Particle in a two-dimensional (x,y) box

Transcription

1 Particle in a two-dimensional (x,y) box Let V(x, y) =, for x < 0 or x > a or for y < 0 or y > b, and = 0, for 0 x a and 0 y b ˆ, 0 inside id the rectangle (separable!) m x m y x y d ( x ) m dx E x ( x ) n1 E n 1,,3, x 1 ma ( ) - d y ( ) E y y m dy E n n 1,,3, mb y nx 1 (x) sin n 1 a a for 0 x a and 0 otherwise n y n(y) sin b b for 0 y b and 0 otherwise n1 x n y nn x, y sin sin 1 a b a b for 0 x a and 0 y b n1 n E total n 1 1,,3 n 1,,3 ma mb

2 Particle in a two-dimensional (x,y) box n x 1 n y x, y sin sin nn 1 a b a b for 0 x a and 0 y b n 1 n E n 1,,3 n 1,,3 total 1 ma mb Note especially!! There are two degrees of freedom with constrained motion and the result is that there are two quantum numbers, n 1 and n. What do the solutions look like? Nodal line ψ=0 (D) (points in 1D) (surface in 3D)

3 1,,3 for Particle in a 1D Box In one dimension, the nodes of the standing debroglie waves are points. The nodal points become lines in D, and surfaces in 3D. The energy levels in 1D are rigorously ordered by the number of nodes. This is not quite the case in higher dimensions, as we will see.

4 Particle in a two-dimensional (x,y) box Notice that we there are two nodal lines, in x and y. What is the total energy of the,1 state? It is the same as the 1, state. What does the,1 state function look like?

5 Particle in a two-dimensional (x,y) box The two state functions correspond to the same energy but they are very The two state functions correspond to the same energy, but they are very different spatially. This is called degeneracy! Here, twofold degenerate. In fact, these two functions are orthogonal!

6 More on degeneracies Consider a square box a by a in extent. E n n x y ma n n x y n x, n y n x + n y 1,1 11 Non degenerate 1, 5 1 1, , , , 8, Doubly degenerate Doubly degenerate Non degenerate Doubly degenerate 3, 13 3 The ψ ij are all orthogonal: 0 unless i j i j

7 Particle in Three Dimensional (a,b,c) Box ˆ V x, y, z m x y z V x, y, z 0 inside the box and outside Separable into three 1D particle in a box problems! n n x y n z E n xnyn z m a b c n 1,,, n 1,,, n 1,,, x y z n x 1 n y nz x, y, z sin sin sin n n n x y z a b c a b c for 0 x a, 0 y b and 0 z c

8 Energy Degeneracy for Particle in 3D box E ma n n n 1 3

9 ConcepTest #1 In a cubic box, what is the degree of the degeneracy if the three quantum numbers n 1, n, n 3 can have the values 1,, 3? A. -fold B. 4-fold C. 6-fold D. 8-fold

10 ConcepTest # In a cubic box, which of the following states is lowest in energy? A. n 1 =3, n =1, n 3 =1 B. n 1 =, n =, n 3 = C. n 1 =3, n =, n 3 =1 D. n 1 =, n =, n 3 =3

11 Three Dimensional Box for Electrons Known as Quantum Dot Quantum dot is "Particle in a Box quantum system Displays size-dependent electronic and optical properties Useful in nanotechnology Sharp emission of light makes them useful light "tags"

12 Micrograph of InGaAs Quantum Dots (0 nm wide, 8 nm high)

13 Fluorescence Microscopy Image of mouse in vivo capillary structure using CdSe-ZnS Quantum Dots

14 Energy Levels of Quantum Dot Depend on Dot Diameter Three dimensional cube 8ma E n n n 1 3 Electron transitions between energy levels have ΔE that depends on diameter

15 Emission Spectrum of Quantum Dots Increasing Quantum Dot Diameter

16 Quantum Dot Problem Assume that light emission from quantum dot particles can be described by a cubic (a=b=c) "particle in a box" We observe light is emitted with wavelengths ranging from red (= 650 nm) to green (= 530 nm). If the observed light emission arises from the transition t (n 1 =3, n =3, n 3 =3) (n 1 =, n =, n 3 =), what is the diameter of the quantum dots giving rise to the red and green emissions? Answer: a = 1.7 nm for red emission a = 1.55 nm for green emission

17 Two particles in an (x,y,z) box (for fun, make them electrons) Electrons have charge e and the Coulomb potential energy between two electrons e 1 and e a distance r 1 apart is e e V r 1 r 1 x x y y z z The Hamiltonian operator is ˆ x, y, z, x, y, z V r V m e x y z x y z box e V box m x y z x y z x x y y z z 0 e inside NOT separable in (x,y,z), so inside (actually, separable in r,,)

18 The Variational Theorem Given any normalized function g(q) satisfying the boundary conditions associated with a Hamiltonian H ˆ, then gq ˆ gq E, 0 where E o is the exact ground state energy. The proof is trivial when we recognize that E 0 0 0, where o is the exact state function for the ground state. Since the set of eigenfunctions f of f ˆ, i, f forms a complete l set, we can expand d g(q) ( ) in terms of i to obtain g(q) = i c i i (q) where c i = g(q) i (q). Thus ˆ ˆ i j i g q g q c q c q c E i i j j i i Since E i E 0 and c i = 1 (since g(q) is normalized), it follows that g(q) ˆ g(q) E 0. The equality only holds when the function g(q) is exactly o. We thus have a systematic method for picking the "best" ground state wave function from any set of functions g i (q). The function g j (q) with the smallest g j (q)ˆ g j (q) will be the "best" approximate ground state wave function possible within the set of functions above. ˆ

19 Object in a (large) 3 D Cube Consider an atom or a molecule placed in a large cube. The boundary conditions mean that the state function must vanish at the edges of the cube. The variational theorem tells us that we can write the state function as a linear combination of particle in a cube state functions involving the coordinates of the nuclei and electrons (as in the two electron example before). We can use the variational theorem to pick the best wave function from a trial set, and keep expanding the set until we obtain something close to the correct * answer. Thus, we know how to get an answer to any problem. Cute, but useless! Everything else we do from now on will be concerned with approximations using the variational theorem and exact or approximate separabilities. *Caveat: The intrinsic spin of particles, the Pauli principle and other relativistic effects have not yet been included.

20 Energy Measurements for particle in a box Let test Normalizing, test, norm Possible results of measurement of E? Only E 1 or E 5 or E 10 Most probable result? E 5 Average result? E E E E ˆ But this discussion has deftly avoided the issue of measurement. So, on request.....