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Representation theory and totally symmetric ground state?

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mrandersdk
#1
Mar8-12, 10:12 AM
P: 230
Hello

My question is about the ground state of vibrations for a solid. I'm working with graphite and have found out that for k=0 (The Gamma symmetry point), the vibrational modes can be decomposed into irreducible represenations in the following way

Vibration = 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g

To calculate which modes are raman active I need to know the ground state symmetry. Is this always the totally symmetric represenation (A1g) even though it does not belong to the vibrational modes? I think i heard someone mention that the ground state needs to be treated seperately for some reason, so that it is not a problem that A1g not is in the vibrational modes.

I've sone the calculation with A1g as the ground state in this case i get that only E2g is raman active (which is the right result)

Doing the calculation with E2g (which has an energy going towards 0 for k=0) I get the wrong result, so I guess A1g must be the groudn state symmetry.

So my questions are:
- Why do we need to treat the ground state seperately?
- If not which is the ground state of graphite?

Additional question:
- I've also read that the ground state usually have the full symmetry of the solid. When is this not the case?
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M Quack
#2
Mar8-12, 11:24 AM
P: 660
The ground state usually is fully symmetric and therefore belongs to the representation A1g, as you have pointed out.

To find Raman active modes, you have to find excited modes that have the right matrix element between the excited mode and the ground state. Since the ground state is fully symmetric, that means that the excited mode has to have the symmetry of the transition operator (see Wiki for selection rule).

I cannot think of any case where the ground state is not fully symmetric. If it wasn't it would break the symmetry, i.e. lower the symmetry of the system to a lower point or space group.

There are, however, space groups where the point symmetry of a given site is lower than the crystal class.
mrandersdk
#3
Mar8-12, 12:45 PM
P: 230
Ok thanks.

I think I have understood how to find the selections rules.

Could you explain why the vibrational modes only describes the exited modes and not the ground state. I guess it is like that because A1g is not contained in the vibrational modes of graphite (2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g).

I thinks it's is logical as you say that the ground state is fully symmetrical (guess you could think of it as all the atoms being at their equilibrium sites, maybe with some caution), but as i said I'm not quit sure why the ground state is not in the problem, let me explain:

If I count the dimensions of the representations 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g i get 12, which I expected because there are 4 atoms in the unit cell each with three degrees of freedom. The representations tells me that I can devide these eigenfunction into sets that transform among each other, but the fully symmetric representation is not present, that is no of the eigenfunctions of the problem has the full symmetry of the problem, and hence is not the ground state. This is what i mean by it seems like the ground state has to be treated seperately.

I guess somehow that the representation theory approach only describes exited modes, but i fail to see why that is?

Hope that explains my problem a bit clearer.

M Quack
#4
Mar9-12, 02:38 AM
P: 660
Representation theory and totally symmetric ground state?

Vibrations are quantized as phonons. Phonons can be excited by light quanta (photons).
In the Raman process, a photon with energyg E0 hits the zero-phonon ground state and excites one phonon at k=0 (because the optical photon carries not momentum worth speaking of) with energy Ep, another photon with energy E0-Ep is emitted.

The selection rule is the matrix element of polarizability between the zero-phonon and 1-phonon states. The zero-phonon state is completly symmetric, therefore you only have to compare the symmetry of the excited state to the symmetry of the polarizability.

If the initial state were asymmetric, you would have to multiply the representation of the ground state with the representation(s) of the polarizability, decompose the result into irreducible representations, and then compare to the final state.

Stokes and anti-Stokes are the same thing, only that you swap initial and final state.

What I cannot find (after googling for 15min) is a decent description and mathematical derivation of the polarizability in terms of symmetry.
mrandersdk
#5
Mar9-12, 03:06 AM
P: 230
You said:
"The selection rule is the matrix element of polarizability between the zero-phonon and 1-phonon states. The zero-phonon state is completly symmetric, therefore you only have to compare the symmetry of the excited state to the symmetry of the polarizability."

I know how to see if the transition matrix element is zero by the symmetries, and i know how to write the polarizability as the irreducible components. I know how to find the possible final states, namely the states i called the vibrational modes.

My question I guess relates to the sentence "zero-phonon state is completly symmetric", I can see that this it how it should be. But how is this possible when the A1g mode is not contained in the vibrational modes, calculated from the group theory? Is the ground state solution not treated by this theory, or should the ground state be in one of the irreducible representations: 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g (which does not include A1g)? Think I have understood the overall scheme of the theory (how to calculate the vibrational modes, selection rules for both Raman and absorption), but I'm just missing that little piece of the puzzle about the ground state.
M Quack
#6
Mar9-12, 05:54 AM
P: 660
Don't know. I would expect at least one of the acoustic branches to have A1g symmetry, so the question should be asked the other way around: Why is A1g not included in the list of phonon modes?

Here is an article that calculates the phonon modes and has a list of theoretical and experimental references

http://dx.doi.org/10.1016/j.ssc.2004.04.042
DrDu
#7
Mar9-12, 06:16 AM
Sci Advisor
P: 3,563
Quote Quote by M Quack View Post
I would expect at least one of the acoustic branches to have A1g symmetry, so the question should be asked the other way around: Why is A1g not included in the list of phonon modes?
http://dx.doi.org/10.1016/j.ssc.2004.04.042
He considers only modes with k=0, which excludes acoustic modes (which become pure translations of the whole crystal in that limit).

The point is that the symmetry labels
Vibration = 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g
describe the excitation of exactly a single vibrational quantum. The ground state corresponds to no quantum at all whence it is not included in that list and has A1g.
To obtain the symmetry of a state with two vibrational quanta one would have to reduce the direct product (2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g)x(2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g) which will contain A1g again (e.g. A2u x A2u=A1g).
M Quack
#8
Mar9-12, 06:56 AM
P: 660
Quote Quote by DrDu View Post
He considers only modes with k=0, which excludes acoustic modes (which become pure translations of the whole crystal in that limit).
I guess I should have known that...
mrandersdk
#9
Mar9-12, 08:41 AM
P: 230
Hmm I'm not quit sure I understand. I have the vibrational modes

Vibration = 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g

Summing the dimensions I get 12 modes, I guess that must be 9 optical and 3 acoustic modes (4 atoms per unit cell should give 12 modes), so to me it seems like the acoustic modes are included.

I have attached a dispersion relation for graphite, it seems like on that picture that one have

Optical = E1u + 2*E2g + 2*B1g + A2u

Acoustic = E1u + A2u

Adding these one gets my result. Could it be that because graphite is non-symmorphic space group it is one of the special cases where A1g is not the ground state but rather E1u or A2u is the ground state? Or is it still because the theory only describes the system with one quanta of energy and thus the list does not include the ground state A1g, even though it seems like the acoustic branches are included in my list?
Attached Thumbnails
graphitedisp.jpg  
DrDu
#10
Mar9-12, 10:51 AM
Sci Advisor
P: 3,563
Quote Quote by mrandersdk View Post
Hmm I'm not quit sure I understand. I have the vibrational modes

Vibration = 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g

Summing the dimensions I get 12 modes, I guess that must be 9 optical and 3 acoustic modes (4 atoms per unit cell should give 12 modes), so to me it seems like the acoustic modes are included.

I have attached a dispersion relation for graphite, it seems like on that picture that one have

Optical = E1u + 2*E2g + 2*B1g + A2u

Acoustic = E1u + A2u
Well, that is correct, translations of the crystal in plane (x y) transform like E1u and along z like A2u. However, they aren't vibrations in the strict sense.
mrandersdk
#11
Mar9-12, 11:43 AM
P: 230
ok I see. Thanks.

Is it then one of these that are the ground state (E1u + A2u aproaches zero energy, so i guess one of them are), or is it still A1g?

If it is one of them, which is it I should use to calculate if a given transition is raman active, and why?

If they are not, how come the ground state is not included in the allowed vibrational modes? I seem to fail to see where the symmetry analysis exclude the ground state of the analysis, if this is the case?
DrDu
#12
Mar9-12, 01:37 PM
Sci Advisor
P: 3,563
I think the Raman tensor transforms as xy, yz , etc. So I would guess it transforms like E2g and some other representations (you have to consider the character table of the group). Cosnider a Stokes transition from the ground state which transforms as A1g, to a final state with 1 vibrational quantum of symmetry X. So
A1g x E2g x X must contain A1g or equivalently E2g x X must contain A1g. Similar analyses hold for the other components of the Raman tensor. E2g is certainly Raman active.
mrandersdk
#13
Mar10-12, 03:40 AM
P: 230
Oh I know how to do the calculation to determine if a given mode is Raman active and also to find how the raman tensor transform from the charactertables.

My problem is I fail to see how A1g can be the ground state symmetry when it is not an allowed vibrational mode (it is not included in Vibration = 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g). From the dispersion relation I've uploaded it seems like one of the modes E1u + A2u is the ground state as they both tend to zero.

So why is it that A1g can be the ground state when it is not included in Vibration = 2 * E1u + 2 * E2g + 2 * A2u + 2 * B1g? Would have thought that the ground state should be included in theese, and I would also have thought that the ground state should be seen in the dispersion relation I've uploaded, but the A1g mode is not included.

Is it because the ground state is no vibration at all, which is not threated in this problem? If this is the case then it is clear that it should have the full symmetry of the problem e.g. A1g. But do the state "no vibration" exist in quantum mechanics?

Hope this makes it more clear what my problem is.

Thank you for taking your time to help me.
DrDu
#14
Mar10-12, 09:58 AM
Sci Advisor
P: 3,563
Quote Quote by mrandersdk View Post
Is it because the ground state is no vibration at all, which is not threated in this problem? If this is the case then it is clear that it should have the full symmetry of the problem e.g. A1g. But do the state "no vibration" exist in quantum mechanics?
Yes, exactly. that's what I wrote in post #7.
After introduction of the normal coordinates, the qm problem reduces to that of a product of harmonic oscillator eigenstates. The ground state is a product of gaussian functions and can easily be seen to be totally symmetric. The states with one quantum correspond (up to a normalization constant) to one where the ground state function is multiplied by q_i, the coordinate of the ith normal coordinate which transforms according to one of the irreps you wrote down.
mrandersdk
#15
Mar10-12, 12:06 PM
P: 230
Thank you so much, I really appreciate it.

You may also have said this in post #7, but just to make sure I ask again: The reason that i only threat one quantum of energy, is that that i'm only doing pertubation theory to the first order, and to it to higher orders I need to make products of the representations I have found in my first order pertubation?
DrDu
#16
Mar10-12, 03:11 PM
Sci Advisor
P: 3,563
That's one reason. Sometimes you can also observe so called "hot" transitions, when e.g. due to high temperatures the initial state is not the ground state but some state with one or several vibrational quanta excited. Then a stokes transition will create one additional quantum and selection rules become more complicated.
mrandersdk
#17
Mar10-12, 06:18 PM
P: 230
thanks, that's all for now, I think.


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