Question

Can you help me?

Answer 1

anyone here good at Solid State Physics??

Answer 2

@ShailKumar hello Sheldon ., are good at Solid state physics?

Answer 3

Nope! :(
However, post your problem. Let me see if I can solve it

Answer 4

oke..,

Answer 5

It will take some time. I'll let you know soon. I think I can solve them. All the parts of question 4 are to be attempted. Right ?

Answer 6

Yes, you are right. Oke

Answer 7

@ShailKumar are you here?? i have a key answer for that

Answer 8

but, i'm still a little bit confused how to derive

Answer 9

@UnkleRhaukus help me :(

Answer 10

Hi Gerry ! I am so sorry for responding you so late. :(
I was messed up with the possible interactions. But here it is the electron-phonon interaction which dominates as the temperature is increased (i.e atomic vibrations start playing important role). The change in interband gap energy is mainly due to the electron-phonon interaction energy. This phenomenon is purely quantum mechanical.

Before I explain the detailed derivation (as also explained in the solution key ), are you introduced to time independent perturbation theory ?

Answer 11

yes i have learned about time independent perturbation theory. a little bit., :),

Answer 12

OK. Assume that you know the exact solutions of the Hamiltonians of electron and phonon i.e you know the eigenvalues and eigenfunctions of electron and phonon.
If one eigenfunction of electron be \[\psi_e\] and that of phonon be \[ \psi_{ph}\] and corresponding eigenvalues be \[ E_e \ and \ E_{ph} \] respectively.
Now if the electron and phonon interact, where should the total wavefunction live ?

Answer 13

Ok..,

the total wavefunction live in hamiltonian

\[H = H_{el} + H_{ph} + H_{el-ph}\]

right?

where \[H_{el-ph}\]
as peturbation

Answer 14

and then ??

Answer 15

Wait...The total wavefunction should live in the outer product space of \[\psi_e \ and \ \psi_{ph}\]

Answer 16

Let's represent it by \[ |\psi_e , \psi_{ph} > \]

Answer 17

ahh yaa., sorry., ok i see...

Answer 18

wait..., \[|\psi_e,\psi_{ph}> \]
or
\[|\psi_e.\psi_{ph}>\]?

Answer 19

That is a way of representation or notation. In standard texts comma is preferred. Don't worry about that. :)

Answer 20

oke ...

Answer 21

Now, as you explained that the interaction Hamiltonian is \[H_{e-ph}\]
Whatever change in the interband gap energy occurring is because of this interaction Hamiltonian. This Hamiltonian depends on temperature because change in temperature results in different vibrational modes of atoms and so in different electron-phonon interaction. It can be proved that this Hamiltonian is independent of time.

Answer 22

So we can use time independent perturbation theory with \[H_{e-ph}\] as the perturbation.

Answer 23

okay, i get it :)

Answer 24

Can you tell me how to find first order energy correction ?

Answer 25

yes i can.., wait

Answer 26

kk

Answer 27

we assume that we have found the eigenvalues and the complete set of normalized eigenfunctions for a hamiltonian Ho
\[H_o \psi_o= E _{o}^{n} \psi_n\]
we imagine that our perturbed system has some additional term in the Hamiltonian the “perturbing Hamiltonian \[H_p\] In our example case that perturbing Hamiltonian would be \[H_{el-ph}\]
then
\[H=H_o+\lambda H_{el-ph}\]
that is , for the solutions of
\[(H_o+\lambda H_{el-ph}) \psi_n=E_n \psi_n\]
where
\[\psi_n = \psi_{n}^{o}+\lambda \psi_{n}^{1} + \lambda^{2} \psi_{n}^{2}+....\]
\[E_n=E_{n}^{o}+ \lambda E_{n}^1+ \lambda^{2} E_{n}^{2} +.....\]

then
\[H \psi_n = E_{n} \psi_n\]
\[(H^o + \lambda H_{1}) [\psi_n^{o} + \lambda \psi_n^1 +....] = [E_{n}^{o}+ \lambda E_{n}^1+ \lambda^{2} E_{n}^{2} +.....] [\psi_n^{o} + \lambda \psi_n^1 +....]\]
and then (look at picture below., :) )

Answer 28

Wait...You don't need to derive the result from the scratch. First order energy correction is the expectation value of perturbed Hamiltonian \[H_{e-ph}\] in the unperturbed state \[|\psi_e , \psi_{ph}>\]

Answer 29

ahh, my bad, sorry, i don't know..,

Answer 30

oke, let see the next.., :)

Answer 31

So, first energy correction comes out to be \[E^{(1)} = <\psi_e, \psi_{ph}|H_{e-ph}|\psi_e, \psi_{ph}>\]

Answer 32

ok

Answer 33

You check the result in the second attachment http://assets.openstudy.com/updates/attachments/53a981eee4b0a819ab11d7a6-gerryliyana-1403789561647-wp_20140626_014.jpg

Answer 34

yes

Answer 35

How to calculate this expectation value ?

Answer 36

One simple way is to see the Hamiltonian in terms of creation and annihilation operators.

Answer 37

This can be found in some standard text books of solid state physics. You can read it here also. http://www.phys.ufl.edu/~pjh/teaching/phz7427/7427notes/ch4.pdf

Answer 38

Do you know creation and annihilation operators ?

Answer 39

hmm..., like hopping from one site to another, right?

Answer 40

:)

Answer 41

Do you know creation and annihilation operators ?

Answer 42

i know

Answer 43

OK. Great! Let's move ahead. Writing these expression in this LaTex is very clumsy :(
Refer to the solution you have attached.
There you see, first order energy correction is zero. Can you verify it ?

Answer 44

why zero?

Answer 45

It's zero because of the orthogonality of eigenfunctions.

Answer 46

hmm., I'm still remembering the the lecture i received.,

Answer 47

So, for instance, creation operator acts on the state \[|\psi_e, \psi_{ph}>\] and results in a new state \[|\psi_e, \psi_{ph+1}>\]
Now this state is orthogonal to \[|\psi_e, \psi_{ph}>\]
and so \[<\psi_e, \psi_{ph}|\psi_e, \psi_{ph+1}>\] = 0

Answer 48

ahh.., ok i get it

Answer 49

raising operator

Answer 50

So, similarly other terms will be zero. and effectively first order energy correction will be zero.

Answer 51

Creation operator is also called raising operator.

Answer 52

yes

Answer 53

Similarly, you van calculate second order energy correction. And you find that it is not zero because in this correction there is a summation and some combinations of states do survive.

Answer 54

For instance, \[<\psi_e,\psi_{ph}|raising \ operator|\psi_e,\psi_{ph-1} >\] is not equal to zero.

Answer 55

Right ?

Answer 56

ok., i get it.., I wonder why only term for phonon wave functions are equal ?

Answer 57

and what is \[j_{n}^{,} = j_n \pm 1\] ?

Answer 58

How do I derive this function?

Answer 59

Because, the creation or annihilation operators are defined to act on phonon eigenfunction only and so they create or annihilate phonons only. Physically, temperature change results in the change of the number of phonons, not electrons !

Answer 60

"How do I derive this function? "
Again, in time independent perturbation theory, this is a standard result for second order energy correction. Just check your notes or text book

Answer 61

Could you find that result ?

Answer 62

i'm trying

Answer 63

Ok. j is the ocupation number i.e number of phonons in momentum state q. So when you act a raising operator on a wavefunction of phonon with j phonons of momentum q, the new wave function will of j+1 phonons with momentum q. For annihilation operator it will be of j-1 phonons with momentum q.

Answer 64

is that the reason why on the left side is j+1 and on the right side is j?

Answer 65

Right!

Answer 66

I have to go now. See you next time. :)

Answer 67

ok see you., thank you very much., i really appreciate it

Answer 68

I also appreciate you for actively participating. :)

Answer 69

Could you absorb what we have discussed till now ?

Answer 70

yes i could

Answer 71

Great!

Answer 72

Could you get the expression for second energy correction in terms of j ?

Answer 73

No.., I'm having trouble with this problem, can you show me and write it down on your paper,, take a photo and upload it, just second order part.., I m just a beginner for this :(.

Answer 74

OK. Sure, I'll do it but I cannot do it now. I don't have data transfer cable. :(
May be in next few hours. :)

Answer 75

okay., no problem, thank you :)

Answer 76

awesome.., thank you., why don't we calculate Ef and Ei??

Answer 77

We calculate but here we don't need to calculate them. They are expectation values of unperturbed Hamiltonian and so we can treat them as a constant. This constant will be absorbed in some other constant. So we don't need to calculate them.

Answer 78

Try to understand these calculations. Any doubt, feel free to ask. :)

Answer 79

Are you comfortable with the calculations ?

Answer 80

Thank you @ShailKumar yes, i'm comfortable., it was great!

Answer 81

Can you help me again. where should I find this equation? (take a look a picture)

Answer 82

Read about 'power radiated by a Hertzian dipole' or 'electric dipole radiation'. For this consult any book on electromagnetism. For example, Griffiths or Jackson.
Or visit http://farside.ph.utexas.edu/teaching/em/lectures/node95.html