{2} = \left[ \begin{array}{ccc} \frac{\hbar^{2}}{I} & 0 & 0\\ 0 & \frac{\hbar^{2}}{I} & 0 \\ 0 & 0 & \frac{\hbar^{2}}{I} \end{array} \right]$$

Also:

$$\alpha \hat{L_{z}} = \alpha \hbar \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right] = \left[ \begin{array}{ccc} \alpha \hbar & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -\alpha \hbar \end{array} \right]$$

Then the result:

$$\hat{H} =\left[ \begin{array}{ccc} \left(\frac{\hbar^{2}}{I} + \alpha \hbar\right) & 0 & 0 \\ 0 & \left(\frac{\hbar^{2}}{I}\right) & 0 \\ 0 & 0 & \left(\frac{\hbar^{2}}{I} - \alpha \hbar \right)\end{array} \right]$$

Bravo. You have successfully completed the first two parts.

Now for part (c). Note that the matrix representing the Hamiltonian is diagonal. What does that mean?

.. finally got there on 3rd attempt!

Um, well are the energies encoded within the Hamiltonian matrix, i.e:

$$E_{1} = \left(\frac{\hbar^{2}}{I} + \alpha \hbar\right) , E_{2} = \left(\frac{\hbar^{2}}{I}\right) , E_{3} = \left(\frac{\hbar^{2}}{I} - \alpha \hbar \right)$$

or something like that?!?

Correct. Since the matrix is already diagonal, you can pick out the energies as you have done.

For future reference: If the matrix is not diagonal, then you will have to diagonalize it first, in which case the resulting eigenvalues will be the energies that you are seeking.

Brill! and thanks for letting me know that.

## 1. What is a rotating molecule and how is it related to angular momentum?

A rotating molecule is a molecule that is in motion and has angular momentum. Angular momentum is a measure of the rotational motion of an object, in this case, a molecule. It is related to the rotation of the molecule around its axis.

## 2. What are angular momentum matrices and how are they used in quantum mechanics?

Angular momentum matrices are mathematical representations of the angular momentum of a rotating molecule. They are used in quantum mechanics to describe the behavior of particles at a quantum level, specifically in relation to their angular momentum. These matrices are used to calculate the energy levels and transitions of molecules.

## 3. How do QM: Angular Momentum Matrices differ from classical mechanics?

In classical mechanics, angular momentum is described as a vector quantity, while in quantum mechanics, it is described as an operator. This means that the angular momentum of a rotating molecule can take on discrete values, rather than continuous values as in classical mechanics. Additionally, in quantum mechanics, the angular momentum of a particle is subject to the uncertainty principle, meaning that the exact position and momentum of a particle cannot be simultaneously known.

## 4. How do QM: Angular Momentum Matrices affect the behavior of molecules?

Angular momentum matrices play a crucial role in understanding the energy levels and transitions of molecules. They determine the allowed energy states and the probability of a molecule transitioning from one state to another. This, in turn, affects the overall behavior and properties of the molecule.

## 5. What are some practical applications of QM: Angular Momentum Matrices?

QM: Angular Momentum Matrices have practical applications in areas such as spectroscopy, where they are used to analyze the energy levels and transitions of molecules. They are also used in quantum chemistry to understand the behavior and properties of molecules. Additionally, these matrices are important in the development of technologies such as lasers and magnetic resonance imaging (MRI).