University of Connecticut OpenCommons@UConn
Chemistry Education Materials Department of Chemistry
6-27-2006 The aP rticle in a Box (and in a Circular Box) Carl W. David University of Connecticut, [email protected]
Follow this and additional works at: https://opencommons.uconn.edu/chem_educ Part of the Chemistry Commons
Recommended Citation David, Carl W., "The aP rticle in a Box (and in a Circular Box)" (2006). Chemistry Education Materials. 12. https://opencommons.uconn.edu/chem_educ/12 The Particle in a Box (and in a Circular Box)
C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: June 27, 2006)
I. SYNOPSIS and, taking the second derivative, we have
d2ψ(x) The particle in a box is the simplest problem in elemen- = −Aω2 cos(ωx) − Bω2 sin(ωx) tary quantum mechanics, and as such is useful in more dx2 places than one can easily enumerate. For instance, the molecular orbital theory is explained easily in terms of which is, of course, particle in a box wavefunctions, easier than using stan- −ω2ψ(x) dard atomic orbitals. Hence, the attention to this par- ticular problem. This means that
¯h2 II. INTRODUCTION − −ω2ψ(x) = Eψ(x) 2m We start with a choice of coordinate systems, which, by or, said in the most straightforward manner, E is related the way, influences the form of the solutions we are going to ω i.e., to get, but not the substance. Here, we choose to use 0
Typeset by REVTEX 2 if n6=m. Of course, when n=m, we have the normalization (a well known result). integral The degeneracy appears when we allow the length of the two sides of the ”box” to become equal. At that Z L 2 point, we can factor the common ` = `x = `y out of this ψn(x)dx 6= 0 0 formula to obtain ¯h2π2 i.e., something other than zero. This last condition is E = n2 + n2 used to choose the (so far) arbitrary constant, A, to force nx,ny 2µ`2 x y the integral to have a value of 1, in accord with the prob- It is clear that the sum of two squares can add up to the abilistic interpretation of the wave function. Since these 2 2 2 2 2 2 are, in the case of the particle in a box, integrals of prod- same value, e.g., 5 +2 = 2 +5 , i.e., nx +ny = reverse. ucts of sines, we have We are seeing a double degeneracy emerging. (In three dimensions, we would have a three fold degeneracy devel- L ı nπ x −ı nπ x 2 oping, when the box became cubical.) Notice that when Z e L − e L A dx nx = ny, we have a loss of degeneracy. 0 2ı
q 2 which are trivial. A turns out to be L if we force the IV. A CIRCULAR BOX integral to be 1. Again, this is explained in every text book. Of course, the orthogonality integrals may also be Squares are nice, but we learn more from circles. Con- evaluated using DeMoivre’s theorem. All of this turns sider a two dimensional particle in a circular box. The out to be extraordinarily elementary. particle is restricted to be within r=R, where R is a constant. This is a polar coordinate problem, since the boundary condition will be that ψ(R, θ) = 0∀θ, i.e., the III. A RECTANGULAR BOX wave function will be required to vanish at the edge of the disk region. When we talk about rectangular and square boxes in The first thing we have to do is transform the two dimensional particle in a box problems, we are get- Schr¨odinger Equation from Cartesian to polar coordi- ting set up to discuss degeneracy. The wave equation nates. We are going from (x, y) → (r, θ). The trans- itself offers no hint of the complexity that is coming, but formation equations are the boundary conditions, which are inextricably bound to the solutions of the Schr¨odinger Equation, are the key x = r cos θ here. y = r sin θ (4.1) Assuming we are working in the x-y space (2 dimen- sions), so ψ(x, y) is what is being sought, we have which says, given r and θ, we can compute x and y. The reverse equations are ¯h2 ∂2ψ ∂2ψ p 2 2 − 2 + 2 = Eψ (3.1) r = x + y 2µ ∂x ∂y y θ = tan−1 (4.2) where x and y are the standard Cartesian coordinates. x The solution to this variable separable differential equa- which says the reverse, i.e., given x and y, we can compute tion is discussed in all texts, and has the form r and θ. ∂ n πx n πy First we need to express is terms of partial deriva- X(x)Y (y) = N sin x sin y ∂x `x,`y ` ` tives with respect to r and θ. From the chain rule we x y have where nx and ny are integer quantum numbers, ranging ∂ ∂r ∂ ∂θ ∂ from 1 to ∞. The box is rectangular if ` 6= ` . N is = + x y `x,`y ∂x ∂x ∂r ∂x ∂θ the normalization factor, which has the form y y θ y r s where we are carefully attempting to indicate what is r 2 2 N = constant during each partial differentiation. We find that `x,`y ` ` x y ! ∂r ∂px2 + y2 1 2x = = = cos θ Once substituted into the original Schr¨odinger Equa- ∂x ∂x 2 r tion, the solution suggests that the energy is given by the y y formula: and 2 2 2 2! −1 y ¯h π nx ny ∂θ ∂ tan x En ,n = + (3.2) = x y 2µ ` ` x y ∂x y ∂x y 3
This latter differential is a little more difficult than the former one was, and we attack it in a special manner, guaranteed to bring a smile of recognition to weary read- ers. Specifically, we use implicit differentiation. Thus dy ydx sin θ sin2 θ d tan θ = − 2 = d = dθ + 2 dθ y x x cos θ cos θ tan θ = x so which is
dy ydx sin2 θ cos2 θ + sin2 θ 1 − = 1 + dθ = dθ = dθ x x2 cos2 θ cos2 θ cos2 θ so, holding y constant, one has
1 ydx dθ y y cos2 θ 1 ydx dθ y y cos2 θ dθ = − → = − → − dθ = − → = − → − 2 2 1 2 2 2 2 1 2 2 cos θ x dx cos2 θ x x cos θ x dx cos2 θ x x
dθ tan θ cos2 θ = − so dx K ∂ ∂ sin θ ∂ = cos θ − dθ tan θ cos2 θ sin θ ∂x y ∂r θ r ∂θ r = − = − dx y r cos θ r
Going the other way i.e., holding x constant, we have ∂ ∂ cos θ ∂ = sin θ + dy dθ ∂y x ∂r θ r ∂θ r = x cos2 θ The second partial derivatives then must be so
2 ∂ sin θ ∂ cos θ ∂θ ∂2 ∂ cos θ − = = ∂r θ r ∂θ r (4.3) x ∂y 2 x ∂x y ∂x
cos2 θ ∂θ = 2 ∂ sin θ ∂ + ∂ cos θ ∂ r cos θ ∂y x ∂ ∂r θ r ∂θ r 2 = (4.4) ∂y x ∂y cos θ ∂θ = Equation 4.3 expands to become r ∂y x
2 ∂ sin θ ∂ ! ∂ sin θ ∂ ! ∂ ∂ cos θ ∂r θ − r ∂θ r sin θ ∂ cos θ ∂r θ − r ∂θ r 2 = cos θ − ∂x y ∂r r ∂θ i.e.,
2 2 2 ∂ 2 ∂ sin θ cos θ ∂ sin θ cos θ ∂ 2 = cos θ 2 + 2 − ∂x y ∂r r ∂θ r ∂r∂θ sin2 θ ∂ sin θ cos θ ∂2 sin θ cos θ ∂ sin2 θ ∂2 + − + + (4.5) r ∂r r ∂r∂θ r2 ∂θ r2 ∂θ2 4
For the y term we have
2 ∂ cos θ ∂ ! ∂ cos θ ∂ ! ∂ ∂ sin θ ∂r + r ∂θ cos θ ∂ sin θ ∂r + r ∂θ 2 = sin θ + (4.6) ∂y x ∂r r ∂θ i.e.,
2 2 2 ∂ 2 ∂ cos θ sin θ ∂ sin θ cos θ ∂ 2 = sin θ 2 − 2 + ∂y x ∂r r ∂θ r ∂r∂θ cos2 θ ∂ cos θ sin θ ∂2 cos θ sin θ ∂ cos2 θ ∂2 + + − + (4.7) r ∂r r ∂r∂θ r2 ∂θ r2 ∂θ2 so, adding the two relevant results we have
z }| { ∂2 ∂2 ∂2 sin θ cos θ ∂ sin θ cos θ ∂2 + = + − ∂y2 ∂x2 ∂r2 r2 ∂θ r ∂θ∂r x y | {z } z }| { sin2 θ ∂ sin θ cos θ ∂2 sin θ cos θ ∂ sin2 θ ∂2 + − + + r ∂r r ∂r∂θ r2 ∂θ r2 ∂θ2 | {z } z }| { cos θ sin θ ∂ sin θ cos θ ∂2 − + r2 ∂θ r ∂r∂θ | {z } z }| { cos2 θ ∂ cos θ sin θ ∂2 cos θ sin θ ∂ cos2 θ ∂2 + + − + (4.8) r ∂r r ∂r∂θ r2 ∂θ r2 ∂θ2 | {z }
which becomes where = 2mE/¯h2. ∂2 1 ∂ 1 ∂2 This is one of the forms of the Bessel differential equa- + + tion. To bring it to standard form, we change variables ∂r2 r ∂r r2 ∂θ2 from r to kρ = r, so ∂ρ ∂ ∂ 1 ∂ V. NOW, THE QUANTUM MECHANICS = = ∂r ∂ρ ∂r k ∂ρ The Schr¨odinger Equation now becomes so, choosing 2 ¯h2 ∂2ψ 1 ∂ψ 1 ∂2ψ k = 1 − + + + zero × ψ = Eψ 2m ∂r2 r ∂r r2 ∂θ2 we have (why zero? because a particle in a box just feels its r1 k = boundaries, i.e., the model is of a particle free to roam (linearly) until it hits a wall.) which is going to be related to Bessel’s equation. We know this equation is variable and we have separable, between r and θ, so we will write the solution 2 00 0 0 2 2 ρ R (ρ) + ρR (ρ) + (ρ − m` )R(ρ) = 0 as It is traditional to solve this equation separately for ±mellθ ψ = Rm` (r)e different values or m, and in fact, it is rare to see solutions which, when we substitute this into the Schr¨odinger for m` > 0 anywhere, since the problem, from the point Equation, gives us of view of quantum mechanics, is quote silly. For m` = 0 we have 2 2 2 ¯h ∂ Rm` (r) 1 ∂Rm` (r) m` 2 00 0 0 2 − + − Rm (r) = ERm (r) ρ R (ρ) + ρR (ρ) + ρ R(ρ) = 0 2m ∂r2 r ∂r r2 ` ` which is and we start the solution by assuming an Ansatz X i 2 00 0 2 2m 2 R(ρ) = aiρ r R (r) + rR (r) + (r − m` )R(r) = 0 ¯h2 i=0 5 leaving the question of the indicial equation to more ad- so vanced study. We then have
0 X i−1 R (ρ) = iaiρ i=1 and
00 X i−2 R (ρ) = (i)(i − 1)aiρ i=2
2 00 X i 2 3 4 ρ R (ρ) → (i)(i − 1)aiρ → 2a2ρ + (3)(2)a3ρ + (4)(3)a4ρ + ··· i=2 0 0 X i 2 3 +ρR (ρ) → iaiρ → a1ρ + 2a2ρ + 3a3ρ + ··· i=1 2 X i+2 2 3 4 +(ρ )R(ρ) → aiρ → a0ρ + a1ρ + a2ρ + ··· i=0 = 0
Gathering terms in the standard manner, we have ematics”, John Wiley and Sons. New York, 1962, page 548). So a2 = −a0/2
((3)(2) + 3)a3 = a1 R R ρ0 = = q a0/2 k 1 ((4)(3) + 4)a = −a = 4 2 ((4)(3) + 4)
It is fairly obvious that, contrary to other Frobenius i.e., Schemes in standard Quantum Chemistry, this one does ρ 2 2mE not lead to quantization through truncation of a power 0 = = series into a polynomial. R ¯h2 Instead, this Bessel Function’s expansion never termi- or, nates, is never truncated, i.e., remains an infinite power series. ¯h2 ρ 2 ¯h2 2.40482 The quantization occurs when the boundary condition E = 0 E = that the wave function vanish when we are at the rim of 0 2m R 0 2m R the enclosure is invoked. where you have specified the value of R when setting up The requirement that the function be zero at the the problem. boundary (r=R), results in requiring that the Bessel The zero’s of the Bessel function act to quantize the function have a zero, call it ρ0. This can be found in energy via the boundary conditions, in a slightly differ- tables of the Bessel Function.A For instance, the first ent mode than what we are used to, but never the less, root, J0(ρ0) occurs at ρ0 = 2.4048 (and 5.5207, 8.6537, we obtain a result which is quite comfortable inside the 11.7915, etc., E. Kreyszig, ”Advanced Engineering Math- context of standard quantum mechanical constructs.