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Quantum Mechanics in Multidimensions

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Quantum Mechanics in Multidimensions Quantum Mechanics in Multidimensions In this chapter we discuss bound state solutions of the Schr¨odinger equation in more than one dimension. In this chapter we will discuss some particularly straightforward examples such as the particle in two and three dimensional boxes and the 2-D harmonic oscillator as preparation for discussing the Schr¨odinger hydrogen atom. The novel feature which occurs in multidimensional quantum problems is called “degeneracy” where different wave functions with different PDF’s can have exactly the same energy. Ultimately the source of degeneracy is symmetry in the potential. With every symmetry, there is a conserved quantity which can be used to “label” the states. For rotationally symmetric potentials, the conserved quantities which serve to label the quantum states are angular momenta. Electron in a two dimensional box By an electron in a two dimensional box, we mean that the potential is zero within the walls of the box and infinity outside the box. For convenience we will place the origin at one corner of the box as illustrated below: y V = V = 0 V = b V = x a V = To go from the one dimensional to the two dimensional time independent SE 1 we simply take ∂2/∂x2 → ∂2/∂x2 + ∂2/∂y2 and obtain for a particle of mass µ: ¯h2 ∂2 ∂2 − + ψ(x, y)+V (x, y) ψ(x, y)=Eψ(x, y)(1) 2µ ∂x2 ∂y2 For the case under discussion here, the electron is confined within the box, where we have V (x, y)=0. We use a separation of variable technique to solve this partial differential equation and try product wave functions of the form ψ(x, y)=X(x) × Y (x). ¯h2 ∂2X ∂2Y − Y + X = EXY (2) 2µ ∂x2 ∂y2 We can divide sides of Eq. (2) by X(x)Y (y) to obtain: 1 ¯h2 ∂2X 1 ¯h2 ∂2Y − + − = E (3) X 2µ ∂x2 Y 2µ ∂y2 1 2 Eq. (3) says that the sum of the first term and second term add to a constant (as opposed to a function). This suggests that each of these terms is equal to a constant or any functional dependence between them cancels when summed to give a constant. However the first term can only be a function of x ; while the second term can only be a function of y. There is therefore no way that the terms can have a non-constant part but sum to a constant. We therefore conclude that E both terms are individual constants. Lets call these constants ( )1 = x,and E E E E ()2 = y,and x + y = . We then have two ordinary differential equations: ¯h2 d2X ¯h2 d2Y − = Ex X,− = Ey Y (4) 2µ dx2 2µ dy2 d2X d2Y = −k2 X, = −k2 Y dx2 x dy2 y 2 h2 ¯ 2 2 where E = Ex + Ey = kx + ky (5) 2µ The solutions of Eq. (5) are sinusoidal products of the form: ψ x, y α k x β k x × γ k y δ k y ( )=( sin ( x )+ cos ( x ))1 ( sin ( y )+ cos ( y ))2 (6) Let us consider the wave function boundary conditions. The wave function must y ψ x, y δ vanish on the line =0,whichmeans ( =0)=0or()2 =0meaning =0. The wave function must vanish on the line x =0whichmeansψ(x =0,y)=0 β or ( )1 =0meaning = 0. We thus know we need sine functions for both x and y: ψ(x, y)=α sin (kx x) × sin (ky y)(7) We next consider the boundary conditions on the top and the right of the well. Since ψ(x = a, y)=0,thesin(kx x) piece must have an argument which is an integral multiple of π which implies kx a = nxπ where nx ∈ 1, 2, 3, .... Because ψ(x, y = b)=0wemusthaveky b = nyπ where ny ∈ 1, 2, 3, ....Wethushave: 2 nx πx ny πy ¯h π2 nx 2 ny 2 ψ(x, y)=α sin ( ) × sin ( )andE(nx,ny)= + a b 2µ a b (8) Degeneracy The particle in the two dimensional box has an energy which is controlled by two integer quantum numbers as opposed to the one dimensional case E = (¯hπ)2 n2/(2µa2) where a single, integer quantum number (n) controls the energy level. We get one integer quantum number per dimension. In the case of the particle in a box, nx + 1 is the number of wave function nodes along x and ny +1 is the number of nodes along y. Each set of quantum numbers {nx,ny} results in a distinguishable wave function. If once selects a square box with a = b,often two different sets of quantum numbers {nx,ny} , with two distinguishable wave functions, will have the same (or degenerate) energies. It easy to confirm the below energy level chart for the case of a square, two dimensional box. 3 (1,5) (5,1) 26E (3,4) (4,3) 25E (2,4) (4,2) 20E (3,3) 18E (1,4) (4,1) 17E a (2,3) (3,2) 13E a (1,3) (3,1) 10E 2 (2,2) 8E ( h π ) E = 2m a2 (1,2) (2,1) 5E (1,1) 2E 0 It looks like the levels are either unique (degeneracy = 1) or pair degenerate (degeneracy = 2), however if we went up further we would see examples of triply degenerate and degeneracy = 4 levels. An example of a quadruple degeneracy is 65(¯hπ)2/(2µa2) since it corresponds to {1, 8} , {8, 1} , {4, 7} ,and{7, 4}.Here is a plot of the ψ∗(x, y)ψ(x, y)forthe{1, 1} ground state. The height of the surface is proportional to the PDF for finding the electron 4 anywhere within the box. The electron tends to populate in the center of the box and has nodes at the wall boundary as expected. Here are plots of ψ∗(x, y)ψ(x, y) for the degenerate excited states {2, 1} and {1, 2}. We note the presence of an extra node in either the x and y directions. It is clear that the PDF for these two cases are related by a 90o rotation about the z axis and hence it is not surprising that they are degenerate with an energy of 5(¯hπ)2/(2µa2). Electron in a two dimensional harmonic oscillator Another fairly simple case to consider is the two dimensional (isotropic) har- monic oscillator with a potential of V (x, y)= 1 µω2 x2 + y2 where µ is the 2 electron mass , and ω = k/µ. The Schr¨odinger equation reads: ¯h2 ∂2ψ ∂2ψ 1 − + + µw2 x2 + y2 ψ(x, y)=Eψ(x, y)(9) 2µ ∂x2 ∂y2 2 Following our treatment of the two dimensional box, we insert a product wave function of the form ψ(x, y)=X(x) Y (y): ¯h2 ∂2X ∂2Y 1 − Y + X + µw2 x2 + y2 XY = EXY (10) 2µ ∂x2 ∂y2 2 5 We can then divide both sides by X(x) Y (y) and group into x dependent and y dependent terms: 1 ¯h2 ∂2X 1 1 ¯h2 ∂2Y 1 − + µw2 x2 + − + µw2 y2 = E (11) X 2µ ∂x2 2 Y 2µ ∂y2 2 1 2 Again Eq. (11) says that the sum of the first term and second term add to a constant, suggesting that each of these terms is equal to a constant. There is no way that the terms can have a non-constant part but sum to a constant since the first term can only depend on x while the second can only depend on y. Calling E E E E E these constants ( )1 = x,and()2 = y,wehave x + y = and the two differential equations: h2 ∂2X h2 ∂2Y ¯ 1 2 2 ¯ 1 2 2 − + µw x X(x)=Ex X(x) , − + µw y Y (Y )=Ey Y (x) 2µ ∂x2 2 2µ ∂y2 2 (12) The Eq. (12) equations are exactly the same as the Schr¨odinger¨rodinger Equa- tion for the harmonic oscillator which we obtained and solved in the chapter on Bound States in One Dimension. The energies are: 1 1 Ex = nx + ¯hω , Ey = ny + ¯hω where nx,ny ∈ 0, 1, 2,.. (13) 2 2 The solutions are: µω 2 Xn (x)=Nn Hn (ξ)exp(−ξ /2) where ξ = x (14) x x x ¯h µω 2 Yn (y)=Nn Hn (η)exp(−η /2) where η = y (15) y y y ¯h Putting this all together we have quantized energies of the form: E =(nx + ny +1)¯hω (16) 6 and wave functions of the form: 2 2 ψ(ξ,η)=N Hnx (ξ)exp(−ξ /2) × Hny (η)exp(−η /2) µω where (ξ,η)= (x, y) (17) ¯h We could write the energy spectrum as E =(n +1) ¯hω where n ∈ 0, 1, 2.... As shown below the degeneracy of the nth level is just equal to n +1. (0,4) (1,3) (2,2) (3,1) (4,0) 5 h ω (0,3) (1,2) (2,1) (3,0) 4 h ω (0,2) (1,1) (2,0) 3 h ω (1,0) (0,1) 2 h ω (0,0) h ω 0 As you can see, the spectrum for the isotropic harmonic oscillator is more degenerate than for the two dimensional square box. We can construct the wave functions using the explicit forms for the Hermite polynomials: Table 1: Hermite polynomials 3 Ho =1 H3 =8ξ − 12 ξ 4 2 H1 =2ξ H4 =16ξ − 48 ξ +12 2 5 3 H2 =4ξ − 2 H5 =32ξ − 160 ξ + 120 ξ The wave functions for these solutions is in Table 1.
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