Symmetric and antisymmetric states of many quantum particles

Symmetric and antisymmetric states describe quantum systems of identical particles.
📚 In this video we define totally symmetric and totally antisymmetric states. Totally symmetric states stay the same when we exchange any two particles, while totally antisymmetric states get an extra minus sign when we exchange any two particles. Their properties make them the only quantum states that can describe systems of identical particles. We will also learn how to build these states, using the so-called symmetrizer and antisymmetrizer operators. Finally, we will define totally symmetric operators, that do not change under the exchange of particles. All these ideas will allow us to study the quantum mechanics of systems of identical particles.
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⏮️ BACKGROUND
Identical particles: • Identical particles in...
Tensor product state spaces: • Tensor product state s...
Permutation operators: • Permutation operators ...
Projection operators: • Projection operators i...
⏭️ WHAT NEXT?
Exchange degeneracy: • Is quantum mechanics "...
Symmetrization postulate: • Bosons and fermions: t...
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Director and writer: BM
Producer and designer: MC

[@armalify]

Many thanks. Getting to the point directly, clearly, and smoothly without distraction from irrelevant bits of information.

[@ProfessorMdoesScience]

That's our intention, so happy to hear it works!

[@dutonic]

I think others will find this helpful: Stop the video at 4:00 and just think before doing any math. What is the symmetrizer actually doing? We are taking the sum of every possible permutation of the state we are operating on. Then all we're doing is dividing that sum by the total number of permutations. This is literally just an average. The symmetrizer is simply creating a new state that is the average of all possible states of the system. The operator will obviously be idempotent (squares to itself) because once we operate on the state and extract a *single average state with only 1 possible permutation*, it doesn't matter if we operate on it again. We're just going to be taking the average of a single state with a single permutation, which naturally returns the same state. No math needed. It's called the symmetrizer because by taking the average of all possible combinations and declaring it as the new system state, we have made all possible permutations exactly the same. So when we swap any two particles the system is the same. It's symmetric.

[@ProfessorMdoesScience]

Thanks for the insight! :)

[@snjy1619]

00:02 Symmetric and antisymmetric states play a fundamental role in quantum mechanics. 01:54 States can be totally antisymmetric or totally symmetric. 03:35 Creation of anti-symmetric states and anti-symmetrizer operator 05:15 S operators are projection operators 07:08 The symmetrizer and anti-symmetrizer create totally symmetric and anti-symmetric states. 09:04 Understanding the interplay between permutations and general operators 11:03 Symmetric observables are defined by their commutation with permutation operators. 13:01 Symmetric and antisymmetric states are fundamental for systems of identical particles.

[@NitinKMSP]

I have a doubt at 11:15. What do you mean by D = AB. Is it a tensor product between A and B? If yes, then at 11:30 how do you exactly introduce PP^(dagger) in between the tensor product of A and B? Can you elaborate?

[@ProfessorMdoesScience]

The key is that PP^(dagger) is equal to the identity, so what we introduce is the identity operator. You can think of the overall identity operator acting on the total space as the tensor product between two identity operators each acting on one of the states. I hope this helps!

[@NitinKMSP]

@@ProfessorMdoesScience Are you saying APⓧP†B is always equal to AⓧB ?

[@sayanjitb]

At the playback time 9:31, While proving transformation of observables under permutation you used P_21(🗡) = P_21 because the operator is hermitian. But operator P^ is not in general hermitian right. Then how to prove for general permutation operator? TIA P.S. 🗡 is delineated as a dagger sign (adjoint)!

[@ProfessorMdoesScience]

This is a very good question: you are correct that for a general permutation of more than 2 particles, then permutation operators are not in general Hermitian. The key to proving this for a general permutation operator is to first decompose it into a product of transpositions. Transpositions are Hermitian, and using this fact then the proof follows similar lines to that of the 2-particle permutation P_21. I hope this helps!

[@user-to3fx2do4d]

Hallo sir ! Unfortunately, in general, neither symmetric nor ant-isymmetric wavefunctions can be said to be eigenfunctions of the Hamiltonian. The wave function for an electron in a hydrogen-like atom with atomic number Z in the ground state is RZ(r)=2(Z/a0)^(3/2)*exp(-Zr/a0). RZ(r) is an eigenfunction of HZ=1/(2m)*p^2-Ze^2/(4πε0r). But RZ(r) is not an eigenfunction of HZ'=1/(2m)*p^2-Z'e^2/(4πε0r), Z'≠Z. Let us consider the case where a hydrogen-type atom with atomic number Z and a hydrogen-type atom with atomic number Z' are sufficiently separated from each other. And each electron in each atom is in the ground state. The anti-symmetric wave function Ψ={RZ(r1)RZ'(r2)-RZ(r2)RZ'(r1)}/2^(1/2) is not an eigenfunction of the Hamiltonian H=1/(2m)*p1^2-Ze^2/(4πε0r1)+1/(2m)*p2^2-Z'e^2/(4πε0r2). It should be an ironclad rule of quantum mechanics that the wave function is an eigenfunction of the Hamiltonian.

[@canyadigit6274]

Awesome!

[@kkhendry3]

thank you so much all your videos are so amazing, clear an informative! I want to clarify one doubt here, based on the last slide - is it accurate to say that if I exchange particles an even number of times, I would not be able to distinguish a totally symmetric and totally antisymmetric state since (eta_a)^2 = +1?

[@ProfessorMdoesScience]

For two exchanges of particles you do indeed get back to the same state for both fermions and bosons. I hope this helps!

[@vaanff1942]

Thank u! Keep going pls, this kind videos r as important as pure mathematical fundaments. This is the way to understand mathematics deeply by the phisycal concepts that r behind of it. This is the better an fluid way to understand physics imo! Keep it up!

[@ProfessorMdoesScience]

Thanks for the kind words!

[@yubeenkim8576]

Awesome!! Thanks for good videos and information!

[@ProfessorMdoesScience]

Glad you like them!

[@sayanjitb]

Dear sir, at 14:02, permutation operator acting on total antisymmetric state was defined by |\psi_-⟩ = n_a * |\psi_-⟩ , where n_a ={1 or -1}. So my question is if n_a is +1 depending on the even permutation, why can't I designate that state to a member of the V_+ subspace? because then eigenvalue equation becomes \hat{P} |\psi_-⟩ = 1*|\psi_-⟩ similar to the total symmetric state equation. TIA

[@ProfessorMdoesScience]

Very good question! The reason for this is that to decide whether a state belongs to V+ or V-, you have to see how it transforms under *all* permutations, not just one permutation. This means that state |psi+> belongs to V+ if Pa|psi+>=|psi+> for *all* Pa, not just one. And similarly, |psi-> belongs to V- if Pa|psi->=|psi-> for all even permutations, and Pa|psi->=-|psi-> for all odd permutations. I hope this helps!

[@johnitaballmer3966]

A nice question and a nice answer. Thank you.

[@arslanullahpractcingphyics8969]

I simply have to say you are outstanding and classic. The way you summarize the information in just 14 minutes. can you please recommend a book for your lectures specifically ?

[@ProfessorMdoesScience]

Thanks for your kind words! We have used a variety of books over the years, and some we like include "Quantum Mechanics" by Cohen-Tannoudji, "Principles of Quantum Mechanics" by Shankar, or "Modern Quantum Mechanics" by Sakurai. If there is interest, we may prepare a few videos reviewing the pros and cons of these (and other) textbooks.

[@arslanullahpractcingphyics8969]

@@ProfessorMdoesScience thank you so much for your kind recommendations , it would be nice idea to give us a recipie in an ordered way. our all support is with you professor. thanks

[@ProfessorMdoesScience]

@@arslanullahpractcingphyics8969 We are working on a website to share more material to complement the videos (including problems+solutions) which will hopefully help. We are hoping to go live with it in a few months' time!

[@johnitaballmer3966]

@@ProfessorMdoesScience WOW! I am looking forward to this.

[@paulbk2322]

Heartfelt thanks for such a nice video. Can we have a video detailing on the Dirac delta function please?

[@ProfessorMdoesScience]

We are hoping to start a whole new series on maths for science and engineering, and the delta function will prominently feature there!

[@workerpowernow]

thanks for another great explanation

[@ProfessorMdoesScience]

Glad you like it, and thanks for your continued support! :)

[@NitinK6]

I understood how we are using rearrangement theorem at 4:18. But I do not understand how we are writing the expression at 4:54. Can you please elaborate?

[@ProfessorMdoesScience]

Just to be clear, do you mean the expression P_alpha S- = S- P_alpha = eta_alpha P_alpha? If so, I would recommend that you give a go to the proof yourself, it is very similar to that for P+, but now we need to remember that the definition of S- has an extra eta_alpha in the sum. Do let me know how you get along and we can take it from there!

[@shruti3721]

i have a question....suppose a particle is bound in a bound state in one dimensional hailtonian, then how symmetric and antisymmetric act??

[@ProfessorMdoesScience]

Note that the concept of symmetric and antisymmetric states refers to multi-particle systems, as it refers to the (anti)symmetry under exchange of particles. If you have a single particle bound in a one-dimensional Hamiltonian, these concepts do not feature. However, an interesting concept in that case is whether the single-particle state is symmetric or antisymmetric under spatial inversion; we discuss these ideas in the following videos: * Basics of parity operator: kzbin.info/www/bejne/haTZiZiKerarr8k * Examples with 1D potentials: kzbin.info/www/bejne/pWnPZ4ywbqp7ptU I hope this helps!

[@getarable]

Incredible work! I had a quick question about projection operators, so you say that by showing that the square of the operator (i.e. operator applied twice) gives back the operator is sufficient to assert the operator is a projection operator. But don't we need to show that if the operator is applied n times it still returns the same operator, how is twice enough to generalize to n times?

[@ProfessorMdoesScience]

Glad you like it! Let P^2=P. Then it immediately follows that P^n=P, because you can always write: P^n=P^2*P^(n-2)=P*P(n-2)=P^(n-1) and then continue in this fashion until you get P^n=P. I hope this helps!

[@motherisape]

Should not symetrizer operator, devide sum of all state by sqrt(N! ) instead of N! ? To make probability equal to one 3:09

[@ProfessorMdoesScience]

This is the standard definition of the symmetrizer/antisymmetrizer. Once you have a symmetrized state, you then often still have to normalize it. We discuss this in the video on the symmetrization postulate: kzbin.info/www/bejne/noC8ZmSvbs6kfbM I hope this helps!

[@motherisape]

@@ProfessorMdoesScience thanks

[@muhammadtanveer1747]

Sir how can we define it in general in bra-ket notation ,i mean \ket{+} represents symmetric and \ket{-} represents antisymmetric, but how can we elaborate it.

[@ProfessorMdoesScience]

In this video I only discuss the most general definition of |psi+> and |psi-> so that any state obeying the definition is such a state. To elaborate on what these states look like in specific cases, I look at the example for 2 and 3 particles in the following video (from about 10:20): kzbin.info/www/bejne/noC8ZmSvbs6kfbM

[@NitinK6]

I am unable to understand what you are saying at 10:16

[@ProfessorMdoesScience]

First, remember that we are writing an operator in the full V1(x)V2 tensor product state space that only acts on one subspace, say V1, in a compact form such that instead of writing A_1(x)1_2 every time, we instead only write A_1. But we can go back to the full form and the expression we are discussing becomes: P_{21} A_1(x)1_2 P_{21}^dagger = 1_1(x)A_2 So the action of the permutation on such an operator is to modify it such that rather than acting on the V1 subspace only, after the permutation is acts on the V2 subspace only. I hope this helps!

[@vperez4796]

Good thing that a Math Guy knows about QM theory.

[@ProfessorMdoesScience]

We are both physicists in fact, so if anything, we try our best with the maths ;)

[@zlaticakaluzna8617]

Hello, I want to know, how we can Prove the Palpha*Salpha, if we use this expression in the next calculation as a fact? It was used in the sum of 1 for all alpha. Thank you.

[@ProfessorMdoesScience]

Thanks for your question, could you please specify what minute in the video you are referring to?

[@zlaticakaluzna8617]

@@ProfessorMdoesScience Yes, of course..around 5:26, where you have green round-square of Palpha*S+

[@ProfessorMdoesScience]

I am using Palpha*S+=S+. The prove of this is in the orange line above. The term (1/N!)*sum_beta Pbeta in the orange line is equal to S+ (by the definition of S+), so the first term in the orange line is Palpha*S+. Then in the first equality of the orange line I move the Palpha inside the sum over beta. Then the key step is in the second equality, which uses the fact that sum_beta Palpha*Pbeta = sum_beta Pbeta = S+. The reason why the sum over Palpha*Pbeta is equal to the sum over Pbeta is due to the re-arrangement theorem, which is explained in the video on permutation operators: kzbin.info/www/bejne/o5jUqaytj7KHoNU I hope this helps!

[@zlaticakaluzna8617]

@@ProfessorMdoesScience YES! Thank you :)

[@sandippaul468]

I don't understand the proof of P^(alpha)S^_=ɳ(alpha)S^_ though. It would really pacify me if you say how. Also if we multiply P^(alpha) with P^(beta) might it not happen that the parity would change? So in case of antisymmetrizer the ɳ values has to be revised, isn't it?

[@ProfessorMdoesScience]

In the video we only discuss the proof of Palpha S+ = S+, and leave this one for the viewer. It works in a similar way, and here are a few steps to get you started: Palpha S- = (1/N!) sum_beta eta_beta Palpha Pbeta. We then remember that Palpha Pbeta = Pgamma, where the parity of Pgamma is the product of the parities of Palpha and Pbeta (eta_gamma=eta_alpha*eta_beta). This means we can re-write the above equation as: Palpha S- = (1/N!)eta_alpha sum_gamma eta_gamma Pgamma = eta_alpha S-, completing the proof. I realize that the comments section is not the best format for this, so I recommend you write it out and I hope this helps!

[@sandippaul468]

@@ProfessorMdoesScience Thank you very much. ALso I have to salute your hard work. A long answer like this just to clarify my doubt, jeez!!

[@soumenkhatua1070]

Thank you sir if I could give 100 likes then I would have peace of mind.

[@ProfessorMdoesScience]

Thanks for your support! :)