Maybe you are missing a step here? At around 6:14 you said that g(E)dE = (something-something)*dE . Where did the dE in the Left-hand side of the equation come from??
Excellent catch! When I wrote g(k), I should have written g(k)*dk. This quantity refers to the total number of states within a shell, so it needs to be written density*length.
@@JordanEdmundsEECS First of all, thanks for the excellent videos! Regarding this question, I don't understand why you should have written g(k)*dk, if you replaced N/L^3 by g(k), which is indeed a density of states.
@@JordanEdmundsEECS i had the same question, wherefrom did g(E) appear on the left hand side? Can you pl elaborate at what point we should add dk? Am not fully clear.
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@Ledger David Thanks for your reply. I got to the site through google and im in the hacking process atm. I see it takes quite some time so I will reply here later when my account password hopefully is recovered.
As mentioned in comments below. The expression for g(k) should not include dk. We can add further view on the derivation on g(k). g(k) = N(k)/dk, where N(k) represents the number of states within the shell, and dk represents the thickness of the shell.
Hi Jordan, I am not quite sure why you mentioned below one of the comment that you should have written g(k)*dk. You wrote about g(k), the density of states, from N/L^3 which is also density of state and which has already had a delta K term in it. I am not sure why still need to incorporate the dk
Great video. Thanks a lot! I didn't catch why do we have 1/8 before. You took infinitesimal resiprocal volume 4PiK^2dK and decided it by volume of of one state in KSpace Pi/L. why do we need 1/8
Think I got it! Since n's are only positive, you only look at 1/8 of the 3D coordinate system (where x,y and z axis are positive). That makes sense to me at least
Hi Jordan, ultimately we want to find the number of electrons in the entire semiconductor cube so if we integral P(E)g(E)dE, does that only give you total number of electrons per volume in the semiconductor?
Ah, yes, that is extremely subtle. The answer is that it depends on how you want to solve the Schrödinger equation. If you assume the solution is a standing wave (a sinewave with nodes at the boundaries), then your spacing is pi/L and you only sum over 1/8th of k-space. If instead you assume the solutions are complex exponentials (which is true, but only a specific combination of them satisfy the boundary conditions), then your spacing is 2pi/L and you count all of k-space.
@@JordanEdmundsEECS could you plz plz explain why it works that way? My book has assumed those conditions and now I am extremely confused and torn between the two proofs
L^3 is the "unit" volume in the actual space, which is exactly what we want. Number of electron states per "unit" volume. I think this is mostly right, if I am wrong, please correct me!
In a sense, yeah, because each state has a different momentum (k), and these are some distance (pi/L) apart from each other. So a single state is (pi/L) away from all the neighboring states. In this sense it occupies a volume of “k-space”. Not a literal volume.
Because solving the Schrodinger equation (and Maxwell’s equations, and virtually all differential equations in circuits) which is nearly impossible in real space becomes trivial in k-space. Also known as frequency space (but now the frequency is in space not time).
sir are we taking a cube instead of a line because while calculating k, we assumed only one axis. But we can do the same for the other two axis as well. Is it correct?
Not at all! A ‘state’, technically speaking, is a solution to the time-independent Schrodinger equation for a 3D quantum well. Intuitively speaking, it’s a particular speed and direction the electron is moving which is allowed by quantum mechanics (and these are discrete).
Great question. A single *state* can have only one electron. But at a single *energy* we can actually have two states, or two spins (these states are called degenerate). This is we need to multiply by two.
The introduction of calculus here assumes the density of states is continuous. This doesn't make sense as amount of states should only be a multiple of pi/l. To my understanding, as electrons are restricted to a wavenumber of npi/l, then the smallest possible wave number is pi/l. Therefore in K space, the number of states is : N= 2(1/K)^3*Vk where Vk is the volume in K space and 2(1/k)^3 is the density of states in k space including spin. For any volume, N can be a non-integer. But the number of states would just be the integer part of N. Now for energy, a similar argument should be used. As only integer values of pi/l for wave number are allowed, then there is only an integer value if Es is allowed for energy. Here Es is the energy corresponding to the wavenumber pi/l. Therefore the number of states at a given energy E should be: N= 2E/Es Where E = (Es+Es+Es+Es+...)=nEs (the factor of 2 accounts for spin) Which in 3D is made of three components (assuming Ex=Ey=Ez for a cube well): E= Sqrt( (uEs)^2+(wEs)^2 + (bEs)^2) = Es(sqrt(u^2+w^2+b^2))= nEs Therefore the density of states for energy is 2/Es= 2 ( 1/(hbar^2k^2/2m))= 4mL^2/hbar^2pi^2. This makes sense as the density of states is not a function of energy, but rather of the quantum well width L. I'm not sure how the density of states (the number of states per unit energy) would increase with energy. That would only be true if L increased. For example, If I had a nanoscale silicon transistor, the density of states should be much smaller and more discreet than that for a block of silicon with a large L value. I'm not sure what's wrong with my understanding, but the calculus approach should be an approximation (that only works for large L values). Meaning there should not be a state that exists between K and K+dK, but rather a state that exists between K and K+ pi/l.
Yes! That's exactly it. This implicitly assumes that L is large enough so that the number of states is huge, and you can even use differential calculus at all. Its typical application is in the "bulk" - crystals larger than ~100nm or so. For nanoscale devices smaller than this, this is not the most accurate approach, but often those devices are only small along 1 dimension (i.e. MQW lasers), in which case you can use this approach with the other 2 dimensions.
g(E) is the density of states per unit volume per unit energy, and g(E)dE is the density of states per unit volume for a small ‘slice’ of energy. I agree the units are confusing.
how g(k)*dk is equal to number of states within a shell. g(k) is no. of states per unit volume to get number of states within a shell you have to multiply by volume and dk is not volume
It is probably easier to explain with g(E) first. g(E) is the volume density of electron states per energy interval. N/L^3 is only the volume density of electron states, not per energy interval (dE). We have to account for this missing dE so we write g(E)dE. More intuitively what this means is that we can determine g(E) when we have some specific value for dE (i.e. what energy interval we are looking at). It is the same principle for g(k)dk except we have just substituted our E with k. Instead of requiring some energy interval we need to know what wavenumber interval we want to look at.
Math murdered with this g(k)=k^2dk. There is an easier derivation that doesn’t commit such an atrocity. Just work your way from E and relate it to the radius of the sphere sqrt(n_x^2+n_y^2+n_z^2).
Maybe you are missing a step here? At around 6:14 you said that g(E)dE = (something-something)*dE . Where did the dE in the Left-hand side of the equation come from??
Excellent catch! When I wrote g(k), I should have written g(k)*dk. This quantity refers to the total number of states within a shell, so it needs to be written density*length.
@@JordanEdmundsEECS had the same question with Rishi....Jordan thank you so much this is really helpfull with my physics courses! :-)
@@JordanEdmundsEECS First of all, thanks for the excellent videos!
Regarding this question, I don't understand why you should have written g(k)*dk, if you replaced N/L^3 by g(k), which is indeed a density of states.
@@JordanEdmundsEECS i had the same question, wherefrom did g(E) appear on the left hand side? Can you pl elaborate at what point we should add dk? Am not fully clear.
@@RoyMustang027 exactly, pl explain
These videos are great. Love the top-down approach. Looking forward to, as Neel said, f(E) derivation
Stephan Flumm Thanks :D I’m looking forward to making it
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@Makai Grady Instablaster ;)
@Ledger David Thanks for your reply. I got to the site through google and im in the hacking process atm.
I see it takes quite some time so I will reply here later when my account password hopefully is recovered.
@Ledger David it worked and I actually got access to my account again. I am so happy!
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As mentioned in comments below. The expression for g(k) should not include dk. We can add further view on the derivation on g(k). g(k) = N(k)/dk, where N(k) represents the number of states within the shell, and dk represents the thickness of the shell.
Hi Jordan, I am not quite sure why you mentioned below one of the comment that you should have written g(k)*dk. You wrote about g(k), the density of states, from N/L^3 which is also density of state and which has already had a delta K term in it. I am not sure why still need to incorporate the dk
Much appreciation for this video. I have been struggling with how the g(E) is obtained in optical properties of solids textbook by mark fox
Great video. Thanks a lot! I didn't catch why do we have 1/8 before. You took infinitesimal resiprocal volume 4PiK^2dK and decided it by volume of of one state in KSpace Pi/L. why do we need 1/8
Thanks :D The 1/8 comes from the fact we are only considering one-eighth of the volume of a sphere (one octant of the 3D state space).
why are we considering only 1/8 of the sphere? Is that a choice or a necessity?
Think I got it! Since n's are only positive, you only look at 1/8 of the 3D coordinate system (where x,y and z axis are positive). That makes sense to me at least
@@Odiskis1 you cleared my doubt as well. Thanks man
Very helpful to me. Thanks a lot
Damn I was suffering with for so long ..finally i get it!!! Thank you so much!😭😭😭😭
Yeah the DOS is a pretty frustrating and challenging subject for most people (myself included). Glad you get it now! :)
In which video do you find f(E) : probablity that the states are occupied
I actually haven’t made that video yet D: definitely in the next month or two
Jordan edmunds : Thanks this videos are really helpful.........👍👍👍
yay :)
Thank you! Very understandable video
Hi Jordan, ultimately we want to find the number of electrons in the entire semiconductor cube so if we integral P(E)g(E)dE, does that only give you total number of electrons per volume in the semiconductor?
Yes, you are correct. And this is often called concentration of charge carriers, which is key to finding current and voltage relationships
Why do we consider the 1/8 spherical shell instead of the 1/8 sphere?
Because we are interested only in the *differential volume* so we can use this derived equation in an integral, not the *total volume*.
Thank you very much sir .
Shouldn't the spacing between k points be 2 pi over L?
Ah, yes, that is extremely subtle. The answer is that it depends on how you want to solve the Schrödinger equation. If you assume the solution is a standing wave (a sinewave with nodes at the boundaries), then your spacing is pi/L and you only sum over 1/8th of k-space. If instead you assume the solutions are complex exponentials (which is true, but only a specific combination of them satisfy the boundary conditions), then your spacing is 2pi/L and you count all of k-space.
@@JordanEdmundsEECS could you plz plz explain why it works that way? My book has assumed those conditions and now I am extremely confused and torn between the two proofs
Super helpful thanks!
Why are we dividing by L^3 ..while the volume under consideration is 4πk^2 delta k
L^3 is the "unit" volume in the actual space, which is exactly what we want. Number of electron states per "unit" volume.
I think this is mostly right, if I am wrong, please correct me!
a state (solution of Schrödinger Equation ) occupies a volume ?
In a sense, yeah, because each state has a different momentum (k), and these are some distance (pi/L) apart from each other. So a single state is (pi/L) away from all the neighboring states. In this sense it occupies a volume of “k-space”. Not a literal volume.
why we used momentum space (k space) when we were doing fine in real space (r space) ?
Because solving the Schrodinger equation (and Maxwell’s equations, and virtually all differential equations in circuits) which is nearly impossible in real space becomes trivial in k-space. Also known as frequency space (but now the frequency is in space not time).
@@JordanEdmundsEECS alright! Thankyou :)
sir are we taking a cube instead of a line because while calculating k, we assumed only one axis. But we can do the same for the other two axis as well. Is it correct?
Yessir, that is correct.
Sir, when the value of “h” is (6.625*10^-34)^3 the result is 0. May I know how to fix this error ?
Sorry for asking a very stupid question but what actually is a state in density of states?
Not at all! A ‘state’, technically speaking, is a solution to the time-independent Schrodinger equation for a 3D quantum well. Intuitively speaking, it’s a particular speed and direction the electron is moving which is allowed by quantum mechanics (and these are discrete).
@@JordanEdmundsEECS This is really great explanation .
A single state can have only one electron. So why did you multiply by 2 ??
Great question. A single *state* can have only one electron. But at a single *energy* we can actually have two states, or two spins (these states are called degenerate). This is we need to multiply by two.
@@JordanEdmundsEECS Thanks 🙌
can we do these things in real space instead of k space because volume in k space little confusing .. :(
No because we use the dispersion relation that connects q and E space to get D(E). So we have to start in q space.
Watching in Muranga university of technology
Where is part 3????? 😰😰😰
The introduction of calculus here assumes the density of states is continuous. This doesn't make sense as amount of states should only be a multiple of pi/l.
To my understanding, as electrons are restricted to a wavenumber of npi/l, then the smallest possible wave number is pi/l.
Therefore in K space, the number of states is :
N= 2(1/K)^3*Vk
where Vk is the volume in K space and 2(1/k)^3 is the density of states in k space including spin. For any volume, N can be a non-integer. But the number of states would just be the integer part of N.
Now for energy, a similar argument should be used. As only integer values of pi/l for wave number are allowed, then there is only an integer value if Es is allowed for energy. Here Es is the energy corresponding to the wavenumber pi/l.
Therefore the number of states at a given energy E should be:
N= 2E/Es
Where E = (Es+Es+Es+Es+...)=nEs
(the factor of 2 accounts for spin)
Which in 3D is made of three components (assuming Ex=Ey=Ez for a cube well):
E= Sqrt( (uEs)^2+(wEs)^2 + (bEs)^2) = Es(sqrt(u^2+w^2+b^2))= nEs
Therefore the density of states for energy is 2/Es= 2 ( 1/(hbar^2k^2/2m))= 4mL^2/hbar^2pi^2.
This makes sense as the density of states is not a function of energy, but rather of the quantum well width L. I'm not sure how the density of states (the number of states per unit energy) would increase with energy. That would only be true if L increased.
For example, If I had a nanoscale silicon transistor, the density of states should be much smaller and more discreet than that for a block of silicon with a large L value.
I'm not sure what's wrong with my understanding, but the calculus approach should be an approximation (that only works for large L values). Meaning there should not be a state that exists between K and K+dK, but rather a state that exists between K and K+ pi/l.
Yes! That's exactly it. This implicitly assumes that L is large enough so that the number of states is huge, and you can even use differential calculus at all. Its typical application is in the "bulk" - crystals larger than ~100nm or so. For nanoscale devices smaller than this, this is not the most accurate approach, but often those devices are only small along 1 dimension (i.e. MQW lasers), in which case you can use this approach with the other 2 dimensions.
i think there is a factor of 2 missing in the last equation g(E)dE
i did not understand from where dE is come because N/L3 is states density so it should be equal to g(E) only
g(E) is the density of states per unit volume per unit energy, and g(E)dE is the density of states per unit volume for a small ‘slice’ of energy. I agree the units are confusing.
how g(k)*dk is equal to number of states within a shell. g(k) is no. of states per unit volume
to get number of states within a shell you have to multiply by volume and dk is not volume
You are absolutely correct, g(k)dk*L^3 is the number of states within a shell. Perhaps I misspoke.
why 1/8 ?
Because we only want to work with positive k in x, y, and z
👍🏼
I watched it once and I didn't take notes.
Pl pl explain why it should be g(k)dk. Am really not following.
It is probably easier to explain with g(E) first. g(E) is the volume density of electron states per energy interval. N/L^3 is only the volume density of electron states, not per energy interval (dE). We have to account for this missing dE so we write g(E)dE. More intuitively what this means is that we can determine g(E) when we have some specific value for dE (i.e. what energy interval we are looking at).
It is the same principle for g(k)dk except we have just substituted our E with k. Instead of requiring some energy interval we need to know what wavenumber interval we want to look at.
Math murdered with this g(k)=k^2dk. There is an easier derivation that doesn’t commit such an atrocity. Just work your way from E and relate it to the radius of the sphere sqrt(n_x^2+n_y^2+n_z^2).
thank you for a fantastic lecture.