I am far from an expert, but I know that as a rookie one can be far from optimal. On the other hand, it could also be that the problem you are trying to do is just too hard (it certainly looks that way). I will look at your notebooks some more. I was hoping to see something besides code. I am not very familiar with condensed matter.

Best,

OL.

Thanks for replying. I don't know how far down the thread you have read.

It's the basic square potential well problem. My tweaks are that I want a generic wave including spin, arbitrarily many identical barriers uniformly spaced ("a" apart and "d" wide) in graphene, with spin orbit interaction and a zeeman field under the potential.

My approach is to first try to replicate some simplified systems (only spin orbit, only one barrier, normal incidence etc) and then to generalise entirely so I can have any number of barriers at any incidence angle and with both SOI and zeeman.

I understand the maths of it (for the most part) but I am having huge trouble translating it to mathematica language cos I am far from fluent. I am making all of the rookie mistakes and it's costing me computing time, memory, and accuracy!

You're right! Delaying everything but the eigenvalue calculations (which are always pretty quick) works great!

Now I need to generalise my equations so I can have arbitrarily many barriers. I'm getting an error I don't recognise.

(apologies if this is a duplicate reply, the thread seems to be bugging out for me)

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The symbolic result - I think so yes.

So I got it to plot the graphs I want. (Well almost. rup and rdown are inverted for k<1 for some reason, but that isn't so important.

But, the way I have it obviously takes more time than is necessary because I make it evaluate symbolically before telling it numbers. As you have seen from my more complex example, I really can't afford to lose that optimization.

How would I have to rewrite this in order to still get those same graphs out in the end?

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Hmmm - how do I avoid this problem then?

Ok. I've had to go back to a simpler example to figure out what the errors are in the more complex one.

So the (almost*) graphs I want are the first three which I get when I evaluate everything separately without delaying (no :=). The graphs I get using := are the second three which are wrong.

I am also having problems explaining to mathematica the a vector k should be k = Sqrt[kx^2 + ky^2] and that ArcTan[ky/ky] is \thetak therefore kx + i ky = k e^(i \thetak) - which is a very longhand maths way of thinking but I understand that and I don't understand computers.

I also can't figure out how to tell mathematica that at the barrier (x=0) then E3[q3] = k = E4[q4]

*these graphs are actually still wrong. But I don't know why. It should be the exact same curve but between 0 and 1 on the y axis.

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Thanks, I'll look into doing those things. As you can tell, I really don't know what I'm at with mathematica!

I even normalized vectors by hand when there is a perfectly good function built in that I've just discovered. :(

I'm going to stop looking at this now as it's past ten here and I have to be up early.

Thank you so much for all your help. I think any errors now are in my own math and I'll have to seek them out. But now I can cos it doesn't take one hundred years to evaluate.

If I have any more problems I know where to come. This forum has been hugely helpful.

Cheers.

Ooh. That will plot Tz.

Unfortunately I just noticed an error in the notebook I sent you

Anz[n_, a_, d_, kx_, l_, g_] := 
 Inverse[L[((n - 1)/2) (a + d), kx]].Smz[a, d, kx, l, 
   g].(MatrixPower[Sz[a, d, kx, l, g], (n - 3)/2]).L[a, kx]

should read

Anz[n_, a_, d_, kx_, l_, g_] := 
 Inverse[L[((n - 1)/2) (a + d), kx]].Smz[a, d, kx, l, 
   g].(MatrixPower[Sz[a, d, kx, l, g], (n - 3)/2]).L[a, kx]

(I had tried to make a simplification that didn't work.)

Changing that the Table[] still evaluates very quickly. And the graph of Tz comes out in 60 seconds.

It is, however not between 1 and 0 which is troubling. I may have more errors than I thought.

That is because I copy and pasted the same thing twice!

Should have read:

Anz[a_, d_, kx_, l_, g_] := 
 Inverse[L[(a + d), kx]].Smz[a, d, kx, l, g].Sz[a, d, kx, l, g].L[a, 
   kx]

should read:

Anz[n_, a_, d_, kx_, l_, g_] := 
 Inverse[L[((n - 1)/2) (a + d), kx]].Smz[a, d, kx, l, 
   g].(MatrixPower[Sz[a, d, kx, l, g], (n - 3)/2]).L[a, kx]

Sorry about that

Hmmm. I guess it should just be a value. It's the transmission probability. It has to be between 0 and 1. But in principle it should just be a number.

When you make the definitions, like you do know, it will evaluate it already and thus 'copying/replacing' the definition that occur in it. Since your definitions are very much nested, it will nest all these functions inside each other symbolically:

Compare the two ways of defining functions (immediate and delayed assignment):

ClearAll[a,b,n]
b[n_]:=n^2
a[m_]:=m b[m]
DownValues[a]
DownValues[b]

ClearAll[a,b,n]
b[n_]=n^2
a[m_]=m b[m]
DownValues[a]
DownValues[b]

Have you tried to evaluate T for some suitable numbers when you have all your definitions done using := (this way the kernel will not die of memory requirements because it does not substitute all the sub-definitions inside each other)

Yes, do all the definitions using := to stop Mathematica from substituting everything inside each other... Could you give me some 'typical' values for the function T you want to calculate?

and the function T itself would also be useful ;)

In your function L[x,kx]... your exponent are k*x not kx. and SL should be:

Sl[a_, kx_, d_] := Simplify[L[2 a + d, kx].Inverse[L[a + d, kx]]]

I presume?

If I remove all the simplify commands, and I run:

Table[Tz[5.0, 0.23, 0.12, k, 0.01, 0.09], {k, DeleteCases[Range[-10.0, 10.0], 0.0]}]

it takes roughly takes 15 seconds. But there might be still an error as it return a 1D vector, not a number. Is that correct. I can further optimize it a lot, because you recalculate Evals and Evecs many times.

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DeleteCases does what is kind-a says: it delete certain cases, in this case 0.0 is removed from the list, as Tz blows up for k=0. No i meant that 'Tz' returns a vector (of 1 value) not just a value. is that correct?