So the question asks me to find the first variation in energy for a system of 2 electrons confined to a 1D space of length L (so an infinite square well) under a potential V(x_1,x_2) = -lambda*delta(x_1-x_2) (Dirac delta). Okay, so my idea is to find the wavefunction and then operate as usual. Okay so the wavefunction has to be anti-symmetric on account of working with fermions (2 spin 1/2 particle system). Assuming they don't interact between each other we can separate in a spatial and spin parts:
Psi = psi(x_1,x)2) * chi(s_1,s_2)
Where psi and chi have opposite parity
Now, we can write psi*(x_1,x_2) = psi(1,2) = psi_1(1)*psi_2(2) (where psi_1,2 are the states of each individual particle). So we write the psi(1,2) with all the sins and square roots, we end up with a sum (or substraction when appropriate) of sins, one for state n_1 and the other for state n_2 (or energy levels E_1 and E_2).
Now it's time to do the perturbative calculations. The issue is that the first correction is written
E_n^1 = < psi_n^{0} | V(1,2) | \psi_n^{0} >
Not sure what to do here because
- I could the symmetric or anti-symmetric coordinate states
- I could pick n_1=n_2 to further simplify the symmetric solution
And so on... How would you answer this? would you do all the calculations? or is there something I'm missing? A friend of mine considered that we put them in the same state and position (with different spins) such that we end up with a sin^4 in the double integral...
Any help is appreciated, thanks in advance!
If I didn't explain this well, let me know and I will provide more details and LaTex equations if necessary.
If we were to randomly select any two electrons, would they actually be identical in terms of their properties, or simply close enough that we could consider them to be identical? Do their properties have a range of values, or a set value?
I'm borrowing the example from this wikipedia article. Let's say I have two indistinguishable bosons, each of which can exist in two states, |0> and |1>. If I put the two particles together in a noisy environment, let them evolve for some time, and then measure their states, I have a 33% probability each of measuring |00>, |01>, or |11>. This makes sense on some level --- the combined system had three possible states, all of the same energy, and a uniform distribution over those states maximizes entropy.
Obviously, this doesn't generalize to normal macroscopic systems. If I put two coins in a cup, shake them for a while, and observe their state, I would expect to find two heads with 25% probability, two tails with 25% probability, and 1 heads and one tails with 50% probability. Of course, macroscopic coins are not indistinguishable, but where exactly does indistinguishably change the statistics so fundamentally? Is it at the time of observation? If I observed the state with a sufficiently low-resolution camera, that could resolve heads and tails, but not distinguish the two coins by any markings, would my observations mimic the quantum-mechanical case? Or is the indistinguishably important while the coins/particles are time-evolving in a noisy environment?
Of course, I know that quantum mechanics are inherently counter-intuitive, and I shouldn't expect my normal intuitions to apply. However, the classical behavior should be a limiting case of the quantum-mechanical behavior, and I don't see where the boundaries between the classical and quantum-mechanical behavior lies. In principle, could one construct two atom-for-atom identical coins, put them in SchrΓΆdinger's box, shake it, and expect two heads with 33% probability?
I am having difficulty identifying what defines a quantum state for a multiple indistinguishable particles in a box, with the focus on fermions.
Assuming no interaction, the wave function for a single particle depends on 3 quantum numbers, say [;n_1, n_2, n_3;].
If the order was [;n_1=1, n_2=2, n_3=1;] for the first particle versus, [;n_1=2, n_2=1, n_3=1;] for the second particle, are these two separate quantum states? I would think so, because the wave-function as a product of sinusoids, which depends on 3 spatial directions, would have different contributions to each direction with changes to [;n_i;].
Perhaps I am confusing 'single-particle states' with quantum states of the system?
Or, is the state defined by the energy level, which in this case, because the energy depends on the sum of squares, the energy is the same for both, and fermions would not be able to share the exact same 3 quantum numbers, irrespective of order.
I am aware of the (anti)symmetrization of the wave-function which needs to be fulfilled, but can you please clarify this in the case of the quantum numbers of two identical particles in a box?
Thanks.
Hi, I resolved a problem where I needed to discuss the degeneration of the system.
My Hamiltonian is that of two fermions (spin 1/2) constrained in an infinite potential well between 0 and a, plus a spin potential of the type V = s1z s2z.
I have chose the basis |n S Sz> where S = S1+S2 and Sz = Sz1+Sz2 and expressed V = Sz^2 - S1z^2 - S2z^2.
|n> is the spatial part.
Now, since I have two fermions my wavefunction should be anti-symmetric so:
I have symmetric spatial part (Ξ¦+) associated to the singlet and anti-symmetric spatial part (Ξ¦-) associated to the triplet.
So far so good.
But I've noticed that Ξ¦+|0 0> and Ξ¦-|1 0> have the same eigenvalues so they are doubly degenerate. Is there any other conclusion that I can come to when I find this result?
Are they symmetric under exchange? (if yes, I'd like to know how I can prove it)
This thing got me perplexed and maybe I shouldn't be.
I know it's only meant as a metaphor but the usual example given is that of an instrument. When I visualize this I think a string is only vibrating based on an exterior force (the pluck let's say), and it eventually slows vibrating or changes pitch. Wouldn't this be changing the particles themselves?
I know any two hydrogen atoms or protons or what have you are functionally identical, but are they all precisely identical? Is there any reason to think that the mass of two protons would vary slightly, or is it fixed in some fundamental way?
Will Hydrogen and anti-Hydrogen be indistinguishable in by their spectral lines? What about larger anti-particles?
I know that for three or more identical particles the symmetry group is the permutation group, but I'm pretty sure that's not the case for only two particles. Why is that so?
Something like any proton is thought to be identical to any other proton. Could a proton be a narrow range of properties (mass, charge, spin, etc), but not necessarily 100% identical.
This was the start of a homework problem. It sounds like it could be the setup to a good physics joke.
I'm kind of stumped here. I'm sure I'm just fundimentally misunderstanding the mechanics of this the problem.
I'm given the state of the two particle system|psi>. There is an operator C, with two eigenstates: |B> and |R>. I'm told that the expectation value of the first particle should be 1/2 for both |B> and |R> and asked to prove that.
C={{1,0}, {0,-1}} (see image if this is confusing)
|B>= (1 0) as a column vector
|R>=(0 1) also as a column vector
I'm coming up with the expectation value as zero, so I was hoping someone could give me an idea of what I'm doing wrong.
Image: https://preview.ibb.co/hJMKuQ/IMG_3540.jpg
