Atomic Term Symbols

An introduction to atomic terms symbols, a bookkeeping method for many-electron wave functions.

[@andresyesidmorenovilla7888]

Mr, from the bottom of my heart and on behalf all confused physics students around the world, thank you.

[@abbyslavens]

I am so thankful for this! You explain this better than my prof, my textbook by Atkins, and all of the other online resources I have looked at.

[@ClawsByCaseyYouTube]

Man I wish you taught at my university....

[@ethangormong7506]

This was a truly exceptional lecture on an otherwise confusing and seemingly arbitrary topic. You have my gratitude!

[@MyGadgetBoy]

Excellent video, clear explanations, great pace. Thank you very much.

[@awesomerahul19]

Great Explanation Sir, Could you please upload more videos on inorganic chemistry Please ?

[@juliamullercraesmeyer2190]

This is THE BEST explanation I found about atomic term symbols. Thank you so much!

[@gaetanoleandropanariello5030]

If I had found this video in the morning, I wouldn't have wasted the entire day. Very clear explanation!

[@deanbitton5766]

I'm really looking forward to watch some more videos of your's. The concept of microstates was never too hard for me, but relating it to the actual quantum numbers (ML, MS etc) was almost impossible. Thank you for making it crystal clear. Lengthy and in depth video

[@alexbohane5959]

Thank you so much for this incredible video! This lecture is by far the clearest explanation I have seen on YT, my only wish would be that you posted more videos as these are so tremendously helpful!!

[@Leticia_Vetrano]

Oh my god, I am so grateful for this wonderful class! You are extremely didactic! I was using McQuarrie to study and I wasn't understanding anything! Thank you very much, Mr! This was really helpful!

[@brainburrito]

Thank you. My textbook was so hard to follow. Trying to read through this subject was like trying to drive through wet cement. This is so much better.

[@MicroageHD]

Textbooks are really horrible when it comes to this indeed.

[@MysteriousSlip]

Man, these are good! I wish you had the whole course uploaded!

[@cathreenacathreena9872]

Dear Dr T. Danial thaaaaaaaaaaaanks from here to the moon and back. so so positive , energetic ...full of tender , extremely giving teacher. wishing u all all the best

[@alifaraghat9727]

You deserve a Nobel prize for teaching me this subject

[@aspie38]

Great! I wish he did more videos. His quantum chemistry explanations would be great!

[@Kaigosu]

Very clear and informative explanations, thank you!

[@dr.abdelmounaimchetoui]

great lecture helped me a lot. Thnak you so much Prof.

[@dhanuddharagamage1313]

Great explanation. Thank you!!!!

[@भवशङ्करदेशिकमेशरणम्]

Thank You Thank You Thank You ....... Thank You So Much Sir..... It was amazing session..... Loved It.... 😍😇🙏 I wish everybody who need it will learn from here....

[@StrangeParticl]

Thank you so much for the great and simple explanation of such a subject

[@tia5558]

This was super helpful. Such a clear and easy to follow explanation. Thank you for this !

[@alexdb2976]

Ty so much Daniel

[@ANWITAROY]

Thank you professor. Indeed a very lucid explanation.

[@lakshithalathpandura6264]

Well Logical Explanation... Great Job... Thank you very much Professor...

[@rajeevk.pathak771]

Lucid. Subtle. Simplified. Palatable. Exciting. Enlightening !!!

[@TheMinecraftSucht]

Really good explanation. You saved me! My exam is in 2 days and I was clueless how this works before I found this :)

[@braket3521]

Hello sir, i have a question. Do complete subshells contribute to the atomic terms? Because when calculating the atomic terms for the n=2 Shell, we have to take into account both 2p and 2s since the n=2 shell is not filled, only the 2s subshell.

[@TDanielCrawford]

Again, sorry to be slow in responding to this, but I didn't see the comment. Completely filled shells have no spin or orbital angular momentum, and thus they contribute nothing to the atomic term symbol. If all shells are completely filled, the term is ^1S.

[@theramkotiindravath]

What an explanation man! Mind blowing!!!!!

[@JulienDrevon]

Thank you sir for your video ! I had a question. I tried to check p2 and p4 equality. When i compute the total spin with 4 electrons i have s1+s2+s3+s4,s1+s2+s3+s4-1,s1+s2+s3+s4-2,....,|s1-s2-s3-s4| ? But if this is true, we have 4 for the maximum value and |-1|=1 for the minimum value. Then we have S=4,3,2,1 ? By applying the Pauli principle we know that 4,3,2 is not possible in this case. We have then only the case where S=1. And for L = l1+l2+l3+l4, ...., |l1-l2-l3-l4| ==> L=4,3,2,1 ? I think i missed understand something in the way to compute L and S... Because i can't find in this example any possible S configuration. Can you help me sir please? Thx in advance

[@TDanielCrawford]

To couple spins (or any type of angular momentum) for several electrons, first couple the spins of just two electrons using the method described in the video, then couple in the next spin of the next electron to the resulting angular momentum from the first two, and then couple the next electron, and so on. For example, if you have two p-type electrons, then you first determine S_12 = s1 + s2, s1+s2-1, ..., and for all the resulting values of S_12, determine S_123 = S12 + s3, S_12+s3-1, ..., etc. I hope this helps!

[@JulienDrevon]

@@TDanielCrawford Yes i think i get it, thank you Sir :)

[@rajatkamboj230]

Best lecture on this topic!

[@jaypatel1449]

does'nt the first three possiblities out of the 15 ways to arrange the 2 electrons in the subshells violating the hunds rule??

[@TDanielCrawford]

Hund's rules are used to estimate the energy ordering of the states *after* you've obtained the terms symbols representing them. Hund's rules don't determine the states themselves.

[@JS-rg7vo]

Thank god, my midterm will only covering single electron systems. Good luck to anyone doing multi electron term symbols

[@samsoft4968]

sir, which text should I follow to get a in-depth view and physical interpretations of term symbols?

[@iqratariq3098]

Thankyou Sir! I really appreciate your efforts and contributions toward real education

[@Rains96]

Super good!

[@hritwikhaldar9272]

What a lecture!! Simply outstanding

[@emilianoalanis3408]

Thank you very much Daniel. Great lecture

[@kishorephysics5928]

wow professor.............................Awesome.................................

[@reginatan9054]

thank you Sir! your explanations were really clear and good! thank you so so much! :)

[@pranjalnobelmukherji1003]

Great explanation professor!!

[@rogerj9425]

This guy is a boss! Great explanation and detailed!

[@FinleyPetit]

Not all heroes wear capes

[@i_amitsharma]

🙏 thanks sir ! you've cleared my doubts... About the allowed and not allowed states.

[@j1o2n3a4s5k6]

I dont get the reasonong for excluding some stuff and including others in the end ? What if we would have begun with the s states ? Seems arbitrary, or did I miss something ?

[@harshbaliyan146]

Simply awsome, thank you!

[@nitink9879]

I have one doubt. At 12:42, you are saying that in case of hydrogen atom, the energy associated with the 2p electron only depends on the principal quantum number 2. But at 15:47 you are saying that there are two groups of wavefunction having different energies corresponding to the same principal quantum number 2. This is confusing me. Please explain.

[@TDanielCrawford]

The energy levels of a hydrogen atom - without considering the spin of the electron - are proportional to -n^2, where n is he principal quantum number. If you include the electron spin, then the spin angular momentum can couple to the orbital angular momentum. For an electron in a p-type orbital, this leads to a *total* angular momentum of either J=3/2 or J=1/2, which are at different energies. In short, if you ignore spin, then the total energy levels of the hydrogen atom depend only on n. If you include spin, then they also depend on the total angular momentum. I hope that helps!

[@kunjalvara6113]

Thanks for the lacture. It is very helpful

[@yashpatley9450]

Sir what about d3 configuration? There are 120 possible quantum states !!! I looked at "partial spin" method in Rita Kakkar's book but I got stuck later !!! I can't figure out why there are two doublet D states and but only one doublet F and P !

[@TDanielCrawford]

The d3 configuration is indeed complicated, but I approach it by working through each possibility, much as I do in this lecture. However, if you understand the process I outline, you can avoid enumerating every possible configuration by taking each potential term symbol and considering whether it's allowed, starting at the highest angular momentum levels and working your way down. Perhaps I should put together another video explaining this configuration.

[@yashpatley9450]

@@TDanielCrawford yes sir I worked out that much. For d3 I got, 4F, 4P, 2H, 2G, and two of 2D, 2F and 2P. The book says there will be only one 2F and 2P and that's right, because only then the degeneracy sums upto 120. But I am unable to comprehend why only one doublet F and P and not two of them as per the method. Thnx 😇

[@ihteshamghani5235]

sir can you be tell me that how to calculate the transition of samarium. which selection rule is used for this transition.

[@ozankarakaya100]

A very clear explanation, thank you so much!

[@joelgeorge5237]

Highly recommended 👏

[@Gio-ps4xi]

this video will save me, tomorrow I have the physical chemistry exam, I still don't understand one thing, when is it necessary to use Ms and Ml? for helium we didn't use them, while for carbon we did, why?

[@humeerabashir8025]

Marvellously explained....thanks

[@aadarshrajsharma1856]

Thanks for your help. You are a life saver..😄

[@ishfaqwani9216]

U r true legend of chemistry sir

[@reemaborkar7741]

it was indeed helpful sir. But I think this is a long cut way. i have come across some Briet method for equivalent electrons. can u upload that video

[@IslamDueren]

why do you put your hand on your back 35:57

[@toufiquejamal3095]

Yes sir,you hoped right I found it very useful.

[@petafoster5084]

Thank you, I did find this very useful :)

[@estherndungu490]

A fantastic video very clear. Thanx!

[@thomasromano9298]

Thank you very much, I found your explanation very helpful and also interesting

[@eriknelson2559]

How can you get from your Hartree microstates to your joint Atomics States? Group likely microstates into anti-symmetric combinations via Slater Determinants and then compute expectation values of S^2, Sz^2, L^2, Lz^2, J^2, Jz^2 ? Also, the fact that the atomic states are not simple product states implies that the electrons combining to create the atomic states are themselves entangled?

[@TDanielCrawford]

The development of the term symbols does not assume simple Hartree products, but instead each microstate is implicitly a Slater determinant comprised of the component orbitals (spin and spatial). Some term symbols are represented by a single Slater determinant, but most others involve linear combinations thereof. (I'm not sure what you mean by "joint atomic states".) Furthermore, even though we are assuming Slater determinants (or linear combinations of a few) represent the atomic states described by a given term symbol, this is only an approximation. A more accurate model would include the effects of dynamic electron correlation.

[@eriknelson2559]

@@TDanielCrawford Is it possible to uniquely identify which Slater determinants are associated with each Term Symbol? And Slater Determinants, as well as combinations of the same, represent Quantum Entangled states, wherein the properties of one electron depend upon those of the others?

[@TDanielCrawford]

@@eriknelson2559 Yes, but not using the approach described in this video. This is often done in electronic structure programs using the unitary group approach whereby one can generate all determinants in a given configuration corresponding to the same S, M_s, L, M_L. (This is analogous to generating symmetry adapted linear combinations of molecular orbitals using point group projection operators or generators.)

[@eriknelson2559]

@@TDanielCrawford Such a method could account for hybrid (e.g. sp3) orbitals? Understand CNO generally "prefer" sp3 and so have tetrahedral symmetry (e.g. methane, ammonia, water). Tangentially, what is physically happening in a dS = dL = dJ = 0 absorption or emission? All of the photon angular momentum is accounted for with an adjustment of M_J ? S,L,J retain the same magnitudes & relative orientations, but the entire atom (or only partially-filled valence sub-shell contributing to the Term Symbol?) "rolls over on its side" vaguely like the planet Uranus tilting over to 90 degrees due to a (hypothesized) collision with another object (representing the "photon") ? Such that M_J changes by +/- 1?

[@TDanielCrawford]

@@eriknelson2559 The specific orbitals that are used is a separate question from how determinants are built. Considering that these are just models, you could use whatever orbitals you like provided they yield the desired accuracy with regard to the target properties (e.g., the structure of a given molecule). I'm afraid I don't understand your section question.

[@hansrajsingh4010]

great explaination sir.

[@fabiogorzonidoro766]

Amazing lecture!!!

[@kanwalmushtaq8282]

Good demonstration and easy explantion 👍

[@cathreenacathreena9872]

what is the book title and version please??

[@sachinsalunke4973]

thank you sir , i expect more videos.really nice lecture

[@ritikthakur4921]

UNDERRATED CHANNEL!

[@samsoft4968]

sir, for d5 system we can have l1+l2+...+l5=2*5=10 but in term symbols we dont see a letter corrresponding to 10 . why? please reply...

[@TDanielCrawfordVT]

Because you cannot have a configuration in which m_l=2 for all five electrons in the d5 configuration.

[@p.pchemistrybyprofessorpra9482]

Shandaar explanation sir

[@fiorellamazzini1401]

He teaches at our university. He's absolutely amazing!

[@ganiesuhailahmad9749]

He is going in a lucid way

[@Dolandtromm]

@@ganiesuhailahmad9749 you are from Kashmir

[@ganiesuhailahmad9749]

@@Dolandtromm Yes,Of course!!!!

[@victoraguilera2267]

how would the terms of 3d10,3d2 y 4p2 ?? plis

[@muraliac2005]

very clear explanation sir...thanking you

[@deepanshu7763]

Thank you so much for this!

[@dexal409]

best video on the subject

[@prabhupadavibes3214]

Sir I think u forgot to tell us about the ground state term symbol

[@luisfernandomejiajiron1828]

Excellent!!!

[@user-ha98]

شكرا لك يا استاذ

[@michalkajan5135]

I dont understand why he count in these configuration with m_l=1 and m_s=0 to triplet P and not to singlet P.

[@TDanielCrawford]

I show first that the ^1D term must appear and that it accounts for five of the 15 possible determinants: (M_L, M_S) = (2, 0), (1,0), (0,0), (-1,0), and (-2,0). The ^3P term accounts for another nine of the determinants: (M_L, M_S) = (1,1), (1,0), (1,-1), (0,1), (0,0), (0,-1). The ^1P term cannot account for the (1,1) and (-1,1) determinants, so it isn't an allowed term symbol for this configuration.

[@braket3521]

true, but what if we started checking which atomic terms are accounted for the other the other way around. Meaning starting from the ^1 S all the way to the ^3 D. then ^3P term would be not allowed. So whats the correct way to deal with this? Is there are a rule that we always have to start from the highest triplet?

[@TDanielCrawford]

@@braket3521 Sorry I didn't see this earlier. The rule is to start from the highest angular-momentum terms because there are fewer ways to account for them based on the available configurations.

[@eriknelson2559]

@@TDanielCrawford Could the singlet-P state exist as some sort of excited state?

[@TDanielCrawford]

@@eriknelson2559 It could certainly arise from an excited electronic configuration, but it's not an excited state in this configuration.

[@warms.8354]

Heck yeah, if I pass my exam I'm gonna go ride a teacup!

[@rabiulchowdhury7904]

great lecture !! thank you so much sir.

[@IslamDueren]

haha what happened on 20:13 and where did your glasses disappear? why did you take them off?

[@stephaniecardinalli5228]

this is a great video - thank you!

[@malyuki39]

you saved my life .. Thanks

[@nUrnxvmhTEuU]

8:58 The (pseudo)vectors L and S can either point the same direction, so the magnitude of J is |L|+|S|, or the opposite direction, so the magnitude of J is |L-S|. But this is vector arithmetics, it has nothing to do with „interference“, which is a wave phenomenon. Using the term interference here is a misnomer.

[@TDanielCrawford]

I don't think the term "interference" is exclusive to describing wave phenomena. My thinking when using this term in the context of angular momentum is akin to two coupled rotating bodies whose angular momenta can enhance or cancel each other out.

[@mobinulhaq50]

Anyone can help me about term symbols of p3 configuration I wrote all 20 now I have no Idea how to write capital S

[@mobinulhaq50]

Molecular Sciences Software Institute Got it finally! Thank you so much...😊

[@恍惚-j3o]

good

[@nicolabellucco5011]

very clear!!!!!!! Thank you so much

[@marcoponts8942]

They only have a two-fold degeneracy in your simple picture you are using for the energy. Maybe specify that when you have for example a magnetic field the two different spin-states are not at all the same energetically.

[@vishalbasu7376]

Sir can u upload videos of quantum chemistry too...please...

[@JS-rg7vo]

this is quantum chemistry. Spectroscopy is an applied field of QM

[@johndb1218]

I LOVE YOU MAN.

[@zms9603]

So clear, thanks A LOT

[@Jeonjongkook7]

شكراً لك

[@ashismodak6481]

Thanks a lot Professor

[@shash2kx]

Fantastic!

[@mubarakayinla]

Thank you so much!!!!!!!!!!!!!!!!!!!!!!!!!

[@moonaliamjad3460]

really liked it.........

[@somnathchavan9909]

what actually mean doublet?????

[@edgarskletnieks5836]

Shalom! Marvelous !