Griffiths' Quantum Mechanics Problem 4.27 : Diatomic particles

[hmparticle9]

Homework Statement

Two particles are attached to the ends of a massless rigid rod of length $a$. The system is free to rotate in three dimensions about the (fixed) centre of mass.

What spectrum would you expect for this system?

Determine the distance between the atoms.

Relevant Equations

$E = \frac{L^2}{2I}$ where $L$ is the angular momentum.

I have included a screenshot of the question as we need to use the figure for the last part of the question anyway.

For part (a) I have said the following:

Let the centre of mass be $R = \frac{m_2 a}{m_1 + m_2}$, then the moment of inertia is $$I = m_1R^2 + m_2(a-R)^2 = \frac{m_1 m_2}{(m_1 + m_2)}a^2.$$ Also, $E = \frac{1}{2} m v^2 = \frac{1}{2m}p^2$, the rotational version is $E = \frac{1}{2I}L^2$. Quantum mechanically, the only values this can take are it's eigenvalues. The eigenvalues of $L^2$ are given in the text. Hence
$$E_n = \frac{h^2}{2I}n(n+1)$$

For b) I have said that the normalised eigenfunctions are the spherical harmonics $Y_n^m$. From the text we have the following equation:
$$L^2 Y_n^m = h^2 n(n+1) Y_n^m$$
where $n = 0, \frac{1}{2}, 1, \frac{3}{2}, 2,... \text{ and } m = -n, -n + 1 , ... , n-1, n.$
This means the degeneracy of of the nth energy level is $2n+1$, surely? as for one eigenvalue we have $2n+1$ different eigenvectors.

For c), this is where I am stuck.

 

[kuruman]

where $n = 0, \frac{1}{2}, 1, \frac{3}{2}, 2,... \text{ and } m = -n, -n + 1 , ... , n-1, n.$

Suppose $n=\frac{1}{2}.$ Can you find an expression for $Y_{1/2}^m(\theta,\phi)$? I didn't think so.

For (c) you need to make a list of the energies that are allowed. That's an energy spectrum. Same as an energy level diagram. It would be helpful to find the distance in energy between adjacent energy levels.

 

[hmparticle9]

I see. No is the answer. The equation that I state originally appeared in the following form:

$$L^2 f_n^m = h^2 n(n+1) f_n^m$$
where $n = 0, \frac{1}{2}, 1, \frac{3}{2}, 2,... \text{ and } m = -n, -n + 1 , ... , n-1, n.$

And then later on, via some other way, it is said that $f_n^m = Y_n^m$ for integer values. A bit sloppy on my part.

 

[kuruman]

I see. No is the answer. The equation that I state originally appeared in the following form:

$$L^2 f_n^m = h^2 n(n+1) f_n^m$$
where $n = 0, \frac{1}{2}, 1, \frac{3}{2}, 2,... \text{ and } m = -n, -n + 1 , ... , n-1, n.$

And then later on, via some other way, it is said that $f_n^m = Y_n^m$ for integer values. A bit sloppy on my part.

OK. See my addendum to post #2. Can you produce an energy level diagram? Note that the zero of energy is arbitrary, so you might as well call the energy of the ground state zero and then find the energies $E_n-E_{\text{ground}}~$ above it.

 

[hmparticle9]

If our ground level is zero then:
$E_n - E_{ground} = E_n = \frac{h^2}{2I}n(n+1)$??

I am looking through my textbook to see where "spectrum" appears. I think we need to link the energy difference to the energy of a photon omitted when our system changes energy levels.

$E_{\gamma} = \frac{h^2}{2I}n(n+1) = hv = 2 \pi \hbar v$

 

[kuruman]

I am looking through my textbook to see where "spectrum" appears.

A spectrum appears in front of your eyes in post #1. Read the first line of part (d). Do you understand how to interpret it?

I think we need to link the energy difference to the energy of a photon omitted when our system changes energy levels.

Yes, you do. Suppose you send a beam of white light (all frequencies equally represented) through a sample of CO. What photon frequencies do you expect to be absorbed? A list of those constitutes "a spectrum."

In this case, it will be a number times $\frac{\hbar^2}{2I}~$ and please, please, please use $\hbar$ (\hbar) not $h$ for the energy expressions and the Hamiltonian. If you don't, your answer in part (d) will be off by a factor of 40 or $(2\pi)^2.$

 

[hmparticle9]

I am still not clear on how to proceed with part (c) @kuruman

 

[JimWhoKnew]

I am still not clear on how to proceed with part (c) @kuruman

@kuruman hinted in #2

It would be helpful to find the distance in energy between adjacent energy levels.

(Very) loosely speaking, the emitted/absorbed photon has spin 1, so only transitions for which $\Delta l= \pm 1$ ( $\Delta n= \pm 1$ in the notation used in OP) are allowed. It is called selection rules.

 

[hmparticle9]

The difference between adjacent energy levels $\Delta E = \frac{\hbar^2}{I}n$. But what in the statement of (c) tells us that we need to look for the difference between adjacent levels?

Spin is covered in the next section, so I don't want to rely on that. Is there another way to explain it? I get your explanation, though :) I have scanned my textbook and I can not find anything that suggests that only unit jumps in energy levels are permitted.

If I set $\Delta E = \frac{\hbar^2}{I}n = E_{\gamma} = \hbar 2 \pi v$ then I get the "answer":
$$v = \frac{\hbar}{2\pi I}n$$
which is the "answer".

But I am not 100% on the reasoning on how we get there.

 

[kuruman]

I am still not clear on how to proceed with part (c) @kuruman

Read about absorption spectroscopy here.
Look at the modified figure below. In view of what you read, describe what it shows.
The horizontal axis is energy in units of cm^-1. Find out how this unit of energy is defined. Then answer the questions following the figure.

Questions to answer:

1. The colored lines show values of energy. What do these energies represent?
2. Write mathematical expressions for each of them.
3. The spacing between adjacent lines appears to be constant. How does your mathematical formulation in your answer to question 2 account for this?

Spin is covered in the next section, so I don't want to rely on that.

You don't have to.

 

[JimWhoKnew]

You don't have to.

You are right. No need to apply selection rules in this case.

 

[kuruman]

I have scanned my textbook and I can not find anything that suggests that only unit jumps in energy levels are permitted.

How about the title "Quantum Mechanics"? The molecule has discrete rotational energy levels $$E_n=\frac{\hbar^2}{2I}n(n+1).$$ Photons are absorbed (or emitted) if they have energies equal to rotational energy level differences, e.g. $$h\nu_{\text{12}}=E_{\text{2}}-E_{\text{1}}=\frac{\hbar^2}{2I}\left[n_2(n_2+1)-n_1(n_1+1)\right].$$It's energy conservation after all.

 

[hmparticle9]

Okay, I am going to answer one of your questions.

1. The dips in energy represent energy loss due to photons of light being omitted. The frequency they are omitted is given by the x-axis position.

From the article that you linked I can summarise:

Molecules absorb radiation most effectively at frequencies that correspond to the energy difference between two of their quantum mechanical energy levels. When a photon's energy matches the gap between a molecule's lower and higher energy states, it can be absorbed, causing the molecule to transition to the higher energy state.

We can replace absorb with emit. But it does not say that they have to be neighbouring states?

 

[kuruman]

We can replace absorb with emit. But it does not say that they have to be neighbouring states?

I agree, it doesn't. Nevertheless, there are transitions between neighbouring states as indicated by the coloured lines in the modified spectrum in post #10. I am asking you to consider them because I want to help you and because I know where this is going. So please answer my questions in post #10.

I will be off line for the next 3-4 hours.

 

[hmparticle9]

1. The only way our sample of CO would lose energy is to give it out via em radiation. Those dips in rotational energy are photons being emitted. But why is it a spike? How/why does the rotational energy recover?

I am really having trouble with 2 and 3. If my answer to 1 is correct the expression for the energy would be $E_n - h \nu_{\text{red}}$

I have read the article a good few times. Sorry, I am trying my best.

 

[kuruman]

I am really having trouble with 2 and 3. If my answer to 1 is correct the expression for the energy would be En−hνred

Look at the picture below. Imagine the vertical to be Intensity of detected radiation and the horizontal axis to be photon energy detected by the detector. Also imagine the blue line to show what you get without the CO sample in place. It is a featureless blob. With the CO in place you get the structure in back with all those dips, four of which are labeled with the coloured dashed lines. These dips show loss of intensity at specific energies.

For example, the red dashed line occurs at about 34 cm^-1. First explain with words, not equations, what energy that is. Then write a mathematical expression for it. Do the same for the orange dashed line and the energy at 38 cm^-1. The distance, in energy, between the orange and red dashed lines is 4 cm^-1. Write another mathematical expression for it.

Do all that first and we'll take from there.

 

[hmparticle9]

Okay. Light is made of photons. Different colour photons have different energies. Intensity is the amount of energy per second per unit area. The red line shows a dip in the intensity of detected radiation. This means that the sample of CO has absorbed energy. Is that correct? @kuruman

 

[kuruman]

That is correct. Let's continue incrementally. Please answer the following three questions. We are doing part (d) now and, in so doing, we will verify part (c).

Question 1
The absorbed energy by the CO is in the form of what?
(A) Kinetic energy $\frac{1}{2}mv^2$
(B) Gravitational potential energy $mgh$
(C) Elastic energy $\frac{1}{2}kx^2$
(D) Thermal energy $kT$
(E) Other. Explain.

Question 2
Given your answer to Question 1, describe the source of the absorbed energy in words.

Question 3
What is an appropriate mathematical expression for your answer in Question 2? This is part (c).

There are more questions to come for part (d).

 

[hmparticle9]

Question 1:
I think it could be A. When the CO absorbs energy it gets excited and so we go from $E_n \rightarrow E_{n+1}$. $E_n$ are the eigenvalues of the rotational kinetic energy. Hence the energy must be kinetic.

 

[kuruman]

Question 1:
I think it could be A. When the CO absorbs energy it gets excited and so we go from $E_n \rightarrow E_{n+1}$. $E_n$ are the eigenvalues of the rotational kinetic energy. Hence the energy must be kinetic.

OK, that's good. What about the answer to Question 2? What is the source of this kinetic energy? What could possibly be the mass and the speed in the expression for the kinetic energy that is the source of excitation?

 

[hmparticle9]

Question 2:
The absorbed energy comes from photons travelling with a certain frequency.

 

[kuruman]

Question 2:
The energy comes from photons travelling with a certain frequency. Photons travel at the speed of light ?

Yes, they do travel at the speed of light and they have zero mass. The energy of a photon is $E=h\nu.$ There is no $\frac{1}{2}mv^2$ to consider. However, photons also have momentum $p=\dfrac{h\nu}{c}.$

At this point I will give you additional hints. Imagine a collection of CO molecules in a container at some temperature $T.$ Classically, they can be viewed as rods that can store energy in three different ways, (a) translational kinetic energy of the centre of mass (CM), (b) rotational kinetic energy and (c) vibrational energy. When you add energy to the gas, say by adding heat to it, you can excite all these modes. Remembering that the average kinetic energy of the CM is $\frac{3}{2}kT$, you can see the mechanism at work here.

CO molecules bump into each other and exchange mechanical energy. Say a molecule gains kinetic energy $\delta K$ after a collision. Where does this added energy go? It can all go in increasing the kinetic energy of the molecule's CM. However, if there is enough energy to excite it to the next rotational state, the molecule can also make the transition to that higher state and only the remaining amount of added energy will increase the kinetic energy of the CM.

Of course, in a collision between molecules when one gains energy, the other loses energy. Suppose that this loss includes a transition to a lower rotational energy state with the emission of a photon. Is the photon likely to make it to the detector? The answer is "no" for two reasons, (a) the photon is emitted over a solid angle of $4\pi$, i.e. in all possible directions, not specifically in the direction to the detector; (b) because this emitted photon has energy equal to the difference between two rotational levels, it is very likely that it will be absorbed by another CO molecule and make the transition to a higher rotational state.

Bottom line: Photons that have energies equal to energy level differences are preferentially taken away from what the detector detects, hence the dips in intensity. There you have it. Now please answer Question 3.

 

[hmparticle9]

Where do you get this information from? I am a mathematician by training and so with regards to physics I only really know classical mechanics (maybe we can discuss later).

It has taken me a few reads, but I understand your post now. But if I was to be asked what the source of the absorbed energy was I would still say "from photons from a light source". Your third paragraph was really interesting. That is, when inter-CO collisions occur, the resulting photon from a reduction in energy level would unlikely make it to the detector. That is why we have dips.

Your "bottom line" is clear as day. Thank you very much.

My expression for Question 3 relies on the differences between energy levels being 1. For instance:
$$E_{\gamma} = E_{n+1} - E_n$$
Is it not possible for a CO molecule in energy state $m$ able to collide with a CO molecule in energy state $n$?

Well. Maybe we can say:
$$2E_{\gamma} = E_{n+2} - E_n$$

 

[kuruman]

My expression for Question 3 relies on the differences between energy levels being 1. For instance:
$$E_{\gamma} = E_{n+1} - E_n$$

Aha! Now we are getting to @JimWhoKnew's point in post #8. A photon carries with it angular momentum $\pm \hbar$ no more, no less, no in-between. So when it is absorbed/emitted by a rotator, it can only change the value of $n$ of that rotator by $\pm \hbar$. That's the last piece of puzzle. Can you put it together now?

Well. Maybe we can say:
$$2E_{\gamma} = E_{n+2} - E_n$$

Maybe we cannot. See what I said above.

Where do you get this information from? I am a mathematician by training and so with regards to physics I only really know classical mechanics (maybe we can discuss later).

This information is traditionally taught in the U.S. at the second year undergraduate level after "Electricity and Magnetism" under the title "Modern Physics" which, in my opinion, has become obsolete and ought to be renamed "20th century physics" because that's what it is. If you wish to discuss this later, you might wish to post in the forum "STEM Educators and Teaching."

 

[hmparticle9]

What I meant when I said
$$2E_{\gamma} = E_{n+2} - E_n$$
is that two photons are emitted, not one photon with twice the energy.

 

[kuruman]

What I meant when I said
$$2E_{\gamma} = E_{n+2} - E_n$$
is that two photons are emitted, not one photon with twice the energy.

Two-photon emission is a second order process meaning that it is much less probable to occur than single photon emission..

 

[hmparticle9]

Okay. Is this another 20th century physics thing? Okay.

So to finish part c):
$$E_{\gamma} = 2 \pi \hbar \nu \text{ and } E_{n+1} - E_n = \frac{\hbar^2}{I} n$$
Hence
$$\nu = \frac{\hbar n}{2\pi I}$$

 

[kuruman]

Okay. Is this another 20th century physics thing? Okay.

So to finish part c):
$$E_{\gamma} = 2 \pi \hbar \nu \text{ and } E_{n+1} - E_n = \frac{\hbar^2}{I} n$$
Hence
$$\nu = \frac{\hbar n}{2\pi I}$$

Stick to $E_{\gamma}=E_{n+1}-E_n=\frac{\hbar^2}{I}n$ and forget the conversion to frequency. I don't know why the abscissa in the plot is labeled "v cm^-1, but I do know that cm^-1 is a unit of energy. 1 cm^-1 is equal to the energy of a photon that has wavelength 1 cm or 0.01 m. You can easily figure out what that is in Joules. Frequency has units of Hz.

Furthermore, if you assume that the abscissa in the figure is energy not frequency, you get the correct answer for part (d). So go with that. Before you do anything else, it would be beneficial if you came up with a reasonably detailed procedure of how you propose to extract the C - O separation from the figure.

 

[hmparticle9]

Okay I just looked at the question again. The distance between the dips looks pretty constant. So we just say:
$$\Delta \nu = \nu_j - \nu_{j-1} = \frac{\hbar}{I} \approx 4 \implies I = \frac{\hbar}{4}.$$
We know $m_1, m_2$. Hence we can get $a$ from the formula
$$I = \frac{m_1 m_2}{m_1 + m_2} a^2$$

 

[kuruman]

Okay I just looked at the question again. The distance between the dips looks pretty constant. So we just say:
Δν=νj−νj−1=ℏI≈4⟹I=ℏ4.
We know m1,m2. Hence we can get a from the formula
I=m1m2m1+m2a2

Can you provide a mathematical expression in the form $a= \dots~$ where the right hand side has either constants that you can look up or numbers that you can read off the figure? Then you need to come up with a numerical value for $a$ in meters and make sure it makes sense.

 

[hmparticle9]

I don't understand. What is wrong with my post #29?
$$ a = \sqrt{\frac{\hbar}{4} \frac{m_1+m_2}{m_1m_2}}$$

 

[kuruman]

I don't understand. What is wrong with my post #29?
$$ a = \sqrt{\frac{\hbar}{4} \frac{m_1+m_2}{m_1m_2}}$$

It's incomplete. You didn't do what the last sentence in part (d) is asking you to do. You need to come up with a number. That's the fun part.

 

[hmparticle9]

I get $2.8 \times 10^{-10} \text{m}$

 

[kuruman]

I get $2.8 \times 10^{-10} \text{m}$

It's the correct order of magnitude but not what I got. Please show explicitly the numbers that you substituted for the symbols in the equation below including units in the SI system, e.g.
$\hbar = 1.05 \times 10^{-34}~\text{J}\cdot \text{s}$, etc. etc.
$$a=\sqrt{\frac{\hbar ^2}{\Delta E}\frac{m_1+m_2}{m_1 m_2}}$$ I hope you are doing this on a spreadsheet. It facilitates troubleshooting.

Note the $\hbar^2$ under the radical. That's not what you have. The correct order of magnitude in your calculation is a fluke because number "4" in your equation is not a pure number; it is an energy difference and has units of energy.

 

[hmparticle9]

Look at post #9. my expression for $a$ is correct

 

[hmparticle9]

$\hbar$ is the same as you. Mass of carbon $2 \times 10^{-26}$ and mass of oxygen $2.656 \times 10^{-26}$

To make the dimensions work out I said:
$$\Delta \nu \approx 4 \text{ cm}^{-1} = 12 \times 10^{10} \text{ s}^{-1}$$

 

[kuruman]

Look at post #9. my expression for $a$ is correct

I looked. $\Delta E$ in that post the energy difference between two adjacent energy levels $$\Delta E_n=E_{n+1}-E_{n}=\frac{\hbar^2}{2I}[n(n+1)-(n-1)n]=\frac{\hbar^2}{I}n$$ and I wholeheartedly agree. That's part (c), it is correct and is behind us.

Now we are in part (d). Note that the energy difference in part (c) depends on the value of $n$ which means that it is not constant. Adjacent energy levels are not equally spaced, therefore their photon energies are not equal. Thus, you cannot invoke the equation in post #9, arbitrarily declare $\Delta E$ in that expression equal to 4 cm^-1 because that is the only energy around and say that you have solved the problem. First you need to understand what this 4 cm^-1 energy represents physically, then find a mathematical expression for it which better be independent of $n$ and then put in that expression the 4 cm^-1 energy.

In short, do what I suggested in post #16

For example, the red dashed line occurs at about 34 cm-1. First explain with words, not equations, what energy that is. Then write a mathematical expression for it. Do the same for the orange dashed line and the energy at 38 cm-1. The distance, in energy, between the orange and red dashed lines is 4 cm-1. Write another mathematical expression for it.

The first mathematical expression mentioned above is $\Delta E_n$ and the answer to part (c). What is the second mathematical expression mentioned in the last sentence?

 

[kuruman]

Mass of carbon $2 \times 10^{-26}$ and mass of oxygen $2.656 \times 10^{-26}$

These numbers are meaningless without units.

 

[hmparticle9]

Are we interested in:
$$\Delta \nu = \frac{\hbar^2}{I}?$$

 

[kuruman]

Are we interested in:
$$\Delta \nu = \frac{\hbar^2}{I}?$$

We are interested in extracting $a$ from the data. You need to have a plan for that. I have already given you an outline in post #16

For example, the red dashed line occurs at about 34 cm^-1. First explain with words, not equations, what energy that is. Then write a mathematical expression for it. Do the same for the orange dashed line and the energy at 38 cm^-1. The distance, in energy, between the orange and red dashed lines is 4 cm^-1. Write another mathematical expression for it.

 

[hmparticle9]

So I said, in words,

"The red line shows a dip in the intensity of detected radiation. This means that the sample of CO has absorbed energy."

And you said that was correct. Okay. So we want a mathematical expression for this energy. Surely this dip in energy has to equal the energy of the photon that has been omitted?

 

[kuruman]

Yes, and there are many such dips at different energies corresponding to transitions between different n values.

 

[kuruman]

OK, just to keep you focused. Shown on the right is the energy level diagram (first 5 energies) for the rotational states. It is drawn to scale. The numbers on the vertical axis are $n(n+1).$ Multiply any number by $\frac{\hbar^2}{2I}$ and you have the energy of that level.

Can you use this diagram to make a list of the energies of the missing photons that appear as dips? How many are there if you limit the transitions to only between shown levels?

 

[hmparticle9]

"Can you use this diagram to make a list of the energies of the missing photons that appear as dips?"

$$0, 2\frac{\hbar^2}{2I}, 6\frac{\hbar^2}{2I}, 12\frac{\hbar^2}{2I}$$

There are 4 transitions.

 

[renormalize]

There are 4 transitions.

Remember that transitions can occur between any pair of levels, so there are more than 4 possible.

 

[kuruman]

Please specify the $n$ numbers between which these transitions occur.
Also

1. Note that a photon with zero energy cannot exist much like an object with zero mass does not exist.
2. Justify what criteria you used to include some transitions but not others.

 

[hmparticle9]

The transitions occur between 5 energy levels, from $n=0$ to $n=4$.
I only included transitions between neighbouring energy levels because #26. I should really include all possible transitions.

Well if we limit transitions between shown energy levels from post #43, then from the 5 shown we need to choose 2. Hence $5 \choose 2$