EqΨq=HΨq
find Eq vs q.
Would someone like to do a more detailed post of this problem? Please let me know if you would like to be added as an author for this blog. Then you can do your own post which makes editing possible, etc.
For this problem, you can assume that you are given a single atom potential as well as the energy of the ground state for that and the ground state wave function. The crystal potential is a sum of single atom potentials with appropriate displacement, and we assume eigenstates in the form given in class, i.e., single atom wave functions with phase factors that depend on q and on the displacement.
Wow. That looks cool!
ReplyDeleteI'm feeling rather stuck on this problem. I feel like there should be a way to sort of pop out the H_0 term and find an E_0 term with that, but I don't quite see how to work that out. Does anyone have some insight?
ReplyDeleteIf I understood you/wikipedia correctly, Dirac notation as it applies to integrals is another way of doing inner products. Following that logic, H0 should pop out by virtue of being the only n=0 term that multiplies ϕ0. Haha am I making any sense at all?
DeleteI might do a write-up tomorrow before class if I have time. I'm thinking that the H0 term just pops out of the integration since it's independent of the Vn≠0(x−na) term, but also because it makes sense that we'd end up with something like Eq=E0+∫(something small) dx because of what we discussed in class... the further out you are from n=0 , the less important the integral is. Admittedly, recollection/intuition tells me I'm headed on the right track, but I'll need some more time and paper to hammer my ideas into place.
ReplyDeleteShoot, I meant to post that as a reply to Mark's comment.
DeleteAlso, the biggest clues Zack gave us (in my opinion) were: 1) Dirac notation, 2) "you decide how many n terms you need to integrate over."
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ReplyDeletePlaying around with the algebra a bit, I've done this:
ReplyDeleteEq<ϕ0(x)|ψq(x)>=<ϕ0(x)|H0|ψq(x)>+∑n≠0V0(x−na)<ϕ0|ψq>
Eq<ϕ0(x)|ψq(x)>=<ϕ0(x)|H0|ψ0(x)>+∑n≠0V0(x−na)<ϕ0|ψq>
Eq<ϕ0(x)|ψq(x)>=<ϕ0(x)|E0|ψ0(x)>+∑n≠0V0(x−na)<ϕ0|ψq>
Eq<ϕ0(x)|ψq(x)>=E0<ϕ0(x)|ψ0(x)>+∑n≠0V0(x−na)<ϕ0|ψq>
Eq<ϕ0(x)|ψq(x)>=E0ϕ20(x)+∑n≠0V0(x−na)<ϕ0|ψq>
Eq=E0ϕ20(x)<ϕ0(x)|ψq(x)>+∑n≠0V0(x−na)
Eq=E0ϕ20(x)ϕ0∞∑n=−∞eiqnaϕ0(x−na)+∑n≠0V0(x−na)
Eq=ϕ0(x)E0∞∑n=−∞eiqnaϕ0(x−na)+∑n≠0V0(x−na)
I'm iffy as to whether I'm allowed to do what I did in step 2 (converting ψ0 into ϕ0), and step 7 (how I converted the bra/ket into a summation), but if those are correct I think this might be in the right direction. Any thoughts?
correction, my worry over step 2 has to do with converting psi_q into psi_0 , not converting psi to phi. As Aaron points out above, it would make sense that we could do this, but I'm not 100% sure.
DeleteOne final thought (I'm compensating for the fact that I'm going to miss class): The final form I have for E_q might make sense. On one hand, as you add in more n factors, you add in more V terms (with smaller contributions). I think this would make sense as more nuclei (basically the n's) would result in a higher energy.
DeleteOn the other hand, as you add in more n terms the denominator in the fraction will grow larger, reducing the initial energy E_0. I think this might also make sense, as more nuclei (spread out as they are) would cause the wavefunction to be spread out over a larger area.
So the two sums would compete as you added more n's, with each new n having less effect on the energy.
Hi!
DeleteHiI have the same up to step 4. Please can you explain how you got phi^2. Thanks !!
Dirac notation implies an integration in this context I think. It is a specific representation of the inner product implied by the brackets. an integration over x...
ReplyDeleteI am not so sure that the Vo(x) terms can come out of the brackets.
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ReplyDeleteI have written out the integrals but am stuck at the moment, this is what I have so far:
ReplyDeleteH0ϕ0=E0ϕ0 but since ϕ0 stands on the left in the bra-ket, would we then have H0ϕ∗0 if we evaluate the bra-kets? And then, is H0ϕ∗0=E0ϕ∗0? I'm not sure if this is true or not.. If we assume that, the integrals would be:
Eq∫∞−∞ϕ∗0ψqdx=E0∫∞−∞ϕ∗0ψqdx+∫∞−∞∑n≠0V0(x−na)ϕ∗0ψqdx
or
(Eq−E0)∫∞−∞ϕ∗0ψqdx=∫∞−∞∑n≠0V0(x−na)ϕ∗0ψqdx
Me and Helen were discussing taking the sum of the neighboring terms into account only (i.e. n = -1,0,1), what Aaron talks about here in the comments above as well. I'm not sure on how to do that though, does anyone have any idea how to do that/go from here?
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DeleteIf we take n=-1,0,1 i think the equation becomes.
DeleteEq∫∞−∞ϕ∗0ψqdx=E0∫∞−∞ϕ∗0ψqdx+∫∞−∞V0(x−a)ϕ∗0e−iqaϕ0dx+∫∞−∞V0(x)ϕ∗0ϕ0dx+∫∞−∞V0(x+a)ϕ∗0eiqaϕ0dx
but I'm not sure where to go from here
we could divide through by ∫∞−∞ϕ∗0ψqdx ?
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DeleteI think you've got it! However, you forgot to omit the n=0 term, so you should drop off the ∫∞−∞V0(x)ϕ∗0ϕ0dx . Doing that, we end up with
DeleteEq∫∞−∞ϕ∗0ψqdx=E0∫∞−∞ϕ∗0ψqdx+∫∞−∞V0(x−a)ϕ∗0e−iqaϕ0dx+∫∞−∞V0(x+a)ϕ∗0e−iqaϕ0dx
I meant, drop off the ∫∞−∞V0(x)ϕ∗0ϕ0dx
DeleteAhh I ran out of time, but there's also a sign error on one of the exponents that was throwing you off, Nellie and Helen! If I can find a way to justify ∫junk\timesV0(n=−1)=∫junk×V0(n=1), it collapses quite nicely!
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