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Featured Post

7518

Recently, @zenterix asked a question related to solving the quantum mechanics of the finite-wall-height particle-in-a-box problem. In the last two decades or so I've been teaching a quantum mechanics course every second year, in which I sketch the method of solution of this problem for the class, I but do not get into the mathematical details. A few times I've started to work on it in Maple, but abandoned it because of lack of time. So I decided to persist this time.

This post is more about some of the challenges and tradeoffs in making it work, hence the title. The title is also a nod to the fact that this problem appears in Engel's textbook [1] in a chapter entitled "Applying the particle in a box model to real-world topics". You don't need to know much about quantum mechanics to understand it. From the mathematical point of view you are solving three ordinary differential equations (in three regions) and stitching the solutions together so that the solution, a wavefunction, is continuous with continuous derivative and tends to zero at plus infinity and minus infinity.

The three regions and the wavefunctions for three eigenvalues (energies) are shown here in the figure, which is the final output of the worksheet. Next is the worksheet and I'll comment further on it below.

(It's a quirk of quantum mechanics that we plot two things with different units (wavefunctions and energy) on the same axis, and worse, we offset one by the other.)

Bound states for particle in finite height box of wall height
V(x) = piecewise(x < -(1/2)*a, V__0, -(1/2)*a <= x and x <= (1/2)*a, 0, (1/2)*a < x, V__0)
I'll call these regions left, box and right respectively.

restart

Schroedinger equation. Here as elsewhere "=0" is implied.

Schroedinger := -`&hbar;`^2*(diff(psi(x), x, x))/(2*m)+V*psi(x)-E*psi(x)

-(1/2)*`&hbar;`^2*(diff(diff(psi(x), x), x))/m+V*psi(x)-E*psi(x)

(1)

Nondimensionalize length as units of box length, X = x/a. Since psi(x)^2*dx is unitless probability, psi(x) has dimensions 1/length^(1/2)so we define a dimensionless Phi = a^(1/2)*psi

tr := {x = a*X, psi(x) = Phi(X)/sqrt(a)}; NDSchroedinger := PDETools:-dchange(tr, Schroedinger, [X, Phi(X)], params = [`&hbar;`, m, a], simplify)

{x = a*X, psi(x) = Phi(X)/a^(1/2)}

 

-((1/2)*`&hbar;`^2*(diff(diff(Phi(X), X), X))+a^2*m*Phi(X)*(E-V))/(a^(5/2)*m)

(2)

This will also change the normalization integrals, e.g., over the box region:

normint := Int(psi(x)^2, x = -(1/2)*a .. (1/2)*a); NDnormint := PDETools:-dchange(tr, normint, [X, Phi(X)], params = [`&hbar;`, m, a], simplify)

Int(psi(x)^2, x = -(1/2)*a .. (1/2)*a)

 

Int(Phi(X)^2, X = -1/2 .. 1/2)

(3)

Outside the box we have 0 <= E and E < V with V = V__0. Define a dimensionless positive constant kappa:

`&kappa;__eqn` := kappa = a*sqrt(2*m*(V__0-E)/`&hbar;`^2); de_outside := simplify(sqrt(a)*kappa^2*(eval(NDSchroedinger, {isolate(`&kappa;__eqn`, m), V = V__0}))/(E-V__0)); outside := rhs(dsolve(de_outside))

kappa = a*2^(1/2)*(m*(V__0-E)/`&hbar;`^2)^(1/2)

 

diff(diff(Phi(X), X), X)-kappa^2*Phi(X)

 

c__1*exp(-kappa*X)+c__2*exp(kappa*X)

(4)

Inside the box we have V = 0 and E > 0. Define a dimensionless positive constant k.

k__eqn := k = a*sqrt(2*m*E/`&hbar;`^2); de_box := simplify(-sqrt(a)*k^2*(eval(NDSchroedinger, {isolate(k__eqn, m), V = 0}))/E); box := rhs(eval(dsolve(de_box), {c__1 = A, c__2 = B}))

k = a*2^(1/2)*(m*E/`&hbar;`^2)^(1/2)

 

diff(diff(Phi(X), X), X)+k^2*Phi(X)

 

A*sin(k*X)+B*cos(k*X)

(5)

We have seven unknowns {A, B, E, c__1(L), c__1(R), c__2(L), c__2(R)}, where for example c__1(L) means Maple's c__1 for the left region. We need seven equations: goes to 0 at x = -infinityand at x = infinity(or Phi(X) wouldn't be square integrable), continuity at the two box boundaries, slopes continuous at the two box boundaries, and normalization. We deal with the ones at `&+-`(infinity)first.

At X = -infinity the wavefunction will be unbounded unless one of the constants goes to zero. We'll rename the other one A__L or A__R. The code here is just to handle the fact that c__1 and c__2 can be the the opposite way around, depending on the session.

`assuming`(['limit(outside, X = -infinity)' = limit(outside, X = -infinity)], [kappa > 0]); left0 := eval(outside, `~`[`=`](`minus`(indets(rhs(%), name), {infinity}), 0)); left1 := eval(left0, `~`[`=`](indets(left0, suffixed(c)), A__L)); `assuming`(['limit(outside, X = infinity)' = limit(outside, X = infinity)], [kappa > 0]); right0 := eval(outside, `~`[`=`](`minus`(indets(rhs(%), name), {infinity}), 0)); right1 := eval(right0, `~`[`=`](indets(right0, suffixed(c)), A__R))

limit(c__1*exp(-kappa*X)+c__2*exp(kappa*X), X = -infinity) = signum(c__1)*infinity

 

A__L*exp(kappa*X)

 

limit(c__1*exp(-kappa*X)+c__2*exp(kappa*X), X = infinity) = signum(c__2)*infinity

 

A__R*exp(-kappa*X)

(6)

Now we have five unknowns {A, A__L, A__R, B, E}. The strategy will be to eliminate A, A__L, A__R and then find the eigenvalues E. Then we will have expressions for all three regions containing the unknown B, which we will find by the normalization condition (The normalization condition fixes the scale of the wavefunction so that the probability of finding the particle somewhere is one.)

Continuity of the wavefunction and its derivative at the left boundary are below
We eliminate A__L and A using this boundary

bcleft := {eval(left1-box, X = -1/2), eval(diff(left1, X)-(diff(box, X)), X = -1/2)}; sol := solve(bcleft, {A, A__L}); left2 := eval(left1, sol); box2 := eval(box, sol)

{A__L*exp(-(1/2)*kappa)+A*sin((1/2)*k)-B*cos((1/2)*k), A__L*kappa*exp(-(1/2)*kappa)-A*k*cos((1/2)*k)-B*k*sin((1/2)*k)}

 

{A = -B*(sin((1/2)*k)*k-cos((1/2)*k)*kappa)/(sin((1/2)*k)*kappa+cos((1/2)*k)*k), A__L = B*k*(sin((1/2)*k)^2+cos((1/2)*k)^2)/(exp(-(1/2)*kappa)*(sin((1/2)*k)*kappa+cos((1/2)*k)*k))}

 

B*k*(sin((1/2)*k)^2+cos((1/2)*k)^2)*exp(kappa*X)/(exp(-(1/2)*kappa)*(sin((1/2)*k)*kappa+cos((1/2)*k)*k))

 

-B*(sin((1/2)*k)*k-cos((1/2)*k)*kappa)*sin(k*X)/(sin((1/2)*k)*kappa+cos((1/2)*k)*k)+B*cos(k*X)

(7)

Eliminate A__R at the right box boundary

bcright := {eval(box2-right1, X = 1/2), eval(diff(box2, X)-(diff(right1, X)), X = 1/2)}; elim := eliminate(bcright, A__R); right2 := eval(right1, elim[1]); energy_eqn := elim[2][]/B

{-B*(sin((1/2)*k)*k-cos((1/2)*k)*kappa)*sin((1/2)*k)/(sin((1/2)*k)*kappa+cos((1/2)*k)*k)+B*cos((1/2)*k)-A__R*exp(-(1/2)*kappa), -B*(sin((1/2)*k)*k-cos((1/2)*k)*kappa)*k*cos((1/2)*k)/(sin((1/2)*k)*kappa+cos((1/2)*k)*k)-B*k*sin((1/2)*k)+A__R*kappa*exp(-(1/2)*kappa)}

 

[{A__R = B*(2*sin((1/2)*k)*cos((1/2)*k)*kappa+2*cos((1/2)*k)^2*k-k)/(exp(-(1/2)*kappa)*(sin((1/2)*k)*kappa+cos((1/2)*k)*k))}, {-2*(sin((1/2)*k)*cos((1/2)*k)*k^2-sin((1/2)*k)*cos((1/2)*k)*kappa^2-2*cos((1/2)*k)^2*k*kappa+k*kappa)*B}]

 

B*(2*sin((1/2)*k)*cos((1/2)*k)*kappa+2*cos((1/2)*k)^2*k-k)*exp(-kappa*X)/(exp(-(1/2)*kappa)*(sin((1/2)*k)*kappa+cos((1/2)*k)*k))

 

-2*sin((1/2)*k)*cos((1/2)*k)*k^2+2*sin((1/2)*k)*cos((1/2)*k)*kappa^2+4*cos((1/2)*k)^2*k*kappa-2*k*kappa

(8)

Now we will get the quantized energies by finding which values will make energy_eqn zero. We recall the relationships of `&kappa;__` and k to energy:

k__eqn; `&kappa;__eqn;`

k = a*2^(1/2)*(m*E/`&hbar;`^2)^(1/2)

 

kappa = a*2^(1/2)*(m*(V__0-E)/`&hbar;`^2)^(1/2)

(9)

The energy scale is set by V__0, so can get a non-dimensionalized energy e = E/V__0 and a non-dimensional parameter b = a*sqrt(2*m*V__0/`&hbar;`^2).

b__eqn := b = a*sqrt(2*m*V__0/`&hbar;`^2); e__eqn := e = E/V__0; eqns := `assuming`([{simplify(eval(eval(k__eqn, isolate(e__eqn, E)), isolate(b__eqn, V__0))), simplify(eval(eval(`&kappa;__eqn`, isolate(e__eqn, E)), isolate(b__eqn, V__0)))}], [a > 0])

b = a*2^(1/2)*(m*V__0/`&hbar;`^2)^(1/2)

 

e = E/V__0

 

{k = (e*b^2)^(1/2), kappa = (-b^2*(e-1))^(1/2)}

(10)

So now for any values of the box length and height, one finds the parameter b describing the problem and solves for the possible e values.

energy_eqn2 := `assuming`([simplify((eval(energy_eqn, eqns))/b^2)], [b > 0, e > 0, e < 1])

2*e^(1/2)*(-e+1)^(1/2)*cos(b*e^(1/2))-2*e*sin(b*e^(1/2))+sin(b*e^(1/2))

(11)

Maple cannot find an analytical solution, so we will find the eigenvalues numerically. The zero solution is unphysical.

solve(energy_eqn2, e)

0, RootOf(-2*(-(_Z^2-b^2)/b^2)^(1/2)*(_Z^2/b^2)^(1/2)*b^2+2*tan(_Z)*_Z^2-tan(_Z)*b^2)^2/b^2

(12)

For any b, read off the possible energies off the vertical line for that b, e.g. there are 3 bound states for b = 9. For other values, choose bval and nsols on the next line appropriately.

bval := 9; nsols := 3; plots:-implicitplot([b = bval, energy_eqn2], b = 0 .. 15, e = 0 .. 1)

9

 

3

 

 

evals := [fsolve(eval(energy_eqn2, b = bval), e = 0 .. 1, maxsols = nsols)]

[0.8115199822e-1, .3189598393, .6884466500]

(13)

In terms of the parameters b and e and normalization constant B, the non-dimensionalized wavefunctions of the 3 regions are as below. The scale factor gives a simpler form and prevents 0/0 problems later in the calculation.

scale := `assuming`([denom(simplify(eval(left2, eqns)))], [positive]); left3 := `assuming`([simplify(eval(scale*left2, eqns))], [positive]); box3 := `assuming`([simplify(eval(scale*box2, eqns))], [positive]); right3 := `assuming`([simplify(eval(scale*right2, eqns))], [positive])

sin((1/2)*b*e^(1/2))*(-e+1)^(1/2)+cos((1/2)*b*e^(1/2))*e^(1/2)

 

B*e^(1/2)*exp((1/2)*b*(-e+1)^(1/2)*(2*X+1))

 

-(sin((1/2)*b*e^(1/2))*(e^(1/2)*sin(b*e^(1/2)*X)-(-e+1)^(1/2)*cos(b*e^(1/2)*X))-cos((1/2)*b*e^(1/2))*(cos(b*e^(1/2)*X)*e^(1/2)+sin(b*e^(1/2)*X)*(-e+1)^(1/2)))*B

 

B*exp(-(1/2)*b*(-e+1)^(1/2)*(-1+2*X))*(e^(1/2)*cos(b*e^(1/2))+(-e+1)^(1/2)*sin(b*e^(1/2)))

(14)

It remains to find the normalization constant B. The three parts of the normalization integral are:

P__L := `assuming`([int(left3^2, X = -infinity .. -1/2)], [positive]); P__box := `assuming`([int(box3^2, X = -1/2 .. 1/2)], [positive]); P__R := `assuming`([int(right3^2, X = 1/2 .. infinity)], [positive])

(1/2)*B^2*e/(b*(-e+1)^(1/2))

 

-(1/4)*B^2*(2*e^(1/2)*(-e+1)^(1/2)*cos(2*b*e^(1/2))-2*e^(1/2)*(-e+1)^(1/2)-2*b*e^(1/2)-2*e*sin(2*b*e^(1/2))+sin(2*b*e^(1/2)))/(b*e^(1/2))

 

(1/2)*B^2*(e^(1/2)*cos(b*e^(1/2))+(-e+1)^(1/2)*sin(b*e^(1/2)))^2/(b*(-e+1)^(1/2))

(15)

Solve for B. There are two (messy) solutions for B of opposite sign; it is conventional to choose the sign such that the ground state has positive values.

Bval := sort([solve(P__L+P__box+P__R = 1, B)])[2]

Assemble it into a single piecewise function.

`&Phi;__all` := eval(piecewise(X < -1/2, left3, `and`(X >= -1/2, X <= 1/2), box3, X > 1/2, right3), B = Bval)

Plot. Plots offset vertically by the energy as usual

Xmax := 1.5; psiscale := .15; colors := [red, blue, magenta]; wavefns := plot([seq(eval(psiscale*`&Phi;__all`, {b = bval, e = evals[i]})+evals[i], i = 1 .. nsols)], X = -Xmax .. Xmax, color = colors); walls := plot([-1/2, y, y = 0 .. 1], color = black), plot([1/2, y, y = 0 .. 1], color = black), plot(1, X = -Xmax .. -1/2, color = black), plot(0, X = -1/2 .. 1/2, color = black), plot(1, X = 1/2 .. Xmax, color = black); evalplot := plot(evals, X = -Xmax .. Xmax, linestyle = dash, color = colors); plots:-display(wavefns, walls, evalplot, axes = boxed, labels = [X, psiscale*Phi+E/V__0])

 

NULL

Download finite_box_non-dim2.mw

Comments
1. The first general comment is about the issue of how much Maple can do. In principle one should be able to give dsolve the three ODEs and the boundary conditions and it should output the result. We are still far away from that. Even for one region, the boundary conditions at infinity are not handled by dsolve. The other extreme is to solve the ODEs for their general solutions, and follow the algebraic manipulations for implementing the boundary conditions as one might do it on paper. Engel for example has a comment "At this point we notice that by dividing the equations in each pair, the coefficients can be eliminated to give [...]". Maple will not easily do that trick or the manipulations that led to the special form on which the trick is applied. Levine's textbook [2] gives a different non-intuitive set of steps.

But Maple can solve complicated equations, so I wanted a more general strategy that avoids as much as possible the requirement to manipulate into special forms. On the other hand, it is tempting (as @zenterix tried) to just give the three general solutions with 6 arbitrary constants and the boundary conditions to Maple's solve, and solve for the six constants. One finds that all six are zero. This us a consequence of the fact that the ODEs are linear, and misses the crucial point that it is an eigenvalue problem. So there must be some sort of step-by-step guidance for Maple and there is a general but non-obvious strategy to solving such problems.

2. The second general comment is that there is a trade-off between presenting the solution steps without any arcane Maple code, and getting to the solution efficiently and programmatically, without too many manual manipulations. One of my suggestions to @zenterix involved manually dividing both sides of an equations by a fairly complicated expression, to which @acer expressed a preference for programmatic solutions. At the time, I mainly did that to keep close the to existing solution, and was thinking that I usually set things up so that is not necessary. But I found here that I do it a lot, and it is a natural consequence of the interactive way I use Maple. Here's two examples:

In the second line of label (4), I originally got a more complicated form of de_outside, then manually figured out the factor to put inside simplify to get the form you see. I do this frequently with Maple, especially when I use dchange.

In the last line of label (8), I manually divided by B to remove it. Had B been in the denominator, I could have removed it programmatically with numer (an operation I use frequently, though it somewhat recklessly ignores denominator zeroes).

Programmatically doing things can be relatively unreadable and disrupt the readability for the non-Maple reader. For example in label (6) there is "extraneous" code to find which coefficient to set to zero so that the solution goes to zero at +/- infinity. This was forced on me since after generating the nice figure I re-ran the worksheet only to have multiple errors. I originally used eval(outside, {c__1 = B__L, c__2 = A__L}); to change Maple's dsolve constants to read the way I wanted. But I realized that dsolve does not reproducibly assign c__1 and c__2 to the same term each time, raising the general issue of: 

3. Session to session reproducibility. If I am only going to use a worksheet myself, I don't usually worry too much about this, since I can usually debug this fairly easily. On the other hand if the worksheet is for others, it needs to be foolproof. I have taken to always sorting the output of solve, e.g., the assignment to bval in the execution group after label (15). The code added for the dsolve constants works, but is non-trivial Maple (suffixed type testing). (Solving this issue for Eigenvectors is yet more difficult.)

4. Some efforts to make the worksheet more readable were hard to do.

- For some reason, Engel chose to call the two constants in the left region B and B'. Entering B*exp(-k*x) + B'*exp(k*x) in 2D leads to B(x)*exp(-k*x) + diff(B(x), x)*exp(k*x). No doubt this can be fixed, but I didn't pursue this.

- I would normally use upper case Psi for the non-dimensionalized psi, but Psi is the name of a special function in Maple. But wait, now we can use "local Phi;" to solve this problem. However diff(Psi(x),x,x) displays as Psi(2,x), so I had to settle for Phi. (SCR submitted.)

- I build up the left wavefunctions incrementally as expressions assigned to left1, left2, etc, which is rather ugly. Why not use arrow procedures so we can use psi(x), D(psi)(x) etc. Aside from some awkwardness in updating such procedures by redefining them as needed, I ran into the following problem:

limit(A*exp(-k*x) + B*exp(k*x), x = infinity) assuming k > 0; gives as expected
signum(B)*infinity;

but

psi := x -> A*exp(-k*x) + B*exp(k*x);
limit(psi(x), x = infinity) assuming k > 0;
returns unevaluated. What happened to full evaluation?

- I'm used to entering hbar as hbar in 1D; it displays as h with the bar through. In 2-D entering hbar^2 is displayed nicely, but internally it is `&hbar;`. We can have the following puzzle:

(hbar/m)^2-`&hbar;`^2/m^2

hbar^2/m^2-`&hbar;`^2/m^2

NULL

- I originally wanted a vertical axis label Psi, E/V__0, but labels = [X, Psi, E/V__0] obviously won't work. It took some time to figure out the answer is to make "Psi, E/V__0" an atomic variable. I finally settled on the more honest, if cryptic, label shown.

5. solve gives an "invalid" solution. Originally I worked with solutions left2 etc (see label (6)) after simplify(eval(left2, eqns)). Every second wavefunction was about 10^9 times higher than the others, which was difficult to debug. It turns out the denominator of those wavefunctions was exactly zero but numerically 10^(-10), meaning they plotted as large-scaled versions of the right shape, without any division-by-zero error. The fix is to remove the denominators by scaling (see label (14)).

This is just a consequence of Maple giving generic solutions, and the only way around it is to be aware of it.

Summary

To solve such problems with Maple requires one to play around in an interactive way. After hacking to a solution, it can take some effort to make it presentable and usable to others. But the final result is pleasing, and is certainly much easier than working through the problem on paper. The ability to manipulate complicated expressions means circumventing tricks that might be needed in manual solutions. 

References
[1] T. Engel, Quantum Chemistry and spectroscopy. Pearson 2019, 4th ed. Sec. 5.1, Prob. 5.3. 
[2] I. N. Levine, Quantum Chemistry. Pearson 2009, 6th ed. Sec. 2.4. 

Featured Post

I'm happy to announce the publication of Volume 5, Issue #1 of Maple Transactions.  You can find it at

mapletransactions.org
 

We have a survey paper by Veselin Jungic and Naomi Borwein on teaching Experimental Mathematics courses as our Featured Contribution.  Many of you will find it interesting and useful.

In the refereed paper section we have a paper on Metaprogramming with Maple and C by Ilias Kotsireas; a paper on fast transposed Vandermonde solving by Hyukho Kwon & Michael Monagan; a paper by David Ulgenes (an honours student in Oslo) on Gamma, Pseudogamma, and Inverse Gamma functions; and a paper by John Campbell on applications of Gosper's nonlocal derangement identity (which, if you don't know that the word "derangement" has a technical mathematical meaning, may give you the wrong impression!).

As usual I've also written something, and I hope you like it: it's about Chladni figures and standing modes in an elliptical drum, and visualizing such in Maple.  It uses Mathieu functions in Maple and noodles a bit about zerofinding (but winds up using fsolve because that's so convenient).
 

Keep the papers coming.  This is the 12th issue of Maple Transactions, and I remind you that it has a "Diamond Class" designation, which means there are no page charges to authors, and the articles are free to read for everyone.  This means that there's some volunteer labour needed, of course: you have to write the articles, and what we want is that you write articles that people in the Maple community actually want to read.

I'd also like to thank the copyeditor, Michelle Hatzel, for her very hard work on this issue.  She's really made a difference, and I think you will be able to see it.   



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