state to put into an \(I_{11}\) calculation so that we can see how bands really form. Let's focus on \(\phi_4 (x)\), using Nellie's notation, which is the fourth state in a 1500 eV deep well. We think it has an energy of -204 eV with respect to zero outside the well. What is \(\phi_4 (x)\)? Can someone find the A and D coefficients in Angstroms^{-1/2} and the k and kappa in Angstrom^{-1}. Help.
PS. If you have trouble getting continuity and smoothness at the same time it means the energy is incorrect. If they do both match, that means the energy is correct!
Addendum on calculating numbers:
For calculating k and kappa, one can multiply through by c to get:
\( \kappa =
\sqrt{\frac{2mc^2(-E)}{\hbar^2 c^2}} \)
where,
\(mc^2 = 0.511 \times 10^6 \) eV
and
\(\hbar c = 1970\) eV-A
check these!
check these!
Then
I would suggest calcuating A, B and D from integrals where everything
is in Angstroms (with A, B and D in Angstrom^{-1/2}. Does that make
sense?
I've been solving for A and D for a while now and I'm wondering if E4 is wrong. If we have \( \sqrt{ \frac{-E}{E+V_0}} = tan(\gamma \sqrt{E+V_0}) \) , where this is gamma, then wolfram provides this as the solution graph:
ReplyDeleteGraph
and zooming in
gives E4 at around -830 eV
Solving for k and kappa:
\(k = 13.26 \AA^{-1} and \kappa = 14.76 \AA^{-1}
which leads to
\( B = 2.33 \AA^{-1/2} and D = -130.79 \AA^{-1/2} \)
and here there is both continuity and smoothness.
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Delete(cleaning up the final results)
ReplyDeletek = 13.26 angstroms^(-1) and kappa = 14.76 angstroms^(-1)
B = 2.33 angstroms^(-1\2) and D = -130.79 angstroms^(-1/2)
the kappa and k seem to large to me.
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DeleteI have been solving for A and D rather than B as we are using the odd states?
DeleteSo i have
\( A=\sqrt{\frac{1}{b-sin(kb)}} \)
and
\(D=Asin(kb)e^{\kappa b} \)
Does this look correct to you?
And I have
Delete\( k=1.85x10^{11} m^{-1} \)
and
\( \kappa = 7.53x10^{10} m^{-1} \)
Hmm. I take that back. They don't see too large anymore. you got continuity and smoothness then for E= -830 eV. Is that perhaps E3??? (because 670 eV above the bottom seems more likely to be E3 to me.
DeleteYes that should be E3!! I think we're getting there.
DeleteI might have done the misstake to not use enough decimals in my 0.02 term, that really changes a lot!
\( E_4 \) is given by \( -\sqrt{\frac{E+V_0}{-E}} = tan \left( \sqrt{0.02eV^{-1}(E+V_0)} \right) \)
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DeleteI'm worried I might be making some trivial mistake now, so here's my work.
Delete\( \psi = Bcos(kx) , = De^{-\kappa x} \)
\( \frac{d\psi}{dx} = -Bksin(kx) , = -\kappa D e^{-\kappa x} \)
so
\( Bk sin(kb) = \kappa D e^{-\kappa b} \) and
\( Bcos(kb) = D e^{-\kappa b} \)
dividing one by the other yields
\( k tan(kb) = \kappa e^{-\kappa b} \) which leads to
\( \frac{\kappa}{k} e^{-\kappa b} = tan(kb) \)
\( \sqrt{ \frac{-E}{E + V_0} } = tan( \sqrt{ \frac{2m b^2}{\hbar^2} (E + V_0) } ) \)
does this seem right? where are you getting the 0.02 eV^-1 from? Looking at your older post where I think you derived this, I plugged the exact numbers you did into wolfram and, while I did get 0.02, the units were J^-1 , not eV^-1
Did I make a mistake somewhere?
Christopher. Your work might be correct. You got continuity and smoothness for a bound state at -830, right? Maybe that is a correct state energy then.
DeleteWhat energies are you getting for other bound states.
That's correct, but that equation gives you the even states (E4 is an odd state).
DeleteI have been thinking a lot about the 0.02 term, here's what I think:
\( \frac{2mb^2}{\hbar^2} = \frac{2*9.1*10^{-31}kg*(0.3*10^{-10}m)^2}{(1.1*10^{-34}J*s)^2} \)
Whatever the result for that calculation is, it should be in 1/J right?
And then I just multiply by \( 1.6*10^{-19} \) to get the unit 1/eV?
I might be doing something wrong?
So you're right I think Christoffer, and you're calculating the energy for E3. If I use more decimals on 0.02 I also get -830eV!
DeleteAfter talking with Zach I'm fairly certain this is actually E3. Turns out that "odd" bound states mean even numbers (E0, E2, E4). I'm making a new post deriving E4.
DeleteYes, cause E1 is the ground state, and E1 is even!
DeleteGr8, thanks!
How about if someone independently calculates E4 for the same well width (0.6 A) and depth 1500 eV.
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ReplyDeleteOne can use the kinetic energies of the states from the infinite square well as a reference.
Delete104 eV, 416 eV, 936 eV, 1664 eV,
The energies for a finite well will each be lower. e.g, rough guesses:
85 eV 340 eV, 800 eV, 1350 eV.
Which, when you put zero outside the well and make the bottom of the well -1500 eV, this become:
-1315, -1160, -700 and -150 eV.
Nellie's numbers are close that and are probably correct. Can someone confirm that?
For a hypothetical state energy of -200 eV, i get kappa = 2.3 A^{-1}. Does that seem right? Are my numbers in the addendum correct?
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ReplyDeleteTo check for continuity and smoothness I have that
ReplyDelete\(D=Asin(kb)e^{\kappa b} \)
and
\(D=\frac{kAcos(kb)e^{\kappa b}}{-\kappa}\)
Therefore,
\( sin(kb) = \frac{cos(kb)}{-\kappa}\)
However, using Nellie's numbers I do not get the above to be equal.
those two k's are different. It is only at b=3 that this needs to work.
Deleteoops. b=0.3 A
DeleteDo you mean the k in sin(kb) and the k in cos(kb) are different? I have put in the values for k, kappa and b into the above and don't get them to be equal.
ReplyDeletei meant the k and the kappa.
Deletethanks just something to do with wolfram alpha limitations. It is not a real concern. The solution is there.
ReplyDeleteI think this is the ground state energy: ~ -1424eV: http://www.wolframalpha.com/input/?i=graph++y+%3D+sqrt+%28+-x%2F%28x%2B1500%29++%29+++and+y+%3D+tan%28+0.154+*+sqrt%28x%2B1500%29++++%29+between+-1450+and+-1400
ReplyDeleteLooking at the numbers we as getting, and how we will use these states to make bands, I am thinking that I made a mistaken in suggesting such a deep well. Let's focus instead on a wider well, 1 Angstrom = 0.1 nm, and let's make it shallower by only requiring 3 bound states. There will be a post on that soon. (You can work on it as well if you like.) Then when we get those state parameters let's use them to calculate the bands emerging from the E1 and E3 states, maybe E2 as well if we get to that. I am guessing that a 1 Angstrom space between wells, center-to-center distance = 2 Angstrom then yes?, will be okay. (Feel free to question that...)
ReplyDeleteUsing the new width we have taken the well depth to be \(V_{0}=300eV\) Using this value we have found
Delete\(E_{1}=-275eV\)
\(E_{2}=-202eV\)
\(E_{3}=-89.3eV\)
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DeleteWhen you say "calculate A, B, and D from integrals", you do mean to say by
ReplyDelete\[ 1 = \int _{- \infty} ^{\infty} | \psi |^2 dx \]
right? Also, is it true that we'd always expect to find the particle inside the well? For example, if the neighboring wells we introduce later on are close enough for tunneling to occur, shouldn't we be taking these integrals over all space (including the "impossible" locations)?
If we only consider the space inside the well, I get
\[ B = \sqrt{ \frac{1}{ \frac{1}{2} a +\frac{1}{2k} \sin{(ka)} } } \approx 1.914 \sqrt{\text{Angstroms}} \\
D = B \exp{( \frac{1}{2} \kappa a )} \cos{ ( \frac{1}{2} k a ) } \approx 12.52 \sqrt{\text{Angstroms}} \]
And my calculations agree with Helen's for \( k \) and \( \kappa \).
Did you calculate the constants using the new depth well or the old one?
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DeleteI have mine as
Delete\(A=\sqrt{\frac{1}{b-sin(kb)}}\)
\(D=\sqrt{\frac{1}{b-sin(kb)}}sin(kb)e^{\kappa b}\)
and
\(B=\sqrt{\frac{1}{b-sin(kb)}}tan(kb)\)
but i haven't put any numbers in yet...
DeleteNellie: This is for the old depth... I wasn't paying close attention.
DeleteHelen: What trig substitutions did you do for the integral? In the correct units, I think my numbers seem reasonable... but I suppose it's a moot point now with a different problem.
Trig substitutions? How would you use that? Those are definite integrals over difficult limits (not multiples of pi or zero to infinity or anything.) Don't you have to do them numerically?
Delete