Two particle systems

[@ILsupereroe67]

You often call "distinguishable particles" two particles whose combined wave function can be written as the product of two separate wave functions, each function only of one particle's position. But it seems to me that has nothing to do with distinguishability, that is rather the two particles being *independent*, (which I think means unentangled).

[@GabrielPohl]

Oh my God! You sound exactly like Eric Foreman from That 70s Show Quoting @andrewdotson

[@AllanMedeiros]

Amazing lecture! Congratulations!!!

[@priteshsrivastava5850]

checking my understanding--- by two fermions dont interact we mean that there was n interaction potential energy term which appeared in the hamiltonian, am I right?

[@岡安一壽-g2y]

Please think about the system consisting of two electrons. Electron 1 is in a hydrogen atom and Electron 2 is in a helium ion He+. H1φA(r1)=EAφA(r1), H2φB(r2)=EBφB(r2). The wave function of this system ψ=φA(r1)φB(r2)-φA(r2)φB(r1) cannot give the energy EA+EB. And ψ is not the eigenfunction of the operator H1+H2. Please calculate it. It's very easy.

[@ILsupereroe67]

18:48 where did that come from? I mean The connection between the two forms of indistinguishability and the internet or half integer spins?

[@theultimatereductionist7592]

Aren't proton equally indistinguishable from other protons, just as electrons are from electrons?

[@sull5307]

The statement is only true for the simplest case, that is H2 molecule. Think about this in this way, if your statement is true for all cases the H-NMR technique wouldn't exists, where you can get different signals from the same proton, H proton.

[@davidhand9721]

Does the exclusion principle additionally force spin up and down particles to spin their phase backward?

[@jitendra4308]

Thanks a lot for the video. I have one fundamental doubt. For constructing the distinguishable two particle wavefunction, you multiplied the individual particle wavefunction. Why can't be it addition of the two single particle wavefunction. Addition would also render the same result for distinguishable two particle wavefunction.

[@giancarloagueci6039]

One answer to this question is the following: Suppose we are dealing with two 1-dimensional spacial wave functions. When we account for the two particle system, the new wave function is now a function of two spacial dimensions, effectively making it a function of two variables. Here it helps to think of the wave functions as vectors. What happens if you add two 1-d vectors together? You get another 1-d vector. What were are actually doing is taking the "tensor product" of the two vectors so that are new vector is now a vector of 2-dimensions. This is not a multiplication of two numbers, but an outer product of vectors that increases dimensions of the variables. So it doesn't make much sense to add wave functions together in this case. Hope that helps.

[@charmendro]

how did they get the normalizarion constant for Boson and Fermion in the last part to be √2/a? do u have to take the state and normalize it to get this constant?

[@BPHSadayappanAlagappan]

I think it's a typo

[@davidhand9721]

Why should we represent distinguishable particles as psi_a(x1) psi_b(x2)? It's just a math trick like Psi = R(r) Y(theta, phi), right? Can you even do that if V depends on both positions?

[@erwinrudolfjosefalexanders1182]

amazing video, i likes the "check yout understanding part", but answers to those questions would be great! thx anyway!

[@Mughal8811]

Erwin Rudolf Josef Alexander Schrödinger Did you solve them by yourself? or found somewhere? Because if you share with me, i can then try to pass the exam next week. thanks

[@nickhuisman4207]

At 28:32 the psi_21 is wrong right? The 2 should be muliplied to the insides of sin(pi*x_1 / a), or shoulden't it?

[@davidhand9721]

I don't understand why adding a second set of coordinates is adequate to describe a two particle system. With one particle, the parameters didn't represent a particle location, it was just the coordinate of interest. And the particle position could only be estimated with the position operator. If we make psi(x1, x2, t), there is no free parameter left to evaluate at. Must we assume one of these particles is at the origin? What does the position operator then calculate? Or any operator? It's not a sensible way to calculate, I don't get it, please help.

[@dreadformer]

you create another vector space basically treating them as in a combination of independent spaces

[@charlieangkor8649]

so the whole universe with 10^86 particles behaves according to a 3x10^86-dimensional cube wavefunction whose each dimension has size of 40 billion light years pixel size of like 10^-30 m and we have to update it every like 10^-30 second. Error: allocation out of memory. No, thanks. I dont feel like I want to understand physics anymore.

[@davidhand9721]

It makes even less sense to represent two indistinguishable particles as the *product* of two individual psi. These particles are supposed to be in the same field, right? Shouldn't they be added like components of a vector? It doesn't compute, please help!! And now they are indistinguishable but swapping them negates the wavefunction? That's not very indistinguishable. Help!

[@JohnVKaravitis]

30:18 s/b "2/a", no?

[@charlieangkor8649]

so if you predict 100 electrons at once, the repeating integral signs in just one equation fill the whole page. lol.

[@sandeepansahoo1292]

indistinguishabalization!

[@abejutuubb]

this is awesome! Thank you a lot!

[@theKiroo]

Hey Brian, thanks for the amazing videos. One question: why is the groundstate of the fermions non-degenerate? If we exchange x1 with x2 we get the same result but with a minus in front right? so in theroy this wouldn't be the same state and thus we would have same energy associated with two different states PS: you're awesome

[@giancarloagueci6039]

This reply may come late but hopefully it will answer the question for any viewers in the future. Take the case of the ground state for fermions, which i will write in a more simplified form: (a,b are constants, x1, x2 are variables) sin(ax1)sin(bx2) - sin(bx1)sin(ax2) = sin(bx2)sin(ax1) - sin(ax2)sin(bx1) = - [sin(ax2)sin(bx1) - sin(bx2)sin(ax1)] so therefore, the wave function with the overall minus sign you refer to is equivalent to the original wave function itself, so there is no degeneracy as there is only one wave function for the system.

[@0Navin0]

thank you!!