Fermion creation and annihilation operators
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- Опубликовано: 6 окт 2020
- 📝 Problems+solutions:
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📚 Second quantization allows us to do quantum mechanics in systems with a variable number of particles. This means that we need to be able to add and remove particles, and the creation and annihilation operators allow us to do this. For fermions, these operators obey a set of anticommutation relations that capture, in a compact manner, all the subtleties associated with the totally antisymmetric nature of fermionic states.
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⏮️ BACKGROUND
Symmetric and antisymmetric states: • Symmetric and antisymm...
Symmetrization postulate: • Bosons and fermions: t...
Occupation number representation: • Occupation number repr...
Fock space: • Fock space: variable n...
⏭️ WHAT NEXT?
Boson creation and annihilation: • Boson creation and ann...
One-body operators: • One body operators in ...
Two-body operators: • Two body operators in ...
Change of basis: • Changing basis in seco...
Hamiltonian in second quantization: • The Hamiltonian in sec...
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Director and writer: BM
Producer and designer: MC
Thanks a lot. I needed a few prerequisites in quantum field operators and second quantization, and these lectures were a real blessing!
Glad that you find them useful!!
pure gold (I'll keep commenting and upvoting. I'll suggest your channel to everyone approaching QM)
Thanks for your support!!
Im glad to tell you that you're officially my QM professor!
I watched almost all the videos from the postulates playlist until here with note talking , writing and studying! And again thanks for this amazing work. Im really waiting for your QFT playlist! ❤
Wow, thanks! Glad you like them!
@Andrea Mora Glad you find our videos helpful!
@@ProfessorMdoesScience my professor made notes from your video and gave us.
@@mrtausif1863 Great! May we ask where you study?
Beautiful. It sometimes feels like you can predict my further question and answer it the very next thing!
Really looking forward to see some of a bit more practical topics (like SSP. DFT would be also very nice to see but I understand that it is a bit too far for now =))
We would definitely like to cover SSP and DFT, we'll see if we have the time... :)
For the last few years I have been pursuing my self paced research in Quantum Computing, and came across something called the Jorda-Wigner mapping for fermionic systems where fermionic states are translated to Qubit representations.
A few things however are not clear to me. Would love to see a detailed video series on Jordan-Wigner mapping and its applications.
Thanks for the suggestion! We do have a long list of topics to cover, but we are hoping to get to quantum computing at some point.
Thank you sir for clearing my doubt
Glad you found the video useful!
when are the next videos coming up? Please upload them soon, they are very useful to understand the concepts clearly.
I aim to publish one per week (there will be a new one tomorrow on how to write one-body operators in second quantization). But if there are any particular areas you are interested in, let me know.
This is really great. Very informative and clear, Thank you for making these videos. @Professor M does Science I would like to know what books or other materials you have used as your guide if you have any (any recommendations would be great). I have used the pink book on Molecular Electronic Structure Theory (Helgaker) for Fermionic treatment. I have a request. I have noticed that most of the books on second quantization go over the nomenclature and the implications but there is little or no space dedicated to showing examples using actual Code. Maybe it would be nice to make a video where some of these concepts are applied to real computations (simple hamiltonian models?) using a specific programming language of your like. I have started playing with the Spin operators (S+, S-, Sz, S^2) in second quantization and think those could also be a good example to show as part of a coding session.
I appreciate these lectures as they are, just a suggestion for future videos maybe. Thanks!
Thanks for the comment! For second quantization specifically, the books I typically use are condensed matter physics books (my area of research), for example "Many-body quantum theory in condensed matter physics" by Bruus and Flensberg.
We are actually already complementing the channel with code (so far only in some of the elementary QM videos). But moving forward, we are planning a series on tight binding theory, which could be a good example of second quantization in action, for example using simple 2-band models. We'll try to get to it as soon as possible :)
Amazing!
Thanks! :)
I would loved to learn QM with your video ! It's the future of pedagogy. Would you make some about Quantizitation of field, path integral, noether theorem, casimir effect,... ? I would be great !
Thanks for the suggestions! These are indeed topics that we hope to cover, although it make take some time to get to some of them.
6:23
I don't get the physical meaning of the order the terms are written: |u_2,u_1> is saying that there's a fermion in state 2 and a fermion in state 1. And |u_1,u_2> is saying that there's a fermion in state 1 and a fermion in state 2. Aren't these saying the exact same thing?
Yes, but there is an extra condition on a fermion state: the overall state needs to be totally antisymmetric. To ensure this, we must include the relative minus sign between these two (otherwise equivalent) states. I hope this helps!
Thank you. Nice video. Is there any difference of the explicit Matrix form of fermion and bosonic creation and annihilation operators? For instance, for Bosonic creation, I understand there will be 1, sqrt(2), sqrt(3), etc. in the one row below diagonal (diagonal -1) position and annihilation in the one row above diagonal (diagonal+1) position. What about for Fermions?
In general, to calculate the entries in the matrix form of an operator A in a given basis {|n>} we need the matrix elements:
@@ProfessorMdoesScience thank you. That helps very much.
Do the occupied quantum states u_i,u_j,u_k represent different fermions with different momenta? What if I move an occupied quantum state to the beginning of the list, does it change the momentum of the moved fermion, or does the momentum of the moved fermion remain the same because it is associated with a particular quantum state? Thank you in advance for your answer
First a clarification: the basis states {u_i} are general in this video, so they do not necessarily correspond to momentum eigenstates. However, you could use the momentum basis to write the Fock states, and then we could think of the different u_i as representing states of fixed (and different) momenta. Either way, the key of the occupation number representation we are using is that it does not specify which fermion is which: all we care about is how many fermions occupy each state, and together with the creation/annihiliation operators this allows us to completely take into account the indistinguishable nature of identical particles in quantum mechanics. It may be useful to take a look at the full "second quantization" series:
ruclips.net/p/PL8W2boV7eVfnSqy1fs3CCNALSvnDDd-tb
I hope this helps!
nice thanks, it was much more constructive if you show the explicit matrix forms.
Thanks for the suggestion!
Really helpful video!
I have a doubt.
We know this (ak+)ak |n1,n2...nk..>
(Where ak+ is "a dagger subscript k" operator)
For N=2 spin chain system,
States are |00> , |01>, |10>, |11>
So how to find the action of femionic operators (ak+)ak on these states? Please help me out.
Spin is a type of angular momentum which typically has a small state space. For example, for spin-1/2 particles (like electrons) we can only have one of two states for each particle spin. We don't yet have videos specifically on spin (hopefully coming soon), but our videos on angular momentum cover the general case, of which spin is just an example. You can find the use of raising and lowering operators in that context here: ruclips.net/video/yGvfqRfw1BE/видео.html
I hope this helps!
Strictly speaking, the last expression in the first slide is not zero. Only occupation number at state u_i is zero since it disappears. Other states are 1. Am I correct, professor?
No, the whole multi-particle state is the null state, not just the occupation number u_i. Remember that, in our notation, if u_i appears in an occupation number state, it means that that single-particle state is occupied by an electron. If a multi-particle state has two instances of u_i, then that would mean that in that many-electron state, two electrons are in the same single-particle state. But is not possible from the Pauli exclusion principle, so the whole multi-particle state must be zero. I hope this helps!
At the time stamp 15:54, just to make sure my understanding, if I go from any random fock state | u_j,.....,u_i,..> to another state | u_i,.....,u_j,..> in order to make it compatible with C^_u(i), why don't I need only one transposition in general because only one swap makes it possible to be operated upon. If this is the case then I think (-1)^p is redundant. could you please look into it!
You are correct, you can also think about this as simply swaping the u_i and u_j states, which gives you a (-1), then applying the number operator, and the swapping them back, which gives you another (-1). Overall, the result is the same, you have (-1) to some even power, which gives +1 overall. Either way works, but yours is more elegant in this case :)
@@ProfessorMdoesScience but that I think would not be right in general ..... since only one transposition would mean a sign change would always be there when we change particles ..
Say we change 3rd particle with 1 doing it the general way we require two exchanges and overall get a positive sign but doing it this way would give us a negative sign ...
I still think this is correct in general. Let's consider a simple three-particle state with a starting state |u_i,u_j,u_k>. In the first scenario we exchange u_k with u_i directly, which introduces a factor of (-1):
-|u_k,u_j,u_i>
If I understand correctly what you are saying, then in the second scenario we first change u_k with u_j, and then u_k with u_i, which gives two factors of (-1)^2=1, and gets us to:
|u_k,u_i,u_j>
Note that here the u_i and u_j are in a different order to the first scenario (hence the difference in sign). To really have comparable states, we need an extra swap (u_i and u_j) which gives an extra (-1). This gets us to:
-|u_k,u_j,u_i>
and now we have the same situation as directly swapping them. I hope this helps!
@@ProfessorMdoesScience oh I didn’t take the last exchange into account , I get it now. ….thanks for the reply
What would be the result of Cu2 |u1> = ?
When we write |u1> we understand that all states other than u1 are unoccupied, and in particular the state u2 is unoccupied. Therefore, if you try to remove a particle in state u2 with c_u2, you will destroy the state because there is no particle in that state. You will get zero.
@@ProfessorMdoesScience thanks
@@ProfessorMdoesScienceAnd... What if we want to calculate C^{+}u2 Cu2 |u1> ? It should give us the number of particles in u2 which is Zero, however, due to Cu2|u1>=|0> we get C^{+}u2 Cu2 |u1> = C^{+}u2 |0> = |u2> that's not the number of particles expected
You are correct that c^{+}u2 cu2 is the number operator and should give us the number of particles in state u2. The mistake in your derivation is that cu2|u1> is not equal to |0>, it is equal to 0. |0> is a ket in state space with zero particles, whereas 0 is the null-ket, this is a subtle but crucial difference.
Just to be absolutely clear. When you act with cu1 on state |u1>, then you remove a particle in state u1 and you end up with a state with zero particles, that is, cu1|u1>=|0>. However, when you act with cu2 on state |u1>, you are trying to remove a particle in state u2, but there is no particle in state u2, so that you kill the state, cu2|u1>=0.