Particle in a Box
V(x) = V(x) =
I II III V(x) = 0 The QM postulates insist that (x) be continuous and single valued. We will solve the Sch. Eqnseppygarately in regions I, II and III and then put the pieces together to obey the above constraints. n22π 2 nh 22 E where n 1231,2,3, … n 2ma22 8ma nxa nnxA sin for 0xa, an d 0 othitherwise ConcepTest #1
Consider the energy difference between the n=1 and n=2 levels for a marble of mass 1 g confined in a one-dimensional box of length 0.10 m. Consider the wavelength that corresponds to a spectral transition between these levels. How does E between n=1 and 2 for the marble compare to that for the e– in the previous ( 4Å, 1D box) problem? How does for this spectral transition for the marble compare to that for th e e– in the previ ous probl em ?
A. larger, larger D. smaller, smaller B. larger, smaller E. smaller, larger C. same, same We now need to see what happens as we add more degrees of freedom to a quantum system. Two independent 1-D particles in two boxes Consider two ppparticles in two separate infinite 1-D wells. Their wave functions are zero outside their own box, so they cannot possibly know about each other.
M1 widths a and b M2
a b From earlier, the energy of each must be quantized . Also, the total energy must be the sum of the iidindiv idua l energ ies. n222 n222 E 1 2 n 1,2,3 n 1,2,3 total22 1 2 22bMa12 M Notice that there are now two quantum numbers, associated with the two degrees of freedom with constrained motion. We will see that the two particle wave function is the product of the individual wave functions. Let’s actually solve this problem Two independent 1-D particles in two boxes Since the two particles do not communicate, I can write
their coordinates as x1 and x2, and express the potential potential energy in the form
V xx12, 0 for 0xa 1 and 0xb 2
for x11 < 0 or x > a
for x22 < 0 or x > b This shows that we can write V xx12, V xab V x 1 2
WiiWriting the Hamil toni an in terms of this potenti ilal, 22 22 ˆ xV xxx 12, V x 22ab 1 2 22MM12xx12 22 22 MM xx Vxabab Vx1212 hx hx 2222 1212 We have set the stage to separate the Schrödinger equation with two variables into two separate equations, each with one variable. Two independent 1-D particles in two boxes We assume a solution of the form
xx12, fx gx 1 2 ,a product form and then can show it to be a gen eral sol uti on The Schrödinger equation is hx h x fx gx gx hxfx fxh x gx Efx gx ab1 a 2 b 12 2 11 1 22 12
Dividing by f(x1)g(x2) and rearranging a bit, hfab h x g x12 fx gx12E This can only be true if each side is a constant. Why?? Let l.h.s. = C Then hfa x Cf11 x , the particle in a (0,a) box Sch equation Similarly, we get 22 , the particle in a (0,b) box Sch equation
hgb x Kg x Two independent 1-D particles in two boxes
The result is that the total energy is the sum of the separately quantized particle in a box energies, and The two particle wave function is the product of the two individual wave functions.
Two degrees of freedom with constrained motion: Two quantum numbers! This result generalizes and is critically important.
Separable!! Additional Dimensions/Particles
M1 M2
Two independent 1-D particles in two boxes widths a and b 22 22 xVxˆ xx , Vx 1222MM22 1 2 12xx12 Separable Two independent partclesparticles in two boxes n222 n222 Exx 1 2 n 1,2,3 n 1,2,3 total22 1 2 22bMa12 M 22ab a bnx x Two quantum numbers 112 2 nn 12, sin sin 12 aand y b n for 0 x 0 Separability of Solutions
If the multi coordinate Hamiltonian operator has the special form (qqq123 , , q n ) =hh12 ( q 1 ) + ( q 2 ) + h3 ( q 3 )+ h +n ( q n ) with each component having the form hqii Tq ii Vq ii then there exists a simple, general solution n (,qq , , q ) q q q 12 n 12()()()123 3 i ()qi i=1 with each k satisfying a 1D Schrödinger-like equation hEkk kk . n and the total energy is the sum of individual components, EE . 1 i Thus whenever the Hamiltonian can be written as a sum, theni becomes a product of one variable terms and E becomes a sum of energies associated with each degree of freedom. The difficulty of obtaining a separable solution scales linearly with the number of degrees of freedom, and any 1D problem is very easy! We will do (almost) anything to achieve separability! Additional Dimensions/Particles
Two non-interacting 1-D particles in a box of width a 22 22 ˆ xVxxx 12, 22 221212 Separable, since we neglectedmm gravitational attraction, (m m )/(x -x )2 xx 1 2 2 1
n222 n222 1 2 Etotal 22 n 1 1,2,3 n 2 1,2,3 22mma a12 Two quantum numbers 22 nx x 11n 2 2 nn xx12, sin sin 12
for 0 x12aa 0a a aand x a Mass m particle in a two-dimensional (x,y) box TTVxy Vxy V(x, y) =xy , for x < 0 or x > a or for y < 0 or y > b, and a = 0 , for 0 x a p and 0 y b m x2 2 b py ,, mm 22 22 22 ˆ xy, 0 inside the rectangle (separable!) 22 22 2 d 2 2 d 2 ()x mm ()y - y()y E x ()xxy 2 E 2m 2 2m dx dy n2 2 2 n 2 2 2 1 2 E n 1,2,3, Ey n 1,2,3, x 2ma 2 1 2mb2 2 (x) 2 nx1 2 ny2 aasin n(y) sin n1 bb aand for 0 x 0 otherwise for 0 y b and 0 otherwise
22 nx1 n2 y nn x, y sin sin 12 ab a b f0for 0 x aandbd 0 y b
n222 n222 E 1 2 n 1,2,3 n 1,2,3 total22mbma22 1 2 Particle in a two-dimensional (x,y) box a 22 nx y x, y sin1 sinn2 m nn b 12 ab a b for 0 xaand 0 y b n222 n222 E 1 2 n 1,2,3 n 1,2,3 total22mbma 22 1 2 Note especially!! There are two degrees of freedom with constrained motion and the result is that there are two quantum numbers, n1 and n2. What do the solutions look like?
Nodal line ψ=0 (2D) (points in 1D) (surface in 3D) 1,2,3 for Particle in a 1D Box
In one dimension, the nodes of the standing deBroglie waves are points. The nodal points become lines in 2D, 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 ss.ee. Particle in a two-dimensional (x,y) box
Consider the 2,1 state as well. Now what is the total energy of the 2,1 state? It is the same as the 1,2 state. Particle in a two-dimensional (x,y) box
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!