Separation of Variables

Physics 342 Lecture 1 Separation of Variables Lecture 1 Physics 342 Quantum Mechanics I Monay, January 25th, 2010 There are three basic mathematical tools we nee, an then we can begin working on the physical implications of Schröinger s equation, which will take up the rest of the semester. So we start with a review of: Linear Algebra, Separation of Variables (SOV), an Probability. There will be no particular completeness to our iscussions at this stage, I want to emphasize those aspects of each of these that will prove useful. Pitfalls an caveats will be aresse as we encounter them in actual problems (where they won t show up, so won t be aresse). Try as one might, it is ifficult to relate these three areas, so we will take each one in turn, starting from the most familiar, an working to the least (that s a matter of taste, so apologies in avance for the wrong orering in anticipation of our section on probability: How many ifferent orerings of these three subjects are there? On average, then, how many people isagree with the current one?). We will begin with separation of variables, a simple technique that you have use recently we will review the basic electrostatic interest in these solutions, an o a few examples of SOV applie to PDE s other than Laplace s equation ( 2 V = 0). 1.1 Separation for the Laplace Problem Separation of variables refers to a class of techniques for probing solutions to partial ifferential equations (PDEs) by turning them into orinary ifferential equations (ODEs). We generally rely on some notion of uniqueness 1 for 1 Uniqueness implying both a solution to the PDE an satisfaction of some bounary conitions. 1 of 11

1.1. SEPARATION FOR THE LAPLACE PROBLEM Lecture 1 our PDE then the logic of SOV rests on the notion that fining a solution, even by this somewhat limite technique, suffices. There are two basic flavors of SOV, aitive an multiplicative in either case, the style of argument is the same, so we ll start with the algebraically simpler version. 1.1.1 Aitive Separation We take the Laplace problem: Fin V (x, y) satisfying 2 V = 0 on some region, with some specifie bounary conitions (for example, in E&M, we woul have an implicit V 0 at spatial infinity conition, an maybe some conucting bounaries where we set V = 0). Working in two imensions (for simplicity), we start with a separation ansatz aitive, in this case: then running this through the Laplacian gives: V (x, y) = V x (x) + V y (y), (1.1) 2 V x x 2 + 2 V y = 0. (1.2) y2 Now for the stanar argument: V x (x) epens only on x just as V y (y) epens only on y they cannot be equal unless each is equal to a constant. Suppose we set V x = α, then we must have V y = α. The solution to these ODEs is: V x (x) = 1 2 α x2 + β x + δ V y (y) = 1 2 α y2 + γ y + ρ. (1.3) for arbitrary constants (β, δ, γ, ρ), to be use in imposing the bounary conition(s). The solution, for any α, can be written as: V (x, y) = 1 2 α (x2 y 2 ) + β x + γ y + κ (1.4) As far as electrostatics goes, this solution oes not match our usual notion of potential it oesn t ie at spatial infinity. That s not always a eal-breaker, but it is true that aitive separation is of limite utility in E&M (it plays a much larger role in, for example, Hamilton-Jacobi theory). Nevertheless, the pattern of: ansatz, argument, solution is the same as for the more immeiately useful, an algebraically involve multiplicative separation. 2 of 11

1.1. SEPARATION FOR THE LAPLACE PROBLEM Lecture 1 1.1.2 Multiplicative Separation The ifference here is in the starting ansatz rather than aitive single variable functions, we take the multiplicative combination: then Laplace s equation becomes V (x, y) = V x (x) V y (y), (1.5) 2 V x x 2 V 2 V y y + V x = 0, (1.6) y2 an if we ivie by V (x, y) itself, we can write this in a form conucive to making the separation argument: V x + V y = 0. (1.7) V x V y The first term epens only on x, the secon only on y, so for a solution satisfying our functional assumptions, we must have both terms equal to a constant, call it α 2 (a convenient choice, as you will see in a moment) for α complex. Then V x = α 2 = V y (1.8) V x V y will certainly have 2 V = 0. The solutions here are familiar: V x = A e α x + B e α x V y = F e i α y + G e i α y. (1.9) Now we see that for a generic complex number, both V x an V y will be combinations of growing an ecaying exponentials, as well as oscillatory sines an cosines. For example, if α is real, V x is growing an ecaying, while V y is pure sine an cosine. For α = i a with a real, the roles are reverse. In general, we use the bounary conitions to choose a convenient expression for α. Example Consier a pair of infinite, groune conucting sheets separate a istance with a conuctor connecting the two sheets hel at V 0 as shown in Figure 1.1. What is the potential in between the plates? We know that the 3 of 11

1.1. SEPARATION FOR THE LAPLACE PROBLEM Lecture 1 potential satisfies Laplace s equation in the region between the plates (no charge in there), an the bounary conitions are clear: ( ) 2 π x V (x = 0, y) = V (x =, y) = 0 V (x, y = 0) = V 0 sin. (1.10) ŷ ˆx V 0 sin ( ) 2πx Figure 1.1: Two parallel plates separate by a istance an hel at V = 0, the connecting plate is hel at V 0 sin(2 π x/). We know the solution will be of the form (1.9), but how to choose α an the constants (A, B, F, G)? There is an aitional, implicit bounary conition we like the potential to go to zero in the open spatial irection, y this tells us that we shoul set α = i a for a IR to get growing an ecaying exponentials for the V y (y) solution: V x = A e i a x + B e i a x V y = F e a y + G e a y (1.11) an, furthermore, we shoul set G = 0 to eliminate the growing exponential. By introucing à = (A+B) an B = i (A B), we can write V x (x) in terms of sine an cosine. Our solution so far, then, is: V x = à cos(a x) + B sin(a x) V y = F e a y V = V x V y. (1.12) Now for the rest of the bounary conitions: take V x (x = 0) = 0 that tells us that à = 0. Our solution is now much simpler, the full potential has the 4 of 11

1.1. SEPARATION FOR THE LAPLACE PROBLEM Lecture 1 form V = Q sin(a x) e a y (1.13) where we have introuce yet another rewrite of combinations of constants (Q = Ã F ). The bounary conition V (x =, y) = 0 is interesting, this tells us that sin(a ) = 0, not a statement about the obvious constant Q, but rather, a constraint on the allowe wavelengths: sin(a ) = 0 a = n π (1.14) for integer n. Then the only a s that satisfy the bounary conition are a = n π, still an infinite family, but labelle by the integer inex n. So, we have V n (x, y) = Q n sin(n π x/) e n π y/, (1.15) an we put the subscript n here to remin us that any integer n provies a solution. This type of bounary conition, one that limits the allowe wavelength, is what leas to the mathematical representation of quantization when applie to the separable solutions of Schröinger s equation. We have one bounary left because Laplace s equation is a linear PDE, sums of solutions are still solutions, an we can make a general solution out of our n-inexe ones. Each term satisfies Laplace s equation an the bounary conitions at x = 0, an y, so the sum oes as well: V (x, y) = ( n π x ) Q n sin n=1 e n π y. (1.16) As for the final bounary conition: V (x, y = 0) = V 0 sin(2 π x/), we can take Q n = 0 for all n except for n = 2. We know that the solution is unique, so this is certainly vali. But it begs the question, what about a less wellaapte riving potential? That is a question for next time, as it is tie to the ecomposition of functions in sin an cosine bases, a linear algebra issue. The final answer: ( ) 2 π x V (x, y) = V 0 sin e 2 π y. (1.17) This solution is just the potential at the y = 0 plate, ecaying into the region between the plates equal contours are shown in Figure 1.2. 5 of 11

0.5 0.4 0.3 0.2 0.1 0.0 0.0 0.2 0.4 0.6 0.8 1.0 V 0 sin ( ) 2πx Figure 1.2: The potential between the plates for the setup shown in Figure 1.1. There is not much more to say about separation of variables the best approach is to o as many examples as you can get your hans on that makes some of the choices that appear unmotivate at first (why shoul we make the integration constant α 2? Why o we pick the y solution to be growing an ecaying?) more reasonable. 1.2 Separation for Other PDEs We will be applying the SOV technique to Schröinger s equation, an that is not going to be much ifferent than the Laplacian example from E&M. The approach is inepenent of PDE, although the PDE must be linear (not strictly speaking true, but the superposition principle we invoke in writing (1.16) relie on linearity) just to show some of the places where SOV comes up, we can look at the Helmholtz equation an the wave equation. 6 of 11

1.2.1 Helmholtz Equation Suppose we consier the same problem as above, but for the Helmholtz PDE: ( 2 + µ 2 ) V = 0 ( ) 2 π x V (x, y = 0, ) = 0 V (x, y = 0) = V 0 sin V (x, y ) 0, (1.18) we are imagining a slight moification to electrostatics (actually amounts to a theory of electrostatics with a massive photon, but forget about that for now). First we ll fin the ODE solutions to the Helmholtz PDE via V (x, y) = V x (x) V y (y): V x + V y + µ 2 = 0. (1.19) V x V y We see immeiately that we will again get oscillatory an exponential solutions, since we want the solution in the increasing y irection to ecay, we ll call V y V y = α 2, with α real. With the constant µ 2 in place, we on t nee to set the x term equal to α 2, we have aitional freeom. We know that the bounary conitions for x are easily impose for sines, so set V x V x = β 2 for β real. Then the constraint we must impose to solve the Helmholtz PDE is: β 2 + α 2 + µ 2 = 0. Imposing the bounary conitions at x = 0, gives back β = n π as before, so that ( n π x ) V x = A sin. (1.20) Now, for the V y equation, we are given µ, an we have β 2 = n 2 π 2 / 2, so the growing an ecaying exponentials have the form: V y = F e n 2 π 2 / 2 µ 2 y + G e n 2 π 2 / 2 µ 2 y. (1.21) There are interesting elements to this solution, most notably, a cutoff in n for which V y is actually oscillatory (n < µ π ) ignoring that for the moment, if we assume the above terms for V y are inee growing an ecaying for n = 2, we have the solution: ( ) 2 π x V (x, y) = V 0 sin e 4 π 2 / 2 µ 2 y, (1.22) which just has a ifferent funamental ecay length when compare to the Laplace form for electrostatics. 7 of 11

1.2.2 The Wave Equation Separation of variables is not limite to purely spatial problems we can mix in time. For the wave equation in one spatial imension (call it x), efine in a meium with funamental spee v, we have 1 v 2 2 u(x, t) t 2 + 2 u(x, t) x 2 = 0. (1.23) Take a multiplicative ansatz: u(x, t) = u x (x) u t (t), then by the now familiar proceure of inputting an iviing by u: u t an again, setting both terms equal to a constant: u t v 2 + u x = 0 (1.24) u t u x v 2 = α 2 u x = α 2. (1.25) u t u x What is interesting here is that both the spatial solution u x an the temporal u t can be oscillatory that is not true for the Laplace equation separable solutions we saw above. Without worrying about bounary conitions, we have the easy solution: u t = A cos(α v t) u x = B cos(α x) (1.26) so we can write u(x, t) = u 0 cos(α x) cos(α v t) of course, just as obvious is the solution û(x, t) = u 0 sin(α x) sin(α v t). Since the wave equation is linear, the sum of these two solutions is also a solution, an it is an instructive one: u(x, t) + û(x, t) = u 0 (cos(α x) cos(α v t) + u 0 sin(α x) sin(α v t)) = u 0 cos(α (x v t)), (1.27) precisely a plane wave. Of course, in general, we woul be given spatial bounary conitions an/or an initial waveform at t = 0. 1.2.3 Poisson s Equation The utility of separation is not limite to source-free equations consier Poisson s equation for electrostatic potential in the presence of source charge ensity ρ: 2 V = ρ ɛ 0. (1.28) 8 of 11

We must be given a ρ that is itself appropriately separable. The simplest possible such istribution woul be a constant, but we can imagine more interesting charge istributions. As an example, suppose we are in spherical coorinates, an we have a spherically symmetric charge ensity: ρ(r, θ, φ) = ρ(r), epening only on our istance from the origin. Poisson s equation becomes, in these variables ( 1 r 2 r ( r 2 V r ) + 1 r 2 sin θ ( sin θ V ) + θ θ 1 r 2 sin 2 θ 2 ) V φ 2 V = ρ(r). ɛ 0 (1.29) The separation ansatz now becomes V (r, θ, φ) = V r (r) V θ (θ) V φ (φ), an we can see what will happen to satisfy the functional epenence assumption, we will have constant V θ an V φ, leaving us with just the raial ODE: 1 r 2 r ( r 2 V r ) ρ(r) =. (1.30) ɛ 0 Suppose we take ρ(r) = ρ 0 a constant, then the solution to the above is: V r (r) = α + β r ρ 0 r 2 6 ɛ 0. (1.31) Notice that the first two terms are just solutions to 2 V = 0 (an overall constant an the potential outsie a spherically symmetric istribution of charge) if we are insie a uniformly charge sphere, with r = 0 inclue in the omain, then we must set β = 0. The constant α, of course, can be set to zero, an we are left with the usual potential for a uniformly charge sphere. 9 of 11

Homework Reaing: Griffiths Introuction to Electroynamics, pp. 127 136. Problem 1.1 Solve Laplace s equation 2 V = 0 using separation: V (x, y) = V x (x) V y (y) for the following two-imensional bounary conitions: V (0, y) = V (, y) = 0, V (x, ) = 0 an V (x, 0) = V 0 sin ( ) 2 π x. y (V = 0) (V = 0) (V = 0) x ( ) 2 π x V 0 sin Physically, we are escribing the electrostatic potential insie a square that has groune lines on three sies, an a spatially-varying potential on the fourth sie. Problem 1.2 The heat equation in one (spatial) imension reas 1 k u(x, t) t = 2 u(x, t) x 2, (1.32) where u(x, t) is the temperature of an object with thermal conuctivity k (a material property). For a complete solution, we must specify two bounary conitions (since this equation is secon orer in x) an an initial conition (the value of u(x, t = 0), for example). 10 of 11

a. Using separation of variables: u(x, t) = u x (x) u t (t), an separation constant α 2, write own the separate heat equation an solve for u x an u t. b. Suppose we have a material that goes from x = 0 to x = we set the temperature on both ens of the material to zero i.e. u(0, t) = u(, t) = 0. Initially, at time t = 0, we have temperature: u(x, 0) = u 0 sin ( ) π x. Using your separation solution, fin the particular solution for this set of bounary an initial conitions. u(x, 0) u 0 ( π x ) u 0 sin x 11 of 11