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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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.

[snjy]

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.

[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.

[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)!

[arslanullahpractcingphyics]

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 ?

[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?

[paulbk]

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

[vaanff]

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!

[kkhendry]

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?

[yubeenkim]

Awesome!! Thanks for good videos and information!

[岡安一壽-gy]

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) 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
is not an eigenfunction of the Hamiltonian
It should be an ironclad rule of quantum mechanics that the wave function is an eigenfunction of the Hamiltonian.

[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

[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?

[NitinK]

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?

[shruti]

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

[muhammadtanveer]

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.

[motherisape]

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

[vperez]

Good thing that a Math Guy knows about QM theory.

[zlaticakaluzna]

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.

[soumenkhatua]

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